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Physics Part I Chapter 6 Work, Energy And Power

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    NCERT Solution For Class 11 Physics Physics Part I

    Work, Energy And Power Here is the CBSE Physics Chapter 6 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Physics Work, Energy And Power Chapter 6 NCERT Solutions for Class 11 Physics Work, Energy And Power Chapter 6 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Physics.

    Question 1
    CBSEENPH110024913
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     Define the term electricity?
    Solution

    Elasticity is that property of the object by virtue of which it regain its original configuration after the removal of the deforming force.

    Question 2
    CBSEENPH110024914
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    Write the SI unit of gravitational constant?
    Solution

    The correct option is C Nm2Kg−2
    Gravitational force: F

    wiredfaculty.com then,

    G = FR2M1M2

    unit of F = N
    unit of mass = Kg
    unit of R = m
    unit of G = Nm2kg−2

    Question 3
    CBSEENPH11017065
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    Is work a scalar or a vector quantity?

    Solution
    Work is dot product of two vectors. i.e. F with rightwards arrow on top times S with rightwards arrow on top.
    And dot product is a scalar quantity. Therefore, work is a scalar quantity. 
    Question 4
    CBSEENPH11017066
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    Define the unit joule.

    Solution
    Work done is said to be one joule if one newton of  force displaces the body through a distance of one meter in the direction of applied  force .
    Question 5
    CBSEENPH11017067
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    What are different units of energy?

    Solution

    The different units of energy are :
    (i) Joule            
    (ii) Erg
    (iii) eV            
    (iv) KWh
    (v) Calorie.

    Question 6
    CBSEENPH11017068
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    What is watt?

    Solution
    SI unit of power is Watt. 
    Question 7
    CBSEENPH11017069
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    Define watt.

    Solution
    Power is said to be one watt if one joule of work is done in one second. 
    Question 8
    CBSEENPH11017071
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    How many joules are there in one eV?

    Solution
    1 eV = 1.6x10–19 J.
    Question 9
    CBSEENPH11017072
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    What is the relation between kWh and Joule?

    Solution
    1kWh = 3.6x106J.
    Question 10
    CBSEENPH11017073
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    What is the dimensional formula of power?

    Solution
    [M1L2T–3] is the dimensional formula of power. 
    Power = Work over time space equals space fraction numerator M to the power of 1 L squared T to the power of negative 2 end exponent over denominator T to the power of 1 end fraction space equals space left square bracket space M to the power of 1 L squared T to the power of negative 3 end exponent right square bracket
    Question 11
    CBSEENPH11017074
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    Does the body at rest possess energy?

    Solution

    A body at rest possess potential energy. 

    Question 13
    CBSEENPH11017076
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    A man is rowing upstream and is at rest w.r.t. shore. Is he doing work w.r.t. shore?

    Solution
    The man is doing zero work because he is at rest w.r.t. shore. Hence, displacement is zero.
    Therefore work done by the man is zero.
    Question 14
    CBSEENPH11017078
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    Is frictional force conservative?

    Solution
    Friction is a non-conservation force because work done by friction over closed path is not equal to zero. 
    Question 15
    CBSEENPH11017079
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    What is the work done by gravitational force in taking the mass m with constant velocity through height h?

    Solution
    Work done by gravitaional force, F = mgh.
    Question 16
    CBSEENPH11017080
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    A mass is tied with a string and whirled in horizontal circle. What is the work done by tension force?

    Solution
    Here, the man is whirled in a horizontal circle by a tied string. Hence, the initial and final points would coincide. Therefore, work done by tension force is zero. 
    Question 17
    CBSEENPH11017082
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    Does the work done by force depend upon the velocity at that instant?

    Solution
    No, work done by force at any instant does not depend upon the velocity at that instant.
    Work is dependent on the force and displacement. 

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    Question 18
    CBSEENPH11017083
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    What is the work done by man against gravity when he is carrying a load on his head and walking on the horizontal road?

    Solution
    The work done by the man carrying a load on his head is zero because the initial and final points of the load is the same. Hence, work is done only if the initial and final points are different. 
    Question 19
    CBSEENPH11017084
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    Are all central forces conservative?

    Solution
    Central forces act along the line joining the centres of two charged or magnetic bodies. Therefore, all central forces are conservative forces. 
    Question 20
    CBSEENPH11017085
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    Work done against what type of force is stored in the form of potential energy?

    Solution
    Work done against conservative forces is stored in the form of potential energy.
    Question 21
    CBSEENPH11017086
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    How does the potential energy of electron change when it jumps from Bohr's first orbit to higher orbit?

    Solution
    When electron jumps from Bohr's first orbit to higher orbit, potential energy of electron will increase.
    Question 22
    CBSEENPH11017087
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    What is the work done by tension in the string when a body tied with string completes half the revolution?

    Solution
    When a body tied to one end of the string is rotated in a circle and it completes half revolution, work done is zero. 
    This is because straight theta space equals space 90 to the power of 0
    Work done, W = Fs cos straight theta space left square bracket space straight theta space equals space 90 to the power of straight o space and space cos space 90 to the power of 0 space equals space 0 right square bracket
    So, Workd done is 0. 
    Question 23
    CBSEENPH11017088
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    Calculate the amount of work done in raising a glass of water weighing 0.5 kg through a height of 20 cm. 

    Solution
    Given, 
    Mass, m = 0.5 kg 
    Height, h = 20 cm = 20/100 m = 1/5 m 
    Acceleration due to gravity, g = 10 m/s2 
    Work done = Force X distance 
    W = mg X h = 0.5 X 10 X 1/5
        = 1 Joule
    Question 24
    CBSEENPH11017089
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    What type of energy is stored in a stretched spring?

    Solution
    Potential energy is stored in a stretched spring. 
    Question 25
    CBSEENPH11017090
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    What is the total energy of the bounded system?

    Solution
    Total energy of the bounded system is negative.
    For a bound system, the sum of potential energy and kinetic energy is less than 0. Hence, the total enery of the bound system will be negative. 
    Question 26
    CBSEENPH11017091
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    What does the area under F (force) versus s (displacement) represent?

    Solution
    Area under Force-displacement graph is equal to the work done by the force F in displacing the body through a distance S.
    Question 27
    CBSEENPH11017093
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    Work done by a human heart is 1.5 J. find the power of human heart. 
    [ Given Heart beats per minute is 72]

    Solution
    Work done = 1.5 J
    No. of heart beats = 72
    total work done by heart in 1 minute = 1.5 cross times space 72 = 108 Joule
    Time taken, t = 1min
    Therefore, 
    Power = fraction numerator Work space done over denominator time space taken end fraction equals space 108 over 60 equals 1.8 space W 
    That is, the power of human heart is 1.8 W. 
    Question 28
    CBSEENPH11017094
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    What is one horse power?

    Solution
    Horsepower is a unit of power and is equal to 550 lb ft/s.
    Question 29
    CBSEENPH11017095
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    What is the value of 1 H.P. in SI units?

    Solution
    1 Horse Power = 746 W .
    Question 31
    CBSEENPH11017097
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    What is Einstein's mass energy relationship?

    Solution
    E = mc2,
    where, 
    c is the velocity of light.
    Question 32
    CBSEENPH11017098
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    What type of energy is stored in a stretched bow?

    Solution
    Potential energy.
    Question 33
    CBSEENPH11017099
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    How does potential energy of system of two protons will change when brought near each other?

    Solution
    When two protons are brought near each other, they are compressed. Since, they are of the same charge, the work done is required to overcome the repulsion between them. Therefore, potential energy of the system of two protons will increase.
    Question 34
    CBSEENPH11017100
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    A heavy body and a light body have same kinetic energy. Which has greater momentum?

    Solution
    Momentum is dependent on mass of the body.
    Therefore, heavy body will have greater momentum.
    Question 35
    CBSEENPH11017101
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    Give an example of non-conservative force.

    Solution
    Frictional force is a non-conservative force.
    If a body is moved from one point to another on a hard surface, work done will against the frictional force will depend on the length of the pat. 
    Question 36
    CBSEENPH11017102
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    Give one example of negative work.
    Solution
    Work done by frictional force is negative.
    When a body is moved over a rough horizontal surface, the motion is opposed by the force of friction. So, work done by frictional force is negative.
    Question 37
    CBSEENPH11017103
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    What is work done by conservation force along the closed path?

    Solution
    Zero, because initial and final points of the path is same for a closed track. 
    Question 38
    CBSEENPH11017104
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    A body undergoes a displacement of space space 2 i with hat on top space straight m under the force 3 j with hat on top straight N. What is the work done by the force?
    Solution

    Work done, W = straight F with rightwards harpoon with barb upwards on top space. space s with rightwards harpoon with barb upwards on top
    Therefore, 
    Work done by force is,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 39
    CBSEENPH11017105
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    The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained– the rocket or the atmosphere?

    Solution
    The heat energy required to burn the casing of rocket is at the cost of kinetic energy of the rocket. 

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    Question 40
    CBSEENPH11017106
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    Is it possible to accelerate the body without doing any work? Give an example.

    Solution
    Yes, it is possible to accelerate the body without doing any work.
    For example, work done by tension force on the stone revolving in circular orbit is zero but tension force accelerates the body.
    Question 41
    CBSEENPH11017107
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    Can the potential energy or kinetic energy be negative?

    Solution
    Potential energy can be negative but kinetic energy cannot be negative.
    Question 42
    CBSEENPH11017108
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    What happens to the centre of mass, if the external forces acting on a system have zero resultant? 

    Solution
    When the external force acting on a system have 0 resultant, the centre of mass will move with a constant velocity. That is, the centre of mass, will not accelerate. 
    Question 43
    CBSEENPH11017109
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    Two bodies of unequal masses have same Kinetic energy. Which one has greater linear momentum?

    Solution

    Given that, the kinetic energy of the two bodies is equal. 
    That is, 
    space space space space space space space space space space 1 half space m subscript 1 v subscript 1 squared space equals space 1 half m subscript 2 v subscript 2 squared space rightwards double arrow space v subscript 2 squared over v subscript 1 squared space equals space m subscript 1 over m subscript 2 space rightwards double arrow v subscript 2 over v subscript 1 space equals space square root of m subscript 1 over m subscript 2 end root space
    Linear momentum is given by, 
    p1 = m1 v1 and p2 = m2 v2 
    Therefore, space space space space space straight p subscript 2 over straight p subscript 1 space equals space fraction numerator m subscript 2 v subscript 2 over denominator m subscript 1 v subscript 1 end fraction space equals space m subscript 2 over m subscript 1. square root of m subscript 1 over m subscript 2 end root space equals space square root of m subscript 2 squared over m subscript 1 squared m subscript 1 over m subscript 2 end root space rightwards double arrow space straight p subscript 2 over straight p subscript 1 space equals space square root of m subscript 2 over m subscript 1 end root space

    That is mass has a direct dependence on linear momentum. 
    Hence, the heavier body will have a greater momentum. 

    Question 44
    CBSEENPH11017110
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    The masses of 1 g and 4 g are moving with equal kinetic energies.Calculate the ratio of the magnitudes of their linear momenta. 

    10J of work is required to take the body from ground to a height 2m. How much energy is required to take the same body from height 2m to 4m?

    Solution

    Given, 
    Mass, m1 = 1 g 
    Mass, m2 = 4 g
    Let v1 and v2 be the velocity of masses m1 and  m2 respectively. 
    Kinetic energy is equal. 
    Therefore, 

    This is the required ratio of linear momentum. 

    Question 45
    CBSEENPH11017111
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    A spring of constant k is stretched by x. What is the potential energy stored in the spring?

    Solution

    P.E of the spring, P.E = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 46
    CBSEENPH11017112
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    How many joules are there in one gravitational unit of work?

    Solution
    In one gravitaional unit of work there are 9.8 J.
    Question 47
    CBSEENPH11017113
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    A spring of constant k is stretched to x1 . What is the additional work required to extend the spring to x2?

    Solution

    The spring is stretched to x1. 
    Again, the spring has been stretched to x2. 
    So, additional work required to extend the spring is given by, 
                           1 half straight k left parenthesis straight x subscript 2 squared minus straight x subscript 1 squared right parenthesis

    Question 48
    CBSEENPH11017114
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    A mass is attached to a spring of constant k and displaced by x from unstretched position and released. Find the velocity of mass when the spring is unstretched?

    Solution

    Velocity of the mass of the spring when it is unstretched is given by, 
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    Question 49
    CBSEENPH11017115
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    At the free ends of an unstretched spring, two masses having equal and opposite charges are attached. What will happen to spring?

    Solution
    The spring will get compressed, at the free ends of an unstretched spring, with two masses of equal and opposite charges. 
    Question 50
    CBSEENPH11017116
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    Water is cooled to form ice. If mass energy equivalence is taken into account, then what is the mass of ice?

    Solution
    Mass of ice is greater than the mass of water. 
    Question 51
    CBSEENPH11017117
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    If the kinetic energy of a moving body increases by 300 %, then by what percentage the momentum will increase?

    Solution

    Let the kinetic energy of the body be K and momentum of the body be p. 
    K.E = 1/2 mv2           ... (1) 
    Given that the K.E increases by 300 %. 
    That is, 
    Increasing factor of K.E = 300X 1/100 = 3 
    Therefore, the new K.E, K' = K + 3K = 4K
    Now, equation (1), we have 

    straight v space equals space square root of fraction numerator 2 straight k over denominator straight m end fraction end root space 
    Initial momentum, p = m v = m X square root of fraction numerator 2 straight K over denominator straight m end fraction end root 
    square root of straight m squared straight X fraction numerator 2 straight K over denominator straight m end fraction end root space equals space square root of 2 Km end root New space momentum comma space straight p apostrophe space equals space square root of 2 straight K apostrophe straight m end root space So comma space fraction numerator straight p apostrophe over denominator straight p end fraction space equals space square root of fraction numerator 2 straight K apostrophe space straight m over denominator 2 Km end fraction end root That space is comma space space space space space space space fraction numerator straight p apostrophe over denominator straight p end fraction space equals space square root of fraction numerator straight K apostrophe over denominator straight K end fraction end root space rightwards double arrow space fraction numerator straight p apostrophe over denominator straight p end fraction space equals space square root of fraction numerator 4 straight K over denominator straight K end fraction end root space rightwards double arrow space fraction numerator straight p apostrophe over denominator straight p end fraction space equals space 2 space rightwards double arrow space straight p apostrophe space equals space 2 straight p Percentage space change space equals space fraction numerator straight p apostrophe space minus space straight p over denominator straight p end fraction space cross times space 100 space percent sign space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 straight p space minus space straight p over denominator straight p end fraction straight X space 100 space percent sign space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 100 space percent sign  
    That is, the momentum will increase by 100%. 

    Question 52
    CBSEENPH11017118
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    What is SI gravitational unit of work?

    Solution
    SI unit of gravitational unit of work is Kg m.
    Question 53
    CBSEENPH11017121
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    Define work.

    Solution
    Work is said to be done if force acting on a body displaces the body through a certain distance in the direction of force applied.
    The work is equal to the product of displacement and component of force in the direction of displacement,
    i.e.          space space space space space space space W equals left parenthesis F cos theta right parenthesis S 
                          equals F S cos theta space equals space F with rightwards arrow on top times S with rightwards arrow on top
    Question 54
    CBSEENPH11017123
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    Define the units of work in different systems of units.

    Solution

    Absolute units:
    (i) SI System - Joule is the unit of work in SI system. 
    Work done is said to be 1 J if a force of IN displaces the body by 1m in the direction of force applied.
    (ii) CGS system – CGS unit of work is erg.
    Work done is said to be 1 erg if a force of one dyne displaces the body by 1 cm in the direction of force applied. 1J = 107 ergs. 
    Gravitational units: 
    (i) SI unit: Gravitational unit of work is kg-m.
    Work done is said to be 1 kg-m if a force of 1 kg weight displaces the body by 1m in the direction of force applied. 
    (ii) CGS: Gravitational unit of work is g-cm.
    Work done is said to be 1 g-cm if 1 g-wt force displaces the body by 1 cm in the direction of force applied.

    Question 55
    CBSEENPH11017125
    Wired Faculty App

    Derive the relation between different units of work.

    Solution
    Relation between joule and erg:
                       1 J = 1 Nm
                           equals space 10 to the power of 5 space dyne space cross times space 100 space cm
                           equals space 10 to the power of 7 space dyne space cm space equals space 10 to the power of 7 space ergs 
    Relation between kg-m and g-cm:
               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
                              equals space 10 to the power of 5 straight g minus cm 
    Relation between Joule and kg-m: 
                1 kg minus straight m equals 9.8 Nm equals 9.8 straight J 
    Relation between erg and g-cm:
    1 g–cm = 980dyne cm
                = 980 erg.
    Question 56
    CBSEENPH11017126
    Wired Faculty App

    Name different units of energy.

    Solution

    The different units of energy are :
    (i) Joule (ii) erg (iii) electron volt (iv) KWt hr (v) Calories.

    Question 57
    CBSEENPH11017127
    Wired Faculty App

    Work is a scalar quantity. Can it have negative value?

    Solution
    Yes, work can be negative.
    Work done by force of friction is negative. When angle between force and displacement is greater than 90°, the work done by the force is negative. 
    Question 58
    CBSEENPH11017131
    Wired Faculty App

    A person is walking on horizontal road with a load on his head. Is he doing any work? Explain.

    Solution

    We know,  
    Work done, W equals F with rightwards arrow on top times S with rightwards arrow on top 
    In this case gravitational force is acting in vertically downward direction and displacement is in horizontal direction.
    Therefore angle betweenspace space F with rightwards arrow on top space a n d space S with rightwards arrow on top is 90 degree. 
                           W equals F s space cos 90 degree space equals space 0
    Hence, work done by the person is zero. 

    Question 59
    CBSEENPH11017132
    Wired Faculty App

    A body of mass m is revolving in circular orbit with constant speed. What is the work done on the body to revolve it?
    Solution
    Centripetal force, which acts along the radius and directed towards the centre is required to keep the body in circular orbit and to revolve with constant speed.
    Instantaneous displacement of particle is always perpendicular to force.
    Hence work required to revolve the body is zero.
    Question 60
    CBSEENPH11017133
    Wired Faculty App

    If two protons are brought near each other, what happens to potential energy of the system?

    Solution
    Two protons carrying the same charge, repel each other.
    Therefore, to bring them near to each other we have to do work against force of repulsion. This work done is converted into potential energy and hence the potential energy increases.
    Question 61
    CBSEENPH11017134
    Wired Faculty App

    Give two illustrations for each of following.
    (a) Positive work
    (b) Negative work
    (c) Zero work.
    Solution

    Positive work:
    (i) Work done by gravity on a free falling body is positive.
    (ii) Work done by applied force when taken vertically up against gravity is positive.

    Negative work:
    (i) Work done by friction force on a moving body is negative.
    (ii) The work done by gravity on a body moving up is negative.

    Zero work:
    (i)Work done by electrostatic force of nucleus on electron revolving in a circular orbit is zero.
    (ii) Work done by tension force in string on a stone whirled in a circle is zero.

    Question 62
    CBSEENPH11017135
    Wired Faculty App

    Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

    Solution
    The gravitational force of the Sun does positive work when the comet moves from apogee to perigee and the gravitational force of the Sun does same amount of negative work when it goes from perigee to apogee.
    Hence net work done by the gravitational force of the Sun on the comet in one complete cycle is zero.
    Question 63
    CBSEENPH11017137
    Wired Faculty App

    A variable force F with rightwards arrow on top left parenthesis x with rightwards arrow on top right parenthesis acts on a body.  Write an expression for the work done in displacing the body from stack x subscript 1 with rightwards arrow on top space to space stack x subscript 2 with rightwards arrow on top. Can we find the work from F versus x graph?
    Solution

    Work done by a variable force is calculated by calculus method.
    Let at any instant, the position of body be x with rightwards arrow on top and the force on the body be F with rightwards arrow on top left parenthesis x with rightwards arrow on top right parenthesis.
    The work done in displacing the body by small displacement stack d x with rightwards arrow on top is,
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Total work in displacing the body from <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  is,
     
    Graphically the work done by a variable force is equal to area under force versus displacement graph.

    Question 64
    CBSEENPH11017138
    Wired Faculty App

    What is the work done in dragging a body in horizontal direction by a force of 40N acting at angle 60° with horizontal through a distance of 24m?

    Solution

    Given that,
    Force space acting comma space straight F space equals space 40 straight N Angle space made space by space the space body space with space the space horizontal comma space straight theta equals 60 degree Distance space moved space by space the space body comma space straight S equals 24 straight m
    So, work done by the body is given by, 
                straight W equals straight F with rightwards arrow on top times straight S with rightwards arrow on top space equals space FScosθ
               space space space space space equals 40 cross times 24 cross times c o s 60 equals 480 straight J

    Question 65
    CBSEENPH11017140
    Wired Faculty App

    Find the work done by force F with rightwards arrow on top space equals space left parenthesis 3 i with hat on top space minus space 2 j with hat on top space plus space k with hat on top right parenthesis straight N in displacing a body of mass 3 kg from straight A left right arrow left parenthesis negative 2 comma space minus 1 comma space 0 right parenthesis space space to space space straight B left right arrow left parenthesis 1 comma space 2 comma space 3 right parenthesis.
    Solution

    Given,
    Force acting on the body, space space F with rightwards arrow on top equals left parenthesis 3 i with hat on top minus 2 j with hat on top plus k with hat on top right parenthesis straight N
    Position space of space straight A space is space given space by comma space straight A left right arrow left parenthesis negative 2 comma space minus 1 comma space 0 right parenthesis Position space of space straight B space is space given space by comma space straight B left right arrow left parenthesis 1 comma space 2 comma space 3 right parenthesis 
    The displacement of particle is,
                       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Work done by force is, 
                         W equals F with rightwards arrow on top times S with rightwards arrow on top
                            equals space left parenthesis 3 i with hat on top minus 2 j with hat on top plus k with hat on top right parenthesis times left parenthesis 3 i with hat on top plus 3 j with hat on top plus 3 k with hat on top right parenthesis
                             = 6J
                                

    Question 66
    CBSEENPH11017142
    Wired Faculty App

    A body of mass 0.24Kg is tied with a string of length 70cm and made to revolve in a circular orbit of radius 70cm on the top of a horizontal table. The table is rough and has coefficient of friction 0.25. Find the work done by tension force in the string and by the force of friction in making one complete revolution.

    Solution
    Work done by tension:
    We know that the tension force is always perpendicular to instantaneous displacement.
    Therefore, the work done by tension force is zero.
    Work done by force of friction:
                    W equals F with rightwards arrow on top. S with rightwards arrow on top space equals space F S cosθ
    We have,
    Force, F equals mu m g equals 0.25 cross times 0.24 cross times 9.8 equals 0.588 straight N
    Distance travelled, S equals 2 πr equals 2 straight pi left parenthesis 0.7 right parenthesis space equals space 4.4 straight m
                         straight theta equals 180 degree
    Therefore,
    Work done by the tension force in the string is,
                       space space space space W equals 0.588 cross times 4.4 cross times left parenthesis negative 1 right parenthesis space equals space minus 2.5872 space straight J
    Question 67
    CBSEENPH11017144
    Wired Faculty App

    A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by space space space space space space space F with rightwards arrow on top space equals space minus i with hat on top plus 2 j with hat on top plus 3 k with hat on top straight N. 
    where <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> are unit vectors along the x-, y- and z-axis of the system respectively.
    What is the work done by this force in moving the body a distance of 4 m along the z-axis?
                     
    Solution

    The force acting on the body is,
           F with rightwards arrow on top space equals space minus i with hat on top plus 2 j with hat on top plus 3 k with hat on top straight N
    As the body is constrained to move along z-axis, therefore the displacement of the body is,
                       s with rightwards arrow on top space equals space 4 k with italic hat on top straight m
    Now the work done by the force is,
             space space W equals F with rightwards arrow on top times s with rightwards arrow on top
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 69
    CBSEENPH11017146
    Wired Faculty App

    In figure (i), the man walks 2m carrying a mass of 15 kg on his hands. In figure (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of 15 kg hangs at its other end. In which case is the work done greater? 

    Solution

    For case (i),
    Force is applied by the man in vertically downward direction. So, the displacement is done in horizontal direction. Therefore work done by man is zero. 
    In case (ii),
    The force required to lift the weight and displacement both are in upward direction. Therefore work is done by the man to lift the weight.
    i.e. W = 15 x 9.8 x 2
             = 294 J.
    Thus work done in case (ii) is greater than in case (i).

    Question 70
    CBSEENPH11017148
    Wired Faculty App

    A uniform chain of length L and mass m is lying on a smooth table and one-nth part of its length is hanging vertically down over the edge of the table. Find the work required to pull the hanging part on to the table.

    Solution

    Let space straight lambda be linear mass density of chain.

    To pull the chain we have to do work against the weight of hanging part of chain.
    Let at any instant length of hanging part be x. 
    Therefore the weight of hanging part of chain is,
                        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The work done in pulling the chain by small distance dx is,
                        d W equals negative lambda x g d x
    Total work done to pull the whole of hanging part of chain is,
                 W equals integral d W space space space space equals space integral subscript L divided by n end subscript superscript 0 minus lambda g x space d x 
                   equals right enclose negative lambda g x squared over 2 end enclose subscript bevelled L over n end subscript superscript 0
                  space space equals fraction numerator lambda g L squared over denominator 2 n squared end fraction space equals fraction numerator m g L over denominator 2 n squared end fraction
                                           

    Question 71
    CBSEENPH11017150
    Wired Faculty App

    A body of mass m is taken up with constant velocity through height h. What is the work done by external force in taking the body up?

    Solution
    Work done by force is given by,
                         W equals F with rightwards arrow on top times S with rightwards arrow on top
    To move the body up with constant velocity, the external force equal to mg must be applied in upward direction.
    Here both the external force and displacement are in same direction.
    Therefore angle between force and displacement is zero.
    Now work done by external force is,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 72
    CBSEENPH11017152
    Wired Faculty App

    Define conservative force and give two examples.

    Solution

    Conservative force is said to be conservative if work done in taking the system from one point to other point is independent of path followed but depends upon initial and final position.
    Examples:
    Electrostatic force, gravitation force.
    Work done by conservative force along closed path is zero.

    Question 73
    CBSEENPH11017153
    Wired Faculty App

    Show that gravitational force is a conservative force.

    Solution

    A force is said to be conservative if work done by the force is independent of the path followed and depends upon the initial and final positions.
    Suppose a body of mass m be taken from A to B along different paths as shown in the figure.
     
    i) Work done if body is taken up straight along Ab, 
                     W1 = -mgh
    ii) Work done by gravity if body is taken up along path ACB, 
     W2 = WAC + WCB 
    straight W subscript AC space equals space straight m space straight g with rightwards harpoon with barb upwards on top. space AC with rightwards harpoon with barb upwards on top space space space space space space space space space space space equals space minus space mg space left parenthesis AC right parenthesis space cos space left parenthesis space 90 space minus space straight theta right parenthesis space space space space space space space space space space space equals space minus mg space AC space sin space straight theta space space space space space space space space space space space space equals space space minus space mg space left parenthesis AB right parenthesis space equals space minus space mgh space straight W subscript BC space equals space straight m space straight g with rightwards harpoon with barb upwards on top. space CB with rightwards harpoon with barb upwards on top space space space space space space space space space equals space minus mg space left parenthesis CB right parenthesis space cos space left parenthesis 90 right parenthesis space equals space 0 Therefore comma space straight W subscript 2 space equals space minus space mgh space  
    iii) Work done by the gravity along the path ADEFGHB, 
    W3 = WAD + WDE + WEF + WFG + WGH + WHB 

     Here comma space the space displacements space DE with rightwards harpoon with barb upwards on top comma space FG with rightwards harpoon with barb upwards on top space and space JB with rightwards harpoon with barb upwards on top space are space perpendicular space to space gravity. space Therefore comma space straight W subscript DE space equals space straight W subscript FG space equals space straight W subscript HB space equals space 0 space comma space and space straight W subscript AD space equals left parenthesis space straight m. straight g with bar on top right parenthesis space. space AD with rightwards harpoon with barb upwards on top space equals space minus space mg space left parenthesis AD right parenthesis space straight W subscript EF space equals space left parenthesis space straight m. straight g with bar on top right parenthesis space. space EF with rightwards harpoon with barb upwards on top space equals space minus space mg space left parenthesis EF right parenthesis space straight W subscript GH space equals space left parenthesis space straight m. straight g with bar on top right parenthesis space. stack GH space with rightwards harpoon with barb upwards on top equals space minus space mg space left parenthesis GH right parenthesis space Therefore comma space straight W subscript 3 space equals space minus space mg space left parenthesis AD space plus space EF space plus space GH right parenthesis space equals space minus space mgh space Since space straight W subscript 1 space equals space straight W subscript 2 space equals space straight W subscript 3 space Therefore comma space gravitational space force space is space conservative. space

    Question 74
    CBSEENPH11017154
    Wired Faculty App

    State important properties of conservative force.

    Solution

    Properties of conservaticve force:
    (i) The work done in moving a body against conservative force is independent of path followed but depends upon initial and final position.
    (ii) The work done in moving a body along closed path against conservative force is zero.

    Question 75
    CBSEENPH11017156
    Wired Faculty App

    Define power. State its SI and cgs units.

    Solution

    The rate of doing work is called power.
    i.e.           P equals limit as increment t rightwards arrow 0 of fraction numerator capital delta W over denominator capital delta t end fraction equals fraction numerator d W over denominator d t end fraction

    Since,       d W equals F with rightwards arrow on top times stack d s with rightwards arrow on top
    ∴               P equals fraction numerator d W over denominator d t end fraction equals F with rightwards arrow on top times fraction numerator stack d s with rightwards arrow on top over denominator d t end fraction equals F with rightwards arrow on top. v with rightwards arrow on top
    The SI unit of power is Watt and cgs unit is erg/s.

    Question 76
    CBSEENPH11017157
    Wired Faculty App

    What is kinetic energy? Obtain an expression for it.

    Solution

    The energy possessed by a body by virtue of its motion is called kinetic energy.
    Let us consider a body of mass m placed on a horizontal smooth surface.
    Apply a constant force F on the body.

    Let initially the body be at rest and after travelling a distance x, it acquires the velocity v.
    Let a be the acceleration produced in the body.
    Using equation of kinematics v2 – u2 = 2as, we get v2 = ax.
    Now, the work done by force is,
    straight W equals F s equals max equals straight m open parentheses 1 half straight v squared close parentheses equals 1 half mv squared 
    Since the work done on the body sets it into motion.
    Therefore kinetic energy of body moving with velocity v is 1 half m v to the power of italic 2 italic.

    Question 77
    CBSEENPH11017158
    Wired Faculty App

    What is potential energy? Give few examples for the potential energy stored in different bodies.

    Solution

    The energy acquired by the body due to its position in conservation field and/or configuration is called potential energy.
    The examples of potential energy are:
    (i) The potential energy stored in the stretched or compressed spring is elastic potential energy due to configuration.
    (ii) The potential energy stored in the mass in gravitational field of the earth is an example of potential energy due to position of mass in gravitational field.
    (iii) A charge placed in the electric field possesses potential energy. This potential energy is electric potential energy and is due to position of charged particle in electric field.
    (iv) A wire when twisted stores potential energy due to configuration.

    Question 78
    CBSEENPH11017159
    Wired Faculty App

    Derive an expression for the potential energy stored in mass m in gravitational field of earth when placed at a height h above ground. Take surface of the earth as reference point for potential energy.

    Solution
    Consider,
    A body of mass 'm' placed at the surface of the earth and moved from surface of the earth O to a point A at a height h from O.
    Let at any instant, the body be at a height x from ground.
    The small amount of work done by gravitational force in displacing the body by dx against gravity is,
    d W space equals space F with rightwards arrow on top. stack d x with rightwards arrow on top space equals space m g with rightwards arrow on top. stack d x with rightwards arrow on top
         equals m g d x cos left parenthesis 180 degree right parenthesis space equals space minus m g d x
    Total amount of work done by gravitational force in displacing the body from ground to height h is,
    W equals integral subscript 0 superscript h minus m g d x equals negative m g h
    Therefore the potential energy stored in the body is mgh.
    Question 79
    CBSEENPH11017160
    Wired Faculty App

    Define conservative force and show that conservative force is equal to negative gradient of potential energy.

    Solution
    When the work done by a force is independent of path followed and depends only on initial and final position, the force is said to be a conservative force. 
    The work done against conservative force is stored in the form of potential energy.
    Change in potential energy of a body is equal to negative of the work done by conservative force.
    Let the body be displaced by dr against conservative force F.
    The change in potential energy dU is,
                                 d U minus F space d r
    rightwards double arrow                         F equals negative fraction numerator d U over denominator d r end fraction

    Sponsor Area

    Question 80
    CBSEENPH11017161
    Wired Faculty App

    Derive an expression for potential energy stored in spring.

    Solution
    Consider, a massless spring attached with mass m at one end and the other end of spring be connected with a rigid wall.
    When we pull the mass towards C, the restoring force directed towards A is set up in spring. Work has to be done against this restoring force in order to displace the mass and hence this work is stored in the form of P.E in the spring. 
    Let at any instant, mass m be at a distance x from A. 
    Restoring force at this instance is, 
                          straight F with rightwards harpoon with barb upwards on top subscript r space equals space minus space k x space
    Therefore, to keep the mas in equilibrium, we have to apply the force Fa equal and opposite to -Fr.
    If the mass is further displaced by dx with rightwards harpoon with barb upwards on top, the amount of work done dW for this dispalcement by applied force is,
    dW space equals space straight F with rightwards harpoon with barb upwards on top subscript straight a space dx with rightwards harpoon with barb upwards on top space space space space space space space equals space straight F subscript straight a space dx space space space space space space space equals space straight k space straight x space dx

    Total amount of work done in order to displace the mass from mean position of A to C is using applied force Fa is, 
    straight W space equals space integral subscript straight A superscript straight C dW space equals space integral subscript 0 superscript straight a straight k space straight x space dx space equals space 1 half kx squared vertical line subscript 0 superscript straight a space equals space 1 half ka squared space
    Work done by applied force against this restoring force is stored in the form of potential energy is, 
                    straight W space equals space straight P space equals space 1 half ka squared
    Question 81
    CBSEENPH11017162
    Wired Faculty App

    Show that during a free fall - total energy (Kinetic + Potential) remains constant.

    Solution
    Let a body of mass m be dropped from point A at a height h from the ground.
     
    At point A: 
    The kinetic energy is,
                    K subscript A italic equals italic 1 over italic 2 m u to the power of italic 2 italic equals italic 1 over italic 2 m left parenthesis 0 right parenthesis squared italic equals 0
    Potential energy is,
                         U subscript A equals m g h
    ∴     Total energy at point A,
        E subscript A equals K subscript A plus U subscript A equals 0 plus m g h equals m g h             ...(1)
    At point B:
    Let in time t, the body reach point B after falling through a distance x.
    Let v be the velocity of body at B. 
    Now,
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    rightwards double arrow                v squared equals 2 g x 
    Kinetic energy at B is,
            K subscript B equals 1 half m v squared equals 1 half m left parenthesis 2 g x right parenthesis equals m g x
     
    Potential energy is,
                        U subscript B equals m g left parenthesis h minus x right parenthesis
    ∴ Total energy at point B is,
    space space space E subscript B equals K subscript B plus U subscript B equals m g x plus m g left parenthesis h minus x right parenthesis equals m g h   ...(2)
    At ground:
    On reaching the ground, let velocity of the body be V.

    ∴            V squared equals 2 g h
    Kinetic energy at ground is,
           K subscript g equals 1 half m V squared equals 1 half m left parenthesis 2 g h right parenthesis equals m g h 
    Potential energy is,
                  U subscript g equals m g left parenthesis 0 right parenthesis space equals space 0 
    ∴ Total energy at point ground is, 
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    Since total energy at A = total energy at B + total energy at ground.
    Therefore total energy during free fall remains conserved. 
     
     
     
    Question 82
    CBSEENPH11017163
    Wired Faculty App

    A body is dropped from a height h. Plot the potential energy curve versus distance fallen by body.

    Solution
    Consider the ground as the reference point.
    Let at any instant, the body fall by distance d.
    Therefore the position of body from ground is,
                            x = h - d

    The potential energy of body at that position is,
                    U = mgx= mg(h-d)
                           = mgh - mgd

    The graph between potential energy Vs. distance fallen by the body is a straight line as shown in figure below.

                
    Question 83
    CBSEENPH11017164
    Wired Faculty App

    When an arrow is shot from a bow, from where does it get kinetic energy?

    Solution
    When a bow is stretched, the rope of the bow acquires elastic potential energy due to its shape. Thus, when the arrow is shot, the potential energy of the rope is converted into kinetic energy of the arrow. 
    Question 84
    CBSEENPH11017165
    Wired Faculty App

    Does the potential energy change due to change in position only?

    Solution

    No, potential energy can also change due to shape.
    For e.g. in the figure below,
                  
    Therefore, the two springs are at the same height.
    Spring in figure (a) is in normal state while in figure (b) spring is in the extended state.
    Potential energy stored in spring (b) is greater than in spring (a).
    Consider that we two masses m1 and m2 separated by distance r. If we increase the mass m1, without changing the position of mass m2, potential energy stored in the mass m2 will increase. 

    Question 85
    CBSEENPH11017166
    Wired Faculty App

    State and prove work-energy relationship.

    Solution

    The Work- Energy Theorem states that the work done on or by the body is equal to the change in its kinetic energy. 
    Consider a force F applied on the body of mass m moving on the horizontal frictionless surface.
    Let after travelling a distances s, the velocity of body change from u to v.
    The small amount of work, dW done by force F in displacing the body by ds is, 
    dW = straight F with rightwards harpoon with barb upwards on top space. space stack d s with rightwards harpoon with barb upwards on top 
          equals space straight m space fraction numerator straight d straight v with rightwards harpoon with barb upwards on top over denominator dt end fraction. space ds with rightwards harpoon with barb upwards on top space equals space straight m space. space dv with rightwards harpoon with barb upwards on top space. space fraction numerator ds with rightwards harpoon with barb upwards on top over denominator dt end fraction equals space straight m space straight v with rightwards harpoon with barb upwards on top. space dv with rightwards harpoon with barb upwards on top equals space mv space dv space  
    dW = mv dv 
    Now, integrating both sides, we have 
    integral dW space equals space integral subscript straight u superscript straight v mv space dv space equals space straight m space integral subscript straight u superscript straight v straight v. space dv straight W space equals space straight m right enclose space straight v squared over 2 end enclose subscript straight u superscript straight v space equals space straight m over 2 left parenthesis space straight v squared space minus space straight u squared right parenthesis space straight W space equals space 1 half mv squared space minus space 1 half mu squared space
    That is, 
    Work = Change in Kinetic energy.

    Question 86
    CBSEENPH11017168
    Wired Faculty App

    State and prove law of conservation of mechanical energy.

    Solution
    Law of conservation of mechanical energy states that the sum of kinetic energy and potential energy of a body at any point remains constant throughout the motion.

    Let a body be displaced from position stack r subscript 1 with rightwards arrow on top space t o space stack r subscript 2 with rightwards arrow on top under the influence of conservative force.
    According to work energy relationship,
                    integral subscript stack straight r subscript 1 with rightwards arrow on top end subscript superscript stack straight r subscript 2 with rightwards arrow on top end superscript F with italic rightwards arrow on top italic. stack d s with italic rightwards arrow on top italic space italic equals italic space K subscript italic 2 italic minus K subscript italic 1                ...(1)
    But, work done on body by conservative force is negative of the change in potential energy.
    i.e.            integral subscript stack straight r subscript 1 with rightwards arrow on top end subscript superscript stack straight r subscript 2 with rightwards arrow on top end superscript F with rightwards arrow on top times stack d s with rightwards arrow on top equals negative left parenthesis U subscript 2 minus U subscript 1 right parenthesis           ...(2)
    From (1) and (2),
                    K subscript 2 minus K subscript 1 equals negative left parenthesis U subscript 2 minus U subscript 1 right parenthesis
    rightwards double arrow             space space K subscript 2 plus U subscript 2 equals K subscript 1 plus U subscript 1
    i.e. Sum of kinetic energy and potential energy is constant throughout the motion.
    Question 87
    CBSEENPH11017169
    Wired Faculty App

    In a thermal station, coal is used for the generation of electricity. Mention how energy changes from one form to another form.

    Solution
    The chemical energy is converted into heat energy, when coal is burnt.
    The heat energy is converted into mechanical energy by heat engine and the mechanical energy is converted into electric energy by dynamo.
    Question 88
    CBSEENPH11017170
    Wired Faculty App

    Can a body have energy without momentum or vice-versa? Explain.

    Solution
    It is possible for a body to have energy without momentum.
    For example, a body at rest in gravitational field possesses the potential energy.
    That is, the potential energy stored in stretched spring.
    But the body cannot have momentum without energy because if the body has momentum, it also possesses the kinetic energy.
    Question 89
    CBSEENPH11017171
    Wired Faculty App

    Two springs of spring constants k1 and k2 are stretched by the same force. First spring undergoes more extension than the second one. In which of the two springs, energy stored will be more?

    Solution

    According to Hook's law we have,
              F = kx
    rightwards double arrow      k = F/x
    Energy stored in stretched spring is,
                     U equals 1 half k x squared equals 1 half F over x x squared equals 1 half F x
    Hence for the same force applied, the energy stored in spring is directly proportional to extension.
    Therefore, energy stored in first spring will be greater than the second spring.

    Question 90
    CBSEENPH11017172
    Wired Faculty App

    What is work done by electrostatic force of nucleus on electron when it revolves around the nucleus?

    Solution
    Work done by electrostatic force is zero. This is because electrostatic force and instantaneous displacement are always perpendicular to each other. [cos 900 = 0]
    Question 91
    CBSEENPH11017173
    Wired Faculty App

    A truck and car are both moving with same kinetic energy and brought to rest by applying brakes. The brakes give equal retarding forces to both the vehicles. Which of the two vehicles come to rest after travelling shorter distance?

    Solution
    When brakes are applied, the frictional forces start acting on the drum of the wheel.
    If F is the force of friction and s is the distance travelled by truck or car, then work done against force of friction is at the cost of kinetic energy of vehicles.
    That is, work done 
    i.e.                    F. s space equals space Kinetic space energy
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    Since kinetic energy and the retarding force on both the vehicles are same, therefore both will come to rest after travelling equal distances.
    Question 93
    CBSEENPH11017176
    Wired Faculty App

    Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

    Solution
    The gravitational force is conservative force.
    We know that, work done by conservative force is independent of path followed, but depends upon initial and final position. Therefore, over the closed path, the work done by conservative force is zero. 
    Question 94
    CBSEENPH11017177
    Wired Faculty App

    An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

    Solution

    For a satellite orbiting the earth, the necessary centripetal force is applied by gravitational pull.
    That is, 
    space space space G fraction numerator M m over denominator r squared end fraction equals fraction numerator m v squared over denominator r end fraction
    rightwards double arrow  left parenthesis m v r right parenthesis v equals G M m equals constant
    Retarding torque is applied by the air friction. Hence, there is a decrease in the angular momentum. Therefore, the speed increases.

    Question 95
    CBSEENPH11017178
    Wired Faculty App

    A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the trolley's floor at the rate of 0.05 kg m/s. What is the speed of the trolley after the entire sand bag is empty?

    Solution
    Sand leaks from the trolley with the same velocity which the trolley has initially.
    Therefore, there is no change in momentum of sand and hence, the trolley.
    Therefore, after the entire bag is emptied, the velocity of the trooley remains the same, i.e., 27 km/hr.
    Question 96
    CBSEENPH11017179
    Wired Faculty App

    The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

    Solution
    In case of rocket, heat energy required for burning is at the cost kinetic energy of rocket.
    Due to air friction the kinetic energy is converted into heat energy which burns up the rocket.
    Question 97
    CBSEENPH11017180
    Wired Faculty App

    In a tug of war, one team is slowly giving way to the other. What work is being done and by whom?

    Solution
    The net force acting on the losing team is the resultant of the forces applied by the two teams. So, it undergoes displacement. Positive work is done by the winning team on the losing team. 
    The force applied by the losing team is zero. This is because the winning team does not move. The losing team does not displace from its position and hence work done by the losing team is zero. 
    Question 98
    CBSEENPH11017181
    Wired Faculty App

    The roads on the mountains are not straight up the slope but wind up gradually. Why?

    Solution
    When the road goes straight up, then inclination of road with horizontal is large. Therefore the limiting friction μmgcos θ will be small and component of weight along the slope wgsin θ will be large. Hence, the vehicle slips on the road. 
    Also, inorder to move the vehicle up the road, large force will be required due to larger component of weight mg sinθ. Thus, high power engines will be required.
    If the slope of the road is decreased, these difficulties can be overcome. Slope of the road is decreased by winding up the road around the mountain. 
    Question 99
    CBSEENPH11017182
    Wired Faculty App

    What is mass-energy relationship? What is its significance?

    Solution
    Einstein unified mass and energy and said that one can be converted into another and the two entities are related by an equation.
    The equation is given by,
                                 E = mc2
    where, m = mass that disappears,
    E = energy that appears, 
    c = velcoity of light in vacuum 
    So, according to modern Physics, mass and energy are not conserved seperately, but are conserved in a single entity called 'mass-energy'.
    Significance of this relation:
    Many unsolved problems has been solved in Physics after the unification of this theory. According to this relation, as c is large; even a small mass difference can produce enormous amount of energy. 


    Question 100
    CBSEENPH11017186
    Wired Faculty App

    If momentum of a body is decreased by 20% without changing the mass of the body, by what percentage kinetic energy of body will change?

    Solution

    Let pi be the initial momentum of body and Ki be the initial kinetic energy of the body.
    If momentum is decreased by 20% then final momentum is,
                   p subscript f equals end subscript p subscript i minus 0.2 p subscript i equals 0.8 p subscript i

    Since, kinetic energy is given by,
                        K equals fraction numerator p squared over denominator 2 m end fraction
    ∴ Final kinetic energy is,
                  K subscript f equals fraction numerator P subscript f superscript 2 over denominator 2 m end fraction equals fraction numerator left parenthesis 0.8 p subscript i right parenthesis squared over denominator 2 m end fraction
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    Now change in Kinetic energy is,
                 increment K space equals space K subscript f minus K subscript i space equals space minus 0 cross times 36 space K subscript i
    Percentage change in kinetic energy is,
                     fraction numerator K subscript f minus K subscript i over denominator K subscript i end fraction 100 space equals space minus 36 %
    Negative sign shows that there is decrease of energy.
                                

    Question 101
    CBSEENPH11017188
    Wired Faculty App

    The velocity of a car is doubled. What is the ratio of percentage increase in kinetic energy to the percentage increase in momentum?

    Solution
    Let u be the initial velocity of the car.
    On doubling its velocity,
    Final velocity becomes 2u.
    Therefore,
    Initial momentum, p subscript i space equals space m u
    Final momentum,space space space p subscript f equals 2 m u
    Percentage increase in momentum is,
                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Initial kinetic energy, K subscript i equals 1 half m u squared 
    Final kinetic energy, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Percentage increase in kinetic energy is,
    space space fraction numerator 4 K minus K over denominator K end fraction cross times 100 % space equals space 300 %
    Ratio of percentage is, 
    Now, 
    fraction numerator Percentage space increase space in space straight K. straight E. over denominator Percentage space increase space in space momentum end fractionequals fraction numerator 300 % over denominator 100 % end fraction equals 3
    Question 102
    CBSEENPH11017190
    Wired Faculty App

    A heavy body and a light body have equal kinetic energies. Which of the two has greater momentum?

    Solution

    We know, kinetic energy is given by,
                         K equals fraction numerator p squared over denominator 2 m end fraction
    rightwards double arrowrightwards double arrow                    p squared equals 2 m K
    or                 p equals square root of 2 m K end root
    That is, we can see that momentum is directly proportional to the square root of the mass.
    Therefore, heavy body will have greater momentum than the light body.

    Question 103
    CBSEENPH11017191
    Wired Faculty App

    An electron and proton have equal momentum. Which has more kinetic and what is the ratio of kinetic energy of electron to that of proton?

    Solution

    Given,
    An electron and proton have   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    We know,
    Kinetic energy, K equals fraction numerator p squared over denominator 2 m end fraction
    Therefore,
    Kinetic energy of the electron, K subscript e equals fraction numerator p squared over denominator 2 m subscript e end fraction
    Kinetic energy of the proton, K subscript p equals fraction numerator p squared over denominator 2 m subscript p end fraction
    Ratio is given by, 
    Now, K subscript e over K subscript p equals fraction numerator p squared divided by 2 m subscript e over denominator p squared divided by 2 m subscript p end fraction equals m subscript p over m subscript e equals 1837 greater than 1
    That is, kinetic energy of electron is greater than proton and ratio of kinetic energy of electron to proton is equal to 1837.

    Question 104
    CBSEENPH11017192
    Wired Faculty App

    A 30gm bullet initially travelling with speed 500m/s penetrates 12 cm into a wooden block. Find the amount of work done to the resistance offered by the block to the bullet.

    Solution
    According to the work-kinetic energy theorem, the kinetic energy of bullet does the work against resistance offered by the block.
    That is,
                         1 half m v squared equals F s
    rightwards double arrow                         F equals fraction numerator m v squared over denominator 2 s end fraction
    Here, we have
    Mass of the bullet, straight m equals 30 gm equals 0.030 kg
    Velocity of the bullet, straight v equals 500 straight m divided by straight s comma
    Distance through which the bullet penetrates the wooden block = <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
              
    ∴        F equals fraction numerator left parenthesis 0.03 right parenthesis cross times left parenthesis 500 right parenthesis squared over denominator 2 cross times 0.12 end fraction equals 3125 straight N 
    Question 105
    CBSEENPH11017193
    Wired Faculty App

    A body is projected horizontally on a rough surface with velocity v and comes to rest after moving a distance S. What is the coefficient of friction?

    Solution
    The surface is rough, therefore kinetic energy of body is used to do work against frictional force.
    Let μ be the coefficient of friction.

    ∴    Force of friction is,
                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    The work done against friction is,
                     W equals mu m g S
    Therefore,
                  space space 1 half m v squared equals mu m g S
    rightwards double arrow                   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>, is the coefficient of friction. 
    Question 106
    CBSEENPH11017194
    Wired Faculty App

    A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1.0m/s and then has the same kinetic energy as the boy. What were original speed of man and boy?

    Solution
    Let M and m be the mass of man and boy respectively.
    Let, the original velocities of man and boy be v1 and v2  respectively.
    we have,
    Mass of the boy is half of mass of man.
    That is,  m equals M over 2
    And K.E of the boy is equal to the K.E of man.
    i.e.,         K subscript m a n end subscript space equals space 1 half K subscript b o y end subscript
    i.e.          1 half M v subscript 1 squared equals open square brackets 1 half open parentheses straight M over 2 close parentheses straight v subscript 2 squared close square brackets
    rightwards double arrow                   straight v subscript 2 equals 2 straight v subscript 1                       ...(1)
    When the speed of man increases by 1m/s, the kinetic energies of both man and boy become same.
    i.e.      1 half straight M left parenthesis straight v subscript 1 plus 1 right parenthesis squared equals open square brackets 1 half open parentheses straight M over 2 close parentheses straight v subscript 2 squared close square brackets 
    rightwards double arrow           2 left parenthesis straight v subscript 1 plus 1 right parenthesis squared equals straight v subscript 2 squared                 ...(2)
    From eqn. (1) and (2),
                         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    On solving we get,
    Velocity of the man, straight v subscript 1 equals 2.41 space straight m divided by straight s
    Velocity of the boy, straight v subscript 2 equals 2 straight v subscript 1 equals 4.82 space straight m divided by straight s
    Question 107
    CBSEENPH11017195
    Wired Faculty App

    A car of mass 1000kg is travelling up a hill of 1 in 49, with engine working at 49kW. Find the greatest speed the car attained up the hill. Find the velocity of car at the instant when the acceleration of car is 2 m/s2. The force of friction is 300N.

    Solution

    Component of weight of car down the hill is,
            m g space sinθ space equals space 1000 cross times 9.8 cross times 1 over 49 equals 200 straight N
    Force of friction on the car acting down the hill =  <pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/82/ca/950e58fb57e60946122e4d4f3f20.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/82/ca/950e58fb57e60946122e4d4f3f20.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php(12): file_put_contents('/home/config_ad...', 'mml=<math xmlns...') #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(48): sys_io_File::saveContent('/home/config_ad...', 'mml=<math xmlns...') #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(112): com_wiris_util_sys_Store->write('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #6 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #7 {main}</pre>
    Total force acting on the car down the slope of hill is,
                        straight F equals mg space sinθ plus straight f equals 500 space straight N
    Let straight v subscript 0 be the maximum speed of the car. 
    Therefore,          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow              49000 space equals space 500 v subscript 0
    rightwards double arrow                      straight v subscript 0 equals 98 space straight m divided by straight s
    Let v be the speed of car when its acceleration is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Force exerted by engine on the car at this moment is,
                               <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Thus,                  straight P equals straight F apostrophe straight v
    i.e.                 49000 = 2500 v
                                  v = 19.6 m/s.

    Question 108
    CBSEENPH11017196
    Wired Faculty App

    A body of mass 5 kg initially at rest is subjected to a force of 20N. What is the kinetic energy acquired by the body at the end of 10 seconds?

    Solution

    Initial velocity of body, u = 0 m/s
    Mass of body, m = 5 kg
    Force on the body, F = 20N
    Time for which force acts on body t  = 10 sec.
    Therefore,
    Acceleration of body, a= 4m/s2

    Now the displacement of body is,
                     straight S equals ut plus 1 half at squared
                      equals 0 plus 1 half cross times 4 cross times left parenthesis 10 right parenthesis squared equals 200 space straight m 
    Work done by force, 
             straight W equals FS equals 20 cross times 200 space equals space 4000 straight J 
    And this work is converted into K.E.
    Hence kinetic energy =  4 KJ 

    Question 109
    CBSEENPH11017197
    Wired Faculty App

    A bullet of mass 25gm moving with velocity 400m/s hits the wall and crosses out of it with speed 150m/s. Find the work done by bullet against friction in crossing the wall.

    Solution
    Work done in crossing the wall is equal to decrease in kinetic energy of bullet. 
    So, work done by the bullet is, 
    space space space space straight W equals 1 half straight m left parenthesis straight u squared minus straight v squared right parenthesis
         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
        space space equals 1718.5 space straight J
    Question 110
    CBSEENPH11017199
    Wired Faculty App

    A body of mass M moving with velocity u suddenly explodes into two fragments. If a fragment of mass m comes to rest, then find the Q value of explosion.

    Solution
    Let v be the velocity of second fragment of mass (M – m).
    According to law of conservation of momentum,
      Mu = (M- m) v
    rightwards double arrow space straight v space equals space open parentheses fraction numerator straight M over denominator straight M space minus space straight m end fraction close parentheses straight u
    Kinetic energy of mass M before explosion is, 
    K1 = 1/2 Mu2 

    Kinetic Energy after explosion is, 
    K2 = 1 half left parenthesis space straight M space minus space straight m right parenthesis thin space straight v squared space equals space 1 half left parenthesis space straight M space minus space straight m right parenthesis space open parentheses fraction numerator straight M over denominator straight M space minus space straight m end fraction straight u close parentheses squared 
        equals space open parentheses fraction numerator straight M over denominator straight M space minus space straight m end fraction close parentheses space 1 half M straight u squared Now comma space straight Q space value space equals space straight K subscript 2 space minus space straight K subscript 1 space equals space open parentheses fraction numerator straight M over denominator straight M space minus straight m end fraction close parentheses space 1 half Mu squared space minus space 1 half Mu squared space equals 1 half open parentheses fraction numerator mM over denominator straight M space minus straight m end fraction close parentheses space straight u squared 
    Question 111
    CBSEENPH11017201
    Wired Faculty App

    An electron and a proton are detected in a cosmic ray experiment, the first with kinetic
    energy 10 keV, and the second with 100 keV. Which is faster, the electron or the
    proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass
    = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).
    Solution

    Given that,
    Kinetic energy, straight K subscript straight e equals 10 KeV space equals space 10 cross times 1.6 cross times 10 to the power of negative 16 end exponent straight J 
                              equals 1.6 cross times 10 to the power of negative 15 end exponent straight J 
    K.E of proton, straight K subscript straight p space equals space 100 KeV space equals space 100 cross times 1.6 cross times 10 to the power of negative 16 end exponent straight J 
                            space space equals 1.6 cross times 10 to the power of negative 14 end exponent straight J 
    K.E is given by, 
                 space space space straight K equals 1 half mv squared 
    rightwards double arrow            straight v equals square root of fraction numerator 2 straight K over denominator straight m end fraction end root 
    ∴ ratio of their speed is given by,
                       space space space straight v subscript straight e over straight v subscript straight p equals square root of straight K subscript straight e over straight K subscript straight p straight m subscript straight p over straight m subscript straight e end root equals square root of 1 over 10 cross times fraction numerator 1.67 cross times 10 to the power of negative 27 end exponent over denominator 9.1 cross times 10 to the power of negative 31 end exponent end fraction end root 
            = 13.54 
    Thus electron is faster than proton by a magnitude of 13.54. 

    Question 112
    CBSEENPH11017238
    Wired Faculty App

    A body of mass 13-28 kg is acted upon by a force as shown in figure. Find the impulse produced by the force over a displacement of 30m. Assume body starts from rest. 

    Solution
    Work done by the force on the body is equal to the area under the graph.
    Work done = Area of rect. 1 + Area of trapezium 2
                       + Area of rect. 3 + Area of trapezium 4. 
           = 16 X 10 + 1/2 (16 + 10) 5+ 10 X 10 + 1/2 ( 10 + 6)5
            = 160 + 65 + 100 + 40
            = 305 J 
    Work done by force appears in the  form of kinetic energy of body. 
    Therefore, 
    W = K.E =  p2 / 2m 
    That space is comma space straight p space equals space square root of 2 mW end root space equals space square root of 2 space cross times space 13.28 space cross times space 305 end root space space space space equals space 90 space Ns space
    Since, initially the body is at rest, so impulse produced by force is equal to final momentum i.e. 90Ns. 
    Question 113
    CBSEENPH11017239
    Wired Faculty App

    The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
    Solution

    Let, 
    Mass of the bob = m
    Length of the string = l 
                       
    At B, P.E of the bob = 0
    At A, P. E of the bob = mgl 
    When bob reaches the point B, 5% of the potential energy is dissipated against air resistance, and rest of 95% is converted in kinetic energy.
    That is,
     
                      1 half mv squared equals 0.95 mgl 
    rightwards double arrow            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
                        equals square root of 2 cross times 0.95 cross times 9.8 cross times 1.5 end root 
                       equals 5.28 space straight m divided by straight s, is the speed with which the boy arrives. 
     
                                

     
    Question 114
    CBSEENPH11017240
    Wired Faculty App

    The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
    Solution

    Given,
    Potential energy for a particle executing linear simple harmonic motion is,
                       space space straight V left parenthesis straight x right parenthesis space equals space 1 half kx squared 
    where,space space space space straight k equals 1 half straight N divided by straight m
                   
    Total energy of particle E = 1J
    Since at extreme position total energy is potential energy,
    ∴                space space space space space space straight V equals 1 half kx squared equals 1 straight J
    rightwards double arrow                straight x equals plus-or-minus square root of 2 over straight k end root equals plus-or-minus square root of 4
    rightwards double arrow                   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Hence the results.

    Question 115
    CBSEENPH11017241
    Wired Faculty App

    A body of mass 2 kg initially at rest moves under the action of an applied horizontal forces of 7N on a table with co-efficient of kinetic friction 0.1. Compute:
    (a) the work done by the applied force in 10 sec.,
    (b) the work done by the friction force in 10 sec.,
    (c) the work done by the net force on the body in 10 sec.,
    (d) the change in kinetic energy of the body in 10 sec and interpret your results. 
    Solution
    It is given that, 
    Mass of the body, m = 2 kg
    Force acting on the body, F = 7N
    Coefficient of friction, space straight mu equals 0.1 
    Time taken, t = 10 sec


    The force of friction between block and surface is,
                           f equals mu space m g space equals space 2 straight N

     The net accelerating force on body is,
                       F - f = 7 - 2 = 5N
    Therefore,
    Acceleration, a equals F over m equals 2.5 m divided by s squared
    Now the displacement undergone by the body in 10 sec is,
                        straight S equals ut plus 1 half at squared
                         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    (a) Work done by applied force in 10 sec.,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    (b) Work done by friction force is,
         space space straight W subscript friction space equals space straight f times straight s space equals space 2 cross times 125 space equals space 250 space straight J
    (c) Work done by net force is,
          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    (d) Change in kinetic energy of body is equal to the work done by net force = 625 J.


    Question 116
    CBSEENPH11017242
    Wired Faculty App

    When a constant force is applied to a body, is power of force constant? If not, how the force should be varied to keep the power of force constant?

    Solution

    We know, Power = Force X velocity.
    When a constant force is applied on a body, it accelerates the body and hence speed of the body will change.
    Therefore, the power of force will also change.
    To keep the power of force constant, F v  should be constant.
    i.e., force should be inversely proportional to the speed of the body.

    Question 117
    CBSEENPH11017243
    Wired Faculty App

    A person A pushes a body of 5 kg placed on rough surface of coefficient of friction 0.3 by a distance of 4m in 15 seconds. Another person B pushes the body of 7 kg on same surface by a distance of 3m in the same time. Who has done more work and has more power?

    Solution
    Person A:
    Mass of body, straight m subscript 1 space equals space 5 space kg 
    Coefficient of friction,space space space straight mu equals 0.3 
    ∴  The force of friction, 
        straight F subscript 1 equals μm subscript 1 straight g equals 0.3 cross times 5 cross times 10 equals 15 straight N 
    Displacement, straight s subscript 1 equals 4 straight m 
    Therefore,
    Work done, straight W subscript 1 equals 15 cross times 4 equals 60 straight J 
    Time taken, straight t subscript 1 equals 15 space sec 
    ∴  Power, straight P subscript 1 equals 60 over 15 equals 4 straight W 
    Person B: 
    Mass of body, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Coefficient of friction,space space straight mu equals 0.3 space 
    ∴   Force of friction, 
             straight F subscript 2 equals μm subscript 2 straight g equals 0.3 cross times 7 cross times 10 equals 21 straight N 
     Displacement,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    ∴  Work done, straight W equals 21 cross times 3 equals 63 straight J 
    Time taken, straight t subscript 2 equals 15 space sec 
    ∴ Power, straight P subscript 2 equals 63 over 15 equals 4.2 space straight W
    Thus, the person B has more energy and power.
    Question 118
    CBSEENPH11017244
    Wired Faculty App

    A body of mass m accelerates uniformly from rest to v1 in time t1. Find the instantaneous power delivered to the body at any instant. 

    Solution
    As the speed of the body increases by v1 in time t1, the acceleration of body becomes, 
                              straight a equals straight v subscript 1 over straight t 
    The instantaneous power delivered to the body by accelerating force is, 
                             straight P space equals space F v space space space space equals m a v space space space space equals m a squared straight t space space space equals straight m open parentheses straight v subscript 1 over straight t subscript 1 close parentheses squared straight t space 
    Question 119
    CBSEENPH11017245
    Wired Faculty App

    An object of mass 4kg falls from height 44.1m and reaches the ground with speed 24.5 m/s. Find the work done by air friction on the object.

    Solution

    Given, 
    Mass of the object, m = 4 kg 
    Height from which the object falls, h = 44.1 m
    Speed with the mass is falling, v = 24.5 m/s
    The work done by gravity on the object, 
     straight W subscript 1 equals mgh equals 4 cross times 9.8 cross times 44.1 
      space space space equals 1728.72 straight J                         

    Increase in kinetic energy of object is, 
    increment straight K equals 1 half mv squared space space space space space space equals 1 half cross times 4 cross times left parenthesis 24.5 right parenthesis squared 
         = 1200.5J 
    Therefore work done by friction is, 
      straight W subscript straight f equals straight W subscript 1 minus increment straight K 
         equals 1728.72 minus 1200.5 
         equals 528.22 space straight J 

    Sponsor Area

    Question 120
    CBSEENPH11017246
    Wired Faculty App

    A rod of length 1m and mass 0.8kg is lying on a horizontal table. To make it vertical, how much work is required?

    Solution
    Given, 
    Length of the rod, l = 1 m 
    Mass, m = 0.8 kg
    When the rod is made vertical from horizontal position, the centre of mass of the rod goes up through height half the length of rod i.e. 0.5m.
    Therefore increase in the potential energy of rod is, 
    ΔU space equals space mgh space space space space space space space equals space 0.8 cross times 9.8 cross times 0.5 straight J 
        equals 3.92 straight J 
    This increase in the potential energy is at the cost of work done on the rod.
    Thus, work done is,
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    Question 121
    CBSEENPH11017247
    Wired Faculty App

    Find the horse power used in pumping 250kg of water per minute from a well 30m deep. If the efficiency of pump is 60%, then also find the horse power of pump.

    Solution

    Given,
    Mass of water pumped per minute, m= 250 kg,
    Depth of the well, d = 30m
    Time taken to pump the water, t = 60 s
    The horse-power used for pumping out the water is, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
     equals 1225 straight J space equals space 1.642 HP 
    Efficiency of pump = 60%
    Therefore, horse power of pump is, 
        straight P subscript 0 equals 1.642 cross times 100 over 60 equals 2.737 space HP

    Question 122
    CBSEENPH11017248
    Wired Faculty App

    A body moves under a force such that momentum of body at any instant is,

    straight p with rightwards arrow on top space equals space straight p space cost space straight i with hat on top space plus space straight p space sint space straight j with hat on top 

    What is the rate of work done by the force?

    Solution
    Since the momentum of the body at any instant is given by,
                 straight p with rightwards arrow on top space equals space straight p space cost space straight i with hat on top space plus space straight p space sint space straight j with hat on top 
    ∴ Force is given by, straight F with rightwards arrow on top equals fraction numerator straight d straight p with rightwards arrow on top over denominator dt end fraction equals negative straight p space sint space straight i with hat on top plus straight p space cost space straight i with hat on top 
    We know, the rate of work done is power P which is given by, 
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
      equals fraction numerator left parenthesis negative straight p space sint space straight i with hat on top plus straight p space cost space straight j with hat on top right parenthesis times left parenthesis straight p space cost space straight i with hat on top plus straight p space sint space straight j with hat on top right parenthesis over denominator straight m end fraction 
      equals fraction numerator negative straight p squared sint space cost space plus space straight p squared space cost space sint over denominator straight m end fraction equals 0 
    ∴  Rate of work done is zero.
    Question 123
    CBSEENPH11017249
    Wired Faculty App

    A body is moving along a straight line by a machine delivering constant power. How does the distance travelled by body will depend upon time?

    Solution

    We have, 

    Power, P equals F v 
              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow  vdv equals straight P over straight m dt 
    Integrating both sides, we have, 
          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    rightwards double arrow          straight v equals dx over dt equals square root of fraction numerator 2 straight P over denominator straight m end fraction end root left parenthesis straight t right parenthesis to the power of 1 divided by 2 end exponent 
                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    On integrating both sides, we have 
            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    rightwards double arrow           straight x equals square root of fraction numerator 2 straight P over denominator straight m end fraction end root space open parentheses 2 over 3 left parenthesis straight t right parenthesis to the power of 3 divided by 2 end exponent close parentheses 
    rightwards double arrow           straight x equals open parentheses square root of fraction numerator 8 straight P over denominator 9 straight m end fraction end root close parentheses straight t to the power of 3 divided by 2 end exponent 
    ∴              straight x proportional to straight t to the power of 3 divided by 2 end exponent 
    x is the distance and t is the time travelled. 

    Question 124
    CBSEENPH11017250
    Wired Faculty App

    What is the minimum amount of energy released in the annihilation of electron positron pair?

    Solution
    Energy released during the annihilation of e–, e+ pair is, 
    space space space space space space straight E equals 2 straight m subscript straight c straight c squared 
          equals 2 cross times 9.1 cross times 10 to the power of negative 31 end exponent cross times left parenthesis 3 cross times 10 to the power of 8 right parenthesis squared space straight J 
          equals fraction numerator 2 cross times 9.1 cross times 9 cross times 10 to the power of negative 15 end exponent over denominator 1.6 cross times 10 to the power of negative 13 end exponent end fraction MeV 
          equals 1.02 space MeV 
    Question 125
    CBSEENPH11017251
    Wired Faculty App

    Estimate the amount of energy released in the nuclear fusion reaction:

              straight H presuperscript 2 subscript 1 plus straight H presuperscript 2 subscript 1 rightwards arrow He presuperscript 3 subscript 2 plus straight n 

    straight m left parenthesis straight H presuperscript 2 subscript 1 right parenthesis equals 2.0141 space amu comma space space straight m left parenthesis He presuperscript 3 subscript 2 right parenthesis space equals space 3.0160 space amu comma

    straight m subscript straight n equals 1.0087 space amu 

    where 1 space amu space equals space 1.661 cross times 10 to the power of negative 27 end exponent kg 

    Express your answer in units of MeV. 

    Solution

    The given reaction is,

                     straight H presuperscript 2 subscript 1 plus straight H presuperscript 2 subscript 1 rightwards arrow He presuperscript 3 subscript 2 plus straight n 
    From the data given, we have
    Total mass of reactant = 2 straight m left parenthesis straight H presuperscript 2 subscript 1 right parenthesis 
                                      = 2(2.0141) amu 
                                       = (4.0282) amu 
    Total mass of product = straight m left parenthesis He presuperscript 3 subscript 2 right parenthesis plus straight m subscript straight n 
                                     = (3.0160) + (1.0087) 
                                     = 4.0247 amu 
    The decrease in mass is, 
        Δm space equals space left parenthesis 4.0282 minus 4.0247 right parenthesis amu 
              = (0.0035) amu 
              = 5.8135 space cross times space 10 to the power of negative 30 end exponent kg 
    The energy released during the reaction is, 
          space space straight E equals Δmc squared 
             equals 5.8135 space cross times space 10 to the power of negative 30 end exponent cross times left parenthesis 3 cross times 10 to the power of 8 right parenthesis squared 
             equals 5.23 cross times 10 to the power of negative 13 end exponent straight J 
             equals fraction numerator 5.23 cross times 10 to the power of negative 13 end exponent over denominator 1.6 cross times 10 to the power of negative 13 end exponent end fraction MeV 
         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
                  

    Question 126
    CBSEENPH11017252
    Wired Faculty App

    Plot the graph between potential energy, potential energy stored in the spring v/s extension.

    Solution

    Potential energy, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Thus, the graph between potential energy stored in the spring and extension will be a parabola. 
                  
                        

    Question 127
    CBSEENPH11017253
    Wired Faculty App

    Two springs of spring constants k1 and k2 are stretched by the same force. First spring undergoes more extension than the second one. Which of the two springs:

    (i) has more value of spring constant

    (ii) stores more energy?

    Solution

    Using Hook's law, 
                             F = kx 
    i.e.,                   space space straight k equals straight F over straight x 
    Since first spring extends more than second,  therefore,  straight k subscript 1 less than straight k subscript 2. 
    Energy stored in stretched spring is, 
              straight U equals 1 half kx squared equals 1 half straight F over straight x straight x squared equals 1 half Fx 
    Hence, for the same force applied, the energy stored in spring is directly proportional to its extension.
    Therefore, energy stored in first spring will be greater than the second spring. 

    Question 128
    CBSEENPH11017254
    Wired Faculty App

    A 9kg block moving on a frictionless surface with a speed of 5m/s collides with a spring placed in horizontal position against the rigid support. If the spring constant is 144 N/m, then find the maximum compression in the spring.
    Solution
    Given, 
    Mass of the block, m = 9 kg
    Speed with which the block is moving, v = 5 m/s
    Spring constant, k = 144 N/m 
    Let x be the maximum compression in the spring.
    The potential energy stored in the spring at maximum compression is at the cost of kinetic energy of moving block.
    Therefore, 
                   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
                          straight x space equals space straight v square root of straight m over straight k end root space space space equals 5 square root of 9 over 144 end root 
                             = 1.25 m
    The spring compresses by 1.25 m. 
    Question 129
    CBSEENPH11017255
    Wired Faculty App

    25 J of work is required to stretch the spring by 2cm. Find the work required to be done to further extend the spring by 4cm.

    Solution
    Given, 
    Amount of work required to stretch the spring = 25 J 
    Let k be the spring constant of the spring.
    As 25J of work is required to stretch the spring by 2cm, thus 
                25 equals 1 half straight k left parenthesis 0.02 right parenthesis squared                            ...(1) 
    Let W work be required to further stretch the spring by 4 cm.
    Therefore, 
              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>               ...(2) 
    Dividing equation (2) by (1), we get 
                   straight W over 25 equals fraction numerator left parenthesis 0.06 right parenthesis squared minus left parenthesis 0.02 right parenthesis squared over denominator left parenthesis 0.02 right parenthesis squared end fraction equals 8 
    i.e.,             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Question 130
    CBSEENPH11017256
    Wired Faculty App

    A spring of force constant 10N/m is lying on a frictionless table with one end fixed and another end attached with 0.1 kg mass. The mass is pushed towards fix end by 8cm and released. Find the velocity of mass when the spring becomes normal.

    Solution
    Given, 
    Spring constant, k = 10 N/m
    The potential energy stored in spring when compressed by 8cm is,
    straight U space equals 1 half kx squared space space space equals 1 half 10 cross times left parenthesis 0.08 right parenthesis squared 
      equals 0.032 straight J 
    Let, v be the velocity of mass when spring becomes normal.
    Applying the law of conservation of energy, we get 
              space space 1 half mv squared equals straight U  
                       straight v space equals square root of fraction numerator 2 straight U over denominator straight m end fraction end root space space space equals square root of fraction numerator 2 cross times 0.032 over denominator 0.1 end fraction end root space space space equals 0.8 straight m divided by straight s
    Question 131
    CBSEENPH11017257
    Wired Faculty App

    A 1 kg block collides with a horizontal weightless spring of constant 36N/m. The block compresses the spring by 4cm. Find the speed of block at the instant of collision:

    (i) if the surface is frictionless

    (ii) if surface is rough and the coefficient of friction of the surface is 0.2.

    Solution

    Mass of the block, m = 1 kg 
    Spring constant, k = 36 N/m 
    Spring compressed by 4 cm. 
    (i)  The potential energy stored in the spring when compresses by 4 cm is, 
          straight U equals 1 half kx squared equals 1 half 36 cross times left parenthesis 0.04 right parenthesis squared equals 0.0288 straight J 
    Let v be the velocity of the block at the time of the collision.
    According to the conservation of energy, the kinetic energy of block at the time of collision is equal to the potential energy in spring at maximum compression.
    i.e.,          1 half mv squared equals 1 half kx squared 
                            straight v equals square root of kx squared over straight m end root equals square root of fraction numerator 36 cross times left parenthesis 0.04 right parenthesis squared over denominator 1 end fraction end root 
                           space space space equals 0.24 straight m divided by straight s, is the speed of the block.

    (ii) Work done in displacing the mass by 4cm on rough surface is, W= μmgS
                                 = 0.2x1x9.8x0.04
                                 = 0.0784J 
    Now, the kinetic energy at the time of collision is equal to sum of the energy lost by block in overcoming friction and potential energy stored in the spring. 
    i.e.      1 half mv apostrophe squared equals straight W plus straight U space space space space space space space space space space space space space equals 0.0784 plus 0.288 
        1 half cross times 1 cross times straight v to the power of apostrophe 2 end exponent equals 0.1072 
                      straight v apostrophe equals 0.463 space straight m divided by straight s , is the speed of the block. 

    Question 132
    CBSEENPH11017258
    Wired Faculty App

    Two bodies A and B of masses 2m and 3m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass 2m moving with velocity v0 hits A along AB elastically as shown in figure. At a certain instant of time when x in the spring is maximum, the velocity v of both the masses is same. Find x and v.

    Solution

    Given that initially, A is at rest and C is moving with velocity v0.
    The masses of A and C are same, therefore when C hits A elastically, it itself gets stopped and A starts moving with velocity v0. There is a transfer of velocity in an elastic collision. 
    Just after collision, A moves with velocity v0.
    B is at rest and spring is normal.
    The momentum of system (bodies A, B and spring) is 2mv0.

    Energy, E =  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> , which is in the form of kinetic energy of A.
    No external force is acting on the system, therefore, the momentum of the system remains conserved. 
    Using conservation of momentum at the instant of maximum compression, 
                    2 mv subscript 0 equals left parenthesis 2 straight m plus 3 straight m right parenthesis straight v 
    rightwards double arrow                  straight v equals 2 over 5 straight v subscript 0 
    Now to find x, again using the principle of conservation of momentum at the instant of compression, we have
               1 half left parenthesis 2 straight m right parenthesis straight v subscript straight sigma superscript 2 equals 1 half left parenthesis 2 straight m plus 3 straight m right parenthesis straight v squared plus 1 half kx squared 
    rightwards double arrow             space space mv subscript 0 superscript 2 equals 5 over 2 straight m open parentheses 2 over 5 straight v subscript 0 close parentheses squared plus 1 half kx squared 
    i.e.,                space space straight x equals square root of 6 over 5 fraction numerator mv subscript 0 superscript 2 over denominator straight k end fraction end root

    Question 133
    CBSEENPH11017259
    Wired Faculty App

    What is collision?

    Solution
    Collision between two bodies is defined as the mutual interaction of the bodies for a short interval of time due to which energy and momentum of the colliding bodies change or gets transferred from one body to another. 
    Question 134
    CBSEENPH11017260
    Wired Faculty App

    Can two bodies collide, without making the physical contact with each other?

    Solution
    Two bodies can collide without making the physical contact with each other. In collision, there is a transfer of energy and momentum but, physical contact is not necessary. 
    Question 135
    CBSEENPH11017261
    Wired Faculty App

    Give an example of collision that takes place without any physical contact.

    Solution
    In Rutherford scattering experiment, the scattering of a particle from the nucleus is the example of collision that takes place without any physical contact.
    In that case, the alpha particles are scattered due to the electrostatic interaction between the alpha particles and the nucleus from a distance. 
    Question 136
    CBSEENPH11017262
    Wired Faculty App

    What are different types of collisions?

    Solution

    Collisions are categorized in tqo types:
    (i) Elastic collision, and 
    (ii) Inelastic collision.

    Question 137
    CBSEENPH11017263
    Wired Faculty App

    What quantities are conserved in an elastic collision?

    Solution
    In an elastic collision, total energy, kinetic energy and momentum remains conserved. 
    Question 138
    CBSEENPH11017264
    Wired Faculty App

    What quantities are conserved in an inelastic collision?

    Solution
    Momentum and total energy remains conserved in inelastic collisions.
    Question 139
    CBSEENPH11017265
    Wired Faculty App

    Two billiard balls collide elastically. Does the total kinetic energy remain conserved during the short time of collision of balls?

    Solution
    No, total kinetic energy is not conserved during the short time interval of collision of balls. This is  because balls during the collision, are deformed and kinetic energy is converted into elastic energy.
    Question 140
    CBSEENPH11017266
    Wired Faculty App

    A body of mass m collides with a wall with speed u and rebounds with speed 0.8u. What type of collision it is?

    Solution
    This is an inelastic collision because the velocity of the ball has changed. 
    Question 141
    CBSEENPH11017267
    Wired Faculty App

    Two bodies approaching towards each other collide and stick together. What type of collision it is?

    Solution
    Perfectly inelastic collision.
    Question 142
    CBSEENPH11017268
    Wired Faculty App

    In what type of collision relative velocity of approach is equal to relative velocity of separation?

    Solution
    For elastic collision, the relative velocity of approach is equal to relative velocity of separation.
    Question 143
    CBSEENPH11017269
    Wired Faculty App

    In what type of collision, kinetic energy is not conserved?

    Solution
    Kinetic energy is not conserved in inelastic collisions. 
    Question 147
    CBSEENPH11017273
    Wired Faculty App

    What types of forces are involved in elastic collision?

    Solution
    Conservative forces are involved in elastic collision. 
    In conservative forces, the work done in moving a particle between two points is independent of the path taken. 
    Question 148
    CBSEENPH11017274
    Wired Faculty App

    What types of forces are involved in an inelastic collision?

    Solution
    Non- conservative forces are involved in inelastic collision. 
    Question 149
    CBSEENPH11017275
    Wired Faculty App

    What is the alone quantity that remains conserved even during collision?

    Solution
    The principle of conservation of momentum is followed in collisions. The total momentum of the colliding bodies remains conserved during a collision.
    Question 150
    CBSEENPH11017276
    Wired Faculty App

    When a ball of mass m moving with velocity u collides elastically with a wall normally, then what is the change in momentum of the ball?

    Solution
    Initial velocity of the ball =u 
    Mass of the ball = m 
    So, initial momentum of the ball = m u 
    After collision, the ball changes it's direction. 
    Therefore, momentum = - mu 
    Therefore, 
    Change in momentum of the ball = mu - (- mu) = 2mu.
    Question 151
    CBSEENPH11017277
    Wired Faculty App

    Can a ball striking the wall with velocity v rebounds with velocity 2v?

    Solution
    No, the velocity of the ball will remain the same and there will be only a change in the direction of the ball. 
    So, the velocity of the ball when it rebounds = - v 
    Question 152
    CBSEENPH11017278
    Wired Faculty App

    Two bodies moving towards each other collide and separate. If there is some rise in temperature of the bodies in the process, then what type of collision will it be?

    Solution
    Given that, there is a rise in temperature. That is, it is an example of an inelastic collision.
    Question 153
    CBSEENPH11017279
    Wired Faculty App

    What is collision? How many types of collisions are there?

    Solution

    A collision between two or more bodies is defined as the mutual interaction of the bodies for a short interval of time due to which energy and momentum of the colliding bodies transfer and change.
    For the transfer of momentum and energy, physical interaction is not necessary. 

    Two types of collisions are: 
    (i) Elastic collision,
    (ii) Inelastic collision.

    Question 154
    CBSEENPH11017280
    Wired Faculty App

    What is difference between elastic and inelastic collision?

    Solution
    In an elastic collision, the total kinetic energy of the system is conserved while in an inelastic collision the total kinetic energy is not conserved.
    Question 155
    CBSEENPH11017281
    Wired Faculty App

    A bomb explodes in the mid air. What happens to its momentum and kinetic energy?

    Solution
    The momentum of the fragments remain conserved but the energy released due to explosion is converted into kinetic energy. Hence, the kinetic energy will increase.
    Question 156
    CBSEENPH11017282
    Wired Faculty App

    When a cracker explodes, how does this effect:
    (a) its total kinetic energy?
    (b) chemical energy ?
    Solution

    (a) When a cracker explodes, its fragments acquire kinetic energy. Initially, the cracker is at rest therefore, initial kinetic energy is zero. Hence, there is an increase in kinetic energy.
    (b) Chemical energy decreases because total energy of system remains constant. Kinetic energy of the fragments is at the cost of chemical energy.

    Question 157
    CBSEENPH11017283
    Wired Faculty App

    Is it possible that a collision may take place without physical contact between two bodies?

    Solution
    Yes, the collision may take place even without physical contact.
    For example, if two protons are shot towards each other, they exert a force on each other without making any physical contact (because electrostatic force is acting at a distance). Hence, collision is said to take place.
    Question 158
    CBSEENPH11017284
    Wired Faculty App

    What is the coefficient of restitution and what is its value for different types of collisions?

    Solution

    The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach. 
    If u1 and u2 be the velocities before collision, and v1 and v2 be the velocities after collision, then 
            straight e equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction 
    When,
     e = 1, for perfectly elastic collision, 
    0 < e < 1, for inelastic collision 
    e = 0, for perfectly inelastic collision

    Question 159
    CBSEENPH11017285
    Wired Faculty App

    What is perfectly inelastic collision? What is the coefficient of restitution for perfectly inelastic collision? Give one example.

    Solution

    The collision is said to be a perfectly inelastic collision if the colliding bodies stick together after the collision.
    Coefficient of restitution for inelastic collision is zero.
    Example: A bullet is fired into the wooden block and remains embedded inside the block.

    Question 160
    CBSEENPH11017286
    Wired Faculty App

    Show that in one dimension elastic collision, velocity of separation is equal to the velocity of approach.

    Solution
    Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along the same straight line.
    Let the two bodies collide and after collision v1 and v2 be the velocities of two masses respectively.
    Before collision:
    Momentum of mass m1 = m1u1 
    Momentum of mass m2 = m2u2
    Total momentum before collision, 
    p1 = m1 u1 + m2u2 
    K.E of mass m1 = 1 half m subscript 1 u subscript 1 squared
    K.E of mass m2 = 1 half m subscript 2 u subscript 2 squared
    Therefore, 
    Total K.E before collision is, 
    K.E = 1 half m subscript 1 u subscript 1 squared + 1 half m subscript 2 u subscript 2 squared
    After collision:
    Momentum of mass m1 = m1v1
    Momentum of mass m2 = m2v2 
    Total momentum after collision is, 
    pf = m1v1 +  m2v2 
    K.E of mass m1 = 1 half m subscript 1 v subscript 1 squared
    K.E of mass m2 = 1 half m subscript 2 v subscript 2 squared
    Total K.E after collision, Kf = 1 half m subscript 1 v subscript 1 squared+ 1 half m subscript 2 v subscript 2 squared 
    Now, according to the law of conservation of momentum, 
    space space space space space straight m subscript 1 straight u subscript 1 space plus space straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus space straight m subscript 2 straight v subscript 2 space rightwards double arrow space straight m subscript 1 space left parenthesis space straight u subscript 1 space minus space straight v subscript 1 right parenthesis space equals space straight m subscript 2 space left parenthesis space straight v subscript 2 space minus space straight u subscript 2 right parenthesis space space space space space space space space space space space space... space left parenthesis 1 right parenthesis thin space According space to space the space law space of space conservation space of space straight K. straight E comma space we space have 1 half straight m subscript 1 straight u subscript 1 squared space plus 1 half straight m subscript 2 straight u subscript 2 squared space equals space 1 half straight m subscript 1 straight v subscript 1 squared space plus space 1 half straight m subscript 2 straight v subscript 2 squared rightwards double arrow space space straight m subscript 1 space left parenthesis straight u subscript 1 squared space minus space straight v subscript 1 squared space right parenthesis space equals space straight m subscript 2 space left parenthesis space straight v subscript 2 squared space minus space straight v subscript 1 squared right parenthesis space space space space space space space space space space space space... space left parenthesis 2 right parenthesis thin space Dividing space left parenthesis 2 right parenthesis space by space left parenthesis 1 right parenthesis comma space we space get space left parenthesis straight u subscript 1 space plus space straight v subscript 1 right parenthesis space equals space left parenthesis space straight v subscript 2 space plus space straight u subscript 2 right parenthesis space rightwards double arrow space straight u subscript 1 space minus space straight u subscript 2 space equals space straight v subscript 2 space minus space straight v subscript 1 space  
    That is, relative velocity of approach is equal to relative velocity of separation.
    Hence the result.
    Question 161
    CBSEENPH11017287
    Wired Faculty App

    Discuss one-dimensional elastic collision. Obtain an expression for their velocities after collision.
    Solution

    In one-dimensional elastic collision, the colliding bodies continue to move along the same straight line after collision. 
    Consider two bodies A and B of masses m1 and m2 moving along the same straight line with velocities u1 and u2 respectively. 
     

    Let the two bodies collide elastically and let and v2 be the velocities of two bodies after collision.
    Since collision is elastic collision, therefore, their momentum and kinetic energy remain conserved. 
    By conservation of momentum, 
                    straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals straight m subscript 1 straight v subscript 1 plus straight m subscript 2 straight v subscript 2 
    rightwards double arrow        straight m subscript 1 left parenthesis straight u subscript 1 minus straight v subscript 1 right parenthesis space equals space straight m subscript 2 left parenthesis straight v subscript 2 minus straight u subscript 2 right parenthesis               ...(1)
    Using the principle of conservation of kinetic energy, we have
                       1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 equals 1 half straight m subscript 1 straight v subscript 1 superscript 2 plus 1 half straight m subscript 2 straight v subscript 2 superscript 2 
       space space space space space space space space straight m subscript 1 left parenthesis straight u subscript 1 superscript 2 minus straight v subscript 1 superscript 2 right parenthesis space equals space straight m subscript 2 left parenthesis straight v subscript 2 superscript 2 minus straight u subscript 2 superscript 2 right parenthesis               ...(2) 
    Dividing (2) by (1), we get, 
    space space space space space space space space space space space space space space space space space space space left parenthesis straight u subscript 1 plus straight v subscript 1 right parenthesis equals left parenthesis straight v subscript 2 plus straight u subscript 2 right parenthesis 
    rightwards double arrow         space space space straight v subscript 2 equals straight u subscript 1 minus straight u subscript 2 plus straight v subscript 1                      ...(3) 
    Substituting (3) by (1), we get 
          straight m subscript 1 left parenthesis straight u subscript 1 minus straight v subscript 1 right parenthesis space equals straight m subscript 2 left parenthesis straight u subscript 1 minus straight u subscript 2 plus straight v subscript 1 minus straight u subscript 2 right parenthesis 
      space space space straight m subscript 1 straight u subscript 1 minus straight m subscript 1 straight v subscript 1 equals straight m subscript 2 straight u subscript 1 minus 2 straight m subscript 2 straight u subscript 2 plus straight m subscript 2 straight v subscript 1 
    This implies,
             left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v subscript 1 equals left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight u subscript 1 minus 2 straight m subscript 2 straight u subscript 2 
    rightwards double arrow                   straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus straight m subscript 2 right parenthesis straight u subscript 1 plus 2 straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    Substituting for straight v subscript 1 in equation (3), we get

    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 

       equals fraction numerator left parenthesis straight m subscript 2 minus straight m subscript 1 right parenthesis straight u subscript 2 plus 2 straight m subscript 1 straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction, is the velocity after collision. 


    Question 162
    CBSEENPH11017288
    Wired Faculty App

    Show that when the two bodies of equal masses collide elastically, then after collision, the bodies interchange their velocities.

    Solution
    When the two bodies collide elastically, the velocities of two bodies after collision are given by 
                 straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus straight m subscript 2 right parenthesis straight u subscript 1 plus 2 straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    and        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Here, we have
                         straight m subscript 1 equals straight m subscript 2 equals straight m space left parenthesis let right parenthesis
    Therefore, 
            <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    and      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 

    Hence the result.
    Question 163
    CBSEENPH11017289
    Wired Faculty App

    Show that when the colliding body is very-very light as compared to target and target is at rest, then colliding body rebounds with same speed.

    Solution
    When the two bodies collide elastically, the velocities of two bodies after collision are given by, 

                  straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus straight m subscript 2 right parenthesis straight u subscript 1 plus 2 straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction
    and         <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Here, we have
                     straight m subscript 1 less than less than less than straight m subscript 2 
    Therefore, 
                       straight m subscript 1 minus straight m subscript 2 almost equal to negative straight m subscript 2 
    and              straight m subscript 1 plus straight m subscript 2 almost equal to straight m subscript 2 
    Thus,  straight v subscript 1 almost equal to fraction numerator negative straight m subscript 2 straight u subscript 1 plus 2 straight m cross times 0 over denominator straight m subscript 2 end fraction almost equal to negative straight u subscript 1 
    Hence, we can see that the colliding body rebounds with the same speed.   

    Question 164
    CBSEENPH11017290
    Wired Faculty App

    A body of mass m1 moving with velocity u collides with a second body of mass m2 at rest. Find the fractional energy retained by mass m1 after collision. Suppose collisioned be elastic.

    Solution
    First body: 
    Mass = m1 
    Velocity = u 
    Second body:
    Mass = m2
    Velocity, u2 = 0
    After collision, the velocity of masses m1 and m2 are given by
                straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus straight m subscript 2 right parenthesis straight u subscript 1 plus 2 straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    and       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Here, we have
                  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>   and   straight u subscript 2 equals 0 
    ∴            straight v subscript 1 equals open parentheses fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses straight u 
    and        straight v subscript 2 equals fraction numerator 2 straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight u 
    Kinetic energy of mass straight m subscript 1 before collision, 
                      straight K. straight E space equals 1 half straight m subscript 1 straight u squared 
    Kinetic energy of mass straight m subscript 1 after collision, 
        straight K. straight E space subscript 1 equals 1 half straight m subscript 1 straight v subscript 1 superscript 2 space space space space space space space space equals 1 half straight m subscript 1 open parentheses fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses squared straight u squared 
    Therefore fractional energy retained by mass straight m subscript 1 is given by, 
    straight K subscript 1 over straight K equals fraction numerator begin display style 1 half space end style straight m subscript 1 space open parentheses begin display style fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction end style close parentheses squared straight u squared over denominator begin display style 1 half space end style straight m subscript 1 space end subscript straight u squared end fraction space space space space space space space space space space equals open parentheses fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses squared
       
                       
       
    Question 165
    CBSEENPH11017291
    Wired Faculty App

    A body of mass m1 moving with velocity u collides elastically with mass m2 at rest. Find the fractional energy transferred to mass m2 after collision. |

    Solution
    Velocity of mass m2 after collision is given by,

                <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Here, we have
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> and straight u subscript 2 equals 0 
    Therefore,
                  straight v subscript 2 equals fraction numerator 2 straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight u 
    ∴ Kinetic energy of mass straight m subscript 2 after collision is,
                              space space space space straight K subscript 2 equals 1 half straight m subscript 2 straight v subscript 2 superscript 2 equals 1 half straight m subscript 2 open parentheses fraction numerator 2 straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses squared straight u squared 
          <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Since initially mass m2 is at rest, therefore kinetic energy transferred to mass m2 is equal to the kinetic energy of mass m2 after the collision. 
    Therefore,
    Fractional energy transferred is, 
    straight K subscript 2 over straight K equals fraction numerator begin display style 1 half end style begin display style fraction numerator 4 straight m subscript 1 squared straight m subscript 2 over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared end fraction end style straight u squared over denominator begin display style 1 half end style straight m subscript 1 straight u squared end fraction equals fraction numerator 4 straight m subscript 1 straight m subscript 2 over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared end fraction 
                      
    Question 166
    CBSEENPH11017292
    Wired Faculty App

    A body of mass m1 moving with velocity u collides with a body of mass m2 at rest. Show that energy transfer in head-on elastic collision is maximum if m1 = m2.

    Solution
    The fractional energy transferred to target in head-on elastic collision when target is at rest is, 
    space space space space space space space space space space space space space space space space straight K subscript 2 over straight K equals fraction numerator 4 straight m subscript 1 straight m subscript 2 over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis squared end fraction 
    Let straight m subscript 2 equals nm subscript 1 
    Therefore, 
        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    rightwards double arrow  straight K subscript 2 equals fraction numerator 4 straight n over denominator left parenthesis 1 plus straight n right parenthesis squared end fraction straight K 
    The energy transfer is maximum if  dK subscript 2 over dn equals 0 
    That is, 
                                straight d over dn open parentheses fraction numerator 4 straight n over denominator left parenthesis 1 plus straight n right parenthesis squared end fraction close parentheses equals 0 
    rightwards double arrow  fraction numerator left parenthesis 1 plus straight n right parenthesis squared begin display style fraction numerator straight d left parenthesis 4 straight n right parenthesis over denominator dn end fraction end style minus 4 straight n begin display style fraction numerator straight d left parenthesis 1 plus straight n right parenthesis squared over denominator dn end fraction end style over denominator left parenthesis 1 plus straight n right parenthesis to the power of 4 end fraction equals 0 
    rightwards double arrow            left parenthesis 1 plus straight n right parenthesis squared left parenthesis 4 right parenthesis minus 4 straight n left square bracket 2 left parenthesis 1 plus straight n right parenthesis right square bracket equals 0 
    We have,  n = 1  
    ∴                     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Hence the result. 
    Question 167
    CBSEENPH11017293
    Wired Faculty App

    A neutron travelling with a velocity v and energy E collides elastically with a nucleus of mass number A. Find the energy retained by the neutron after collision.

    Solution
    The fractional energy retained by colliding mass is, 
    straight K subscript 1 over straight K equals open parentheses fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses squared 
    Here, we have
    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>,  and    
    K.E = E 
    Therefore, 
                straight K subscript 1 over straight E equals open parentheses fraction numerator straight m subscript 1 minus Am subscript 1 over denominator straight m subscript 1 plus Am subscript 1 end fraction close parentheses squared equals open parentheses fraction numerator 1 minus straight A over denominator 1 plus straight A end fraction close parentheses squared 
                  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> , is the energy retained by the neutron after collision. 
    Question 168
    CBSEENPH11017294
    Wired Faculty App

    Discuss two-dimensional elastic collision.

    Solution
    Consider two bodies A and B of masses m1 and m2 moving along the same straight line along X-axis with velocities and u2 respectively as shown in the figure.


    Let, after collision two bodies move with velocities v1 and v2 making angles θ1 and θ2 with initial direction of motion i.e. with X-axis.
    Since collision is an elastic collision, therefore, their momentum and kinetic energy remain conserved.
    Momentum is a vector quantity, therefore the momentum along X-axes as well as along Y-axes will conserve individually.
    By conservation of momentum along X-axis, 
    space space space space space space space space space space space space space straight m subscript 1 straight u subscript 1 straight x end subscript plus straight m subscript 2 straight u subscript 2 straight x end subscript equals straight m subscript 1 straight v subscript 1 straight x end subscript plus straight m subscript 2 straight v subscript 2 straight x end subscript 
    rightwards double arrow   straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals straight m subscript 1 straight v subscript 1 cosθ subscript 1 plus straight m subscript 2 straight v subscript 2 cosθ subscript 2  ...(1)
    By conservation of momentum along Y-axis,
           straight m subscript 1 straight u subscript 1 straight y end subscript plus straight m subscript 2 straight u subscript 2 straight y end subscript equals straight m subscript 1 straight v subscript 1 straight y end subscript plus straight m subscript 2 straight v subscript 2 straight y end subscript
    rightwards double arrow  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
    rightwards double arrow           straight m subscript 1 straight v subscript 1 sinθ subscript 1 equals straight m subscript 2 straight v subscript 2 sinθ subscript 2                 ...(2)
    By conservation of kinetic energy,
      1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 equals 1 half straight m subscript 1 straight v subscript 1 superscript 2 plus 1 half straight m subscript 2 straight v subscript 2 superscript 2       ...(3)
    Here, in two-dimension collision, there are four variables v1 v2, θ1 and θ2 to solve, but we have only three equations from this analytical study.
    Hence, we cannot find the values of all the variables by analytical method.
    To solve the problem, one variable is measured experimentally and the remaining three are calculated from the above three equations.

    Question 169
    CBSEENPH11017295
    Wired Faculty App

    Discuss one dimensional inelastic collision. Obtain the expression for their velocities after collision. Discuss the special cases.

    Solution

    Consider two bodies A and B of masses m1 and m2 moving along the same straight line with velocities u1 and u2 respectively.
    Let the two bodies collide inelastically and let v1 and v2 be the velocities of two bodies after collision and e be the coefficient of restitution. 

    In an inelastic collision, only law of conservation holds while law of conservation of energy does not do so. 
    By conservation of momentum,
            straight m subscript 1 straight v subscript 1 plus straight m subscript 2 straight v subscript 2 equals straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2             ...(1) 
    Also by definition of coefficient of restitution, 
                     straight e equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction 
    rightwards double arrow            straight v subscript 2 minus straight v subscript 1 equals eu subscript 1 minus eu subscript 2                   ...(2)
    Solving (1) and (2), we get 
             straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus em subscript 2 right parenthesis straight u subscript 1 plus left parenthesis 1 plus straight e right parenthesis straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction      ...(3) 
    and    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>      ...(4) 
    Special cases: 


    Case(i):  
    Two bodies are of equal masses. 
    Putting,space space straight m subscript 1 equals straight m subscript 2 equals straight m in equation (3) and (4), we get
    rightwards double arrow     space space space straight v subscript 1 equals fraction numerator left parenthesis 1 minus straight e right parenthesis straight u subscript 1 plus left parenthesis 1 plus straight e right parenthesis straight u subscript 2 over denominator 2 end fraction                ...(5) 
    and     space space straight v subscript 2 equals fraction numerator left parenthesis 1 plus straight e right parenthesis straight u subscript 1 plus left parenthesis 1 minus straight e right parenthesis straight u subscript 2 over denominator 2 end fraction                ...(6) 
    Case (ii):
    When straight m subscript 2 is at rest.

    Putting straight u subscript 2 equals 0 in equation (3) and (4), we get 
    rightwards double arrow               straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus em subscript 2 right parenthesis straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction                  ...(7)
    and              straight v subscript 2 equals fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction                    ...(8) 
    If,<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> then 
    rightwards double arrow space space straight v subscript 1 equals fraction numerator left parenthesis 1 minus straight e right parenthesis straight u subscript 1 over denominator 2 end fraction  and straight v subscript 2 equals fraction numerator left parenthesis 1 plus straight e right parenthesis straight u subscript 1 over denominator 2 end fraction 
    If,space space space space straight m subscript 1 greater than greater than straight m subscript 2 comma then 
    rightwards double arrow   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>    and    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    If,  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> then 
    rightwards double arrow   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>  and     straight v subscript 2 equals fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 straight u subscript 1 over denominator straight m subscript 2 end fraction
    This is the required expression for velocities after the collision. 

    Question 170
    CBSEENPH11017296
    Wired Faculty App

    Discuss one dimensional perfectly inelastic collision of two bodies. Obtain the expression for their velocities after collision. Discuss the special cases.

    Solution
    Consider two bodies A and B of masses m1 and m2 moving along the same straight line with velocities u1 and u2 respectively.
    Let the two bodies undergo perfect inelastic collision.
    In a perfect inelastic collision, the two bodies stick together, therefore the relative velocity of separation of two bodies after the collision is zero.
    Let v be the velocity of combined masses after collision.
    By conservation of momentum, 
        <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>              ...(1) 
    rightwards double arrow              <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>            ...(2) 
    Special cases:
    Case (i):
    Two bodies are of equal masses. 
    Putting space space straight m subscript 1 equals straight m subscript 2 equals straight m left parenthesis let right parenthesis in equation (2), we get 
    rightwards double arrow               straight v equals fraction numerator straight u subscript 1 plus straight u subscript 2 over denominator 2 end fraction 
    Case(ii): When straight m subscript 2 is at rest. 
    Putting straight u subscript 2 equals 0 in equation (2), we get 
    rightwards double arrow              straight v equals fraction numerator straight m subscript 1 straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction , is the velocity after collision.            
    Question 171
    CBSEENPH11017297
    Wired Faculty App

    A body of mass m1 moving with velocity u1 collides with another body of mass m2moving with velocity u2 along the same straight line. Two bodies after collision stick together. Find the loss of energy in the collision.

    Solution

    Velocity of combined masses after collision is given by, 
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Total kinetic energy before collision is, 
                     straight K equals 1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 
    Kinetic energy after collision is, 
                 space space space straight K apostrophe equals 1 half left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v squared 
                      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
                       = 1 half fraction numerator left parenthesis straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 right parenthesis squared over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    Loss of energy is given by, 
    straight K minus straight K apostrophe space space equals space 1 half straight m subscript 1 space straight u subscript 1 superscript 2 space end superscript plus 1 half straight m subscript 2 space straight u subscript 2 superscript 2 minus 1 half fraction numerator left parenthesis straight m subscript 1 space straight u subscript 1 plus straight m subscript 2 space straight u subscript 2 right parenthesis squared over denominator straight m subscript 1 plus straight m subscript 2 end fraction
               
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

    Question 172
    CBSEENPH11017298
    Wired Faculty App

    A sphere of mass m moving with velocity u collides with stationary sphere of same mass. If e is the coefficient of restitution, then find the velocities of two masses after collision.

    Solution

    Let v1 and v2 be the velocities of colliding body and target body after collision respectively.
    Applying conservation of momentum, 
                   straight m space straight u plus straight m cross times 0 equals straight m space straight v subscript 1 plus straight m space straight v subscript 2  
    rightwards double arrow                  straight v subscript 1 plus straight v subscript 2 equals straight u                    ...(1) 
    Using the definition of coefficient of restitution, 
                          straight e equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u end fraction 
    rightwards double arrow          straight v subscript 2 minus straight v subscript 1 equals straight e space straight u                           ...(2) 
    Solving (1) and (2), we get
               space space straight v subscript 2 equals fraction numerator straight e plus 1 over denominator 2 end fraction straight u  and  straight v subscript 1 equals fraction numerator straight e minus 1 over denominator 2 end fraction straight u

    Question 178
    CBSEENPH11017304
    Wired Faculty App

    What are the conditions so that transfer of kinetic energy is maximum during collision?

    Solution

    Conditions for maximum transfer of Kinetic Energy are: 
    (i) Collision should be elastic collision.
    (ii) Collision should be head on.
    (iii) Mass of target and striking bodies should be equal.
    (iv) Target should be at rest.

    Question 179
    CBSEENPH11017305
    Wired Faculty App

    Show that when a ball collides elastically against the wall, it rebounds back with the same speed.

    Solution

    We know that velocity of colliding body after collision is, 
                 straight v subscript 1 equals fraction numerator left parenthesis straight m subscript 1 minus straight m subscript 2 right parenthesis straight u subscript 1 plus 2 straight m subscript 2 straight u subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    Here,
             <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Therefore,
                    straight m subscript 1 minus straight m subscript 2 almost equal to negative straight m subscript 2 
    and          straight m subscript 1 plus straight m subscript 2 almost equal to straight m subscript 2 
    ∴   <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 


    Hence the result.

    Question 180
    CBSEENPH11017306
    Wired Faculty App

    A small cake of wet cement thrown at a corner in a wall remains glued to the wall. What happens to its initial energy?

    Solution
    Part of the initial energy, i.e., kinetic energy is used to deform the cake. The remaining energy gets converted into heat and sound etc.
    Question 181
    CBSEENPH11017307
    Wired Faculty App

    What can be the maximum speed with which a body colliding with stationary target rebound? When does this happens? 
    Solution
    The maximum speed of rebound is the speed with which the colliding body strikes the target. This happens when the target is massive as compared to the colliding body and the collision is elastic. 
    Question 182
    CBSEENPH11017308
    Wired Faculty App

    Explain how can the fast moving neutrons be quickly slowed down by passing through water or heavy water.

    Solution
    Water or heavy-water contains protons. When fast moving neutrons pass through water they collide with protons during which a part of kinetic energy of neutrons is transferred to protons and hence neutrons get slowed down. 
    Question 183
    CBSEENPH11017309
    Wired Faculty App

    A molecule in a gas container hits a horizontal wall with speed 200 m/s at angle 30° with normal and rebounds with same speed. Is momentum of molecule conserved in the collision? Is the collision elastic or inelastic? 

    Solution

                            

    Momentum is a vector quantity, therefore, though its magnitude remains constant, its direction changes, hence the momentum will change. After collision, the direction of motion and hence momentum changes, therefore, momentum of molecule will not remain conserved.
    Since the speed of molecule remains constant, its kinetic energy remains conserved. Therefore collision is an example of elastic collision.

    Question 184
    CBSEENPH11017310
    Wired Faculty App

    The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

    Solution
    The collision of bobs is elastic and masses of both the bobs are equal. Bob B is at rest, therefore, after collision of bob A with bob B, bob A comes to rest. Therefore, the bob A will not rise after the collision. 
    Question 185
    CBSEENPH11017311
    Wired Faculty App

    Answer carefully, with reasons:

    In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?

    Solution
    No, kinetic energy during the short time of collision of balls is converted into potential energy.
    Question 186
    CBSEENPH11017312
    Wired Faculty App

    Answer carefully, with reasons:

    Is the total linear momentum conserved during the short time of an elastic collision of two balls?

    Solution
    Yes, because during the collision there is no external force on the colliding bodies. Hence, according to the law of conservation of linear momentum, the total linear momentum of the system will remain conserved.
    Question 187
    CBSEENPH11017313
    Wired Faculty App

    Answer carefully, with reasons:

    What are the answers to (a) and (b) for an inelastic collision?

    Solution
    In an inelastic collision, the linear momentum of the system remains conserved but not the kinetic energy.
    Question 188
    CBSEENPH11017314
    Wired Faculty App

    Answer carefully, with reasons:

    If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

    Solution
    Elastic.
    In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
    Question 189
    CBSEENPH11017315
    Wired Faculty App

    Which of the following potential energy curves cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centers of the balls, R is the radius of each ball. 

    Solution
    If collisions are elastic collisions then potential energy is a function of position only.
    Potential Energy is zero after the two colliding bodies make contact with each other and are now separate. 
    The potential energy of a system of two masses is inversely proportional to the separation between them.
    In the given case, with decrease in separation, potential energy increases after two bodies make the contact with each other.
    i.e.          straight V equals 0,             for straight r greater or equal than 2 straight R, 
     straight V equals increases space as space straight r space decrease comma space <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Therefore, 
    It will become zero (i.e., V(r) = 0) when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball.
    The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions.
    Hence, they do not describe the elastic collisions between them.
    Possible graph is, 
                      
    Question 193
    CBSEENPH11017319
    Wired Faculty App

    A spherical ball of mass 0.3kg moving with velocity 1.2m/s on a smooth horizontal table collides elastically with another ball of mass 0.15kg at rest. Assuming the collision to be perfectly elastic, find the speed of the two balls after collision.

    Solution

    Given,
    Mass of the first body, straight m subscript 1 equals 0.3 kg comma
    Mass of the second body, <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    Initial velocity, straight u subscript 1 equals 1.2 space straight m divided by straight s comma
    Final velocity, straight u subscript 2 equals 0

    Let straight v subscript 1 and straight v subscript 2 be the velocities of two bodies after collision.
    Therefore, using conservation of momentum and kinetic energy, we have
            0.3 cross times 1.2 plus 0.15 cross times 0 equals 0.3 straight v subscript 1 plus 0.15 straight v subscript 2  
    rightwards double arrow                    <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>                     ...(1) 
    By conservation of kinetic energy,
        space space 1 half cross times 0.3 space left parenthesis 1.2 right parenthesis squared plus 1 half space left parenthesis 0.15 right parenthesis space left parenthesis 0 right parenthesis squared equals 1 half 0.3 space straight v subscript 1 superscript 2 plus 1 half 0.15 space straight v subscript 2 superscript 2
                                       <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    On solving (1) and (2), we get 
     <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>   and straight v subscript 2 equals 1.6 space straight m divided by straight s
    v1 and v2 is the required speed of the two balls after collision. 

    Question 194
    CBSEENPH11017320
    Wired Faculty App

    A body P strikes with another body Q of mass four times and at rest. The striking body starts moving with velocity 2m/s. Find the velocity with which P hits Q.

    Solution
    Let P hit Q with velocity u.
    The velocity of Q after collision is given by, 
              space space straight v equals fraction numerator 2 straight m subscript 1 straight u over denominator straight m subscript 1 plus straight m subscript 2 end fraction 
    Therefore,
    Initial velocity, straight u equals fraction numerator straight v left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis over denominator 2 straight m subscript 1 end fraction equals fraction numerator 2 left parenthesis straight m plus 4 straight m right parenthesis over denominator 2 straight m end fraction equals 5 straight m divided by straight s
    Question 195
    CBSEENPH11017321
    Wired Faculty App

    A body P moving with velocity 10m/s strikes elastically with another body Q moving in the opposite direction with speed 12m/s. The body P after collision rebounds with velocity 9m/s. Find the velocity of Q.

    Solution

    Given, 
    Velocity of body P, straight u subscript straight p equals 10 space straight m divided by straight s,     
    Velocity of body Q, straight u subscript straight Q equals negative 12 straight m divided by straight s  
    Velocity of body P, space straight v subscript straight p equals negative 9 straight m divided by straight s,
    Velocity of body Q, straight v subscript straight Q = ?

    We know that relative velocity of approach is equal to relative velocity of separation, 
    i.e.        space space space straight u subscript straight p minus straight u subscript straight Q equals straight v subscript straight Q minus straight v subscript straight P 
    rightwards double arrow      <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre> 
    That is,
                           straight v subscript straight Q equals plus 13 straight m divided by straight s 

    Question 196
    CBSEENPH11017322
    Wired Faculty App

    A body of mass 2kg moving with velocity 3 m/s collides head on with 3kg moving with a velocity of 4m/s in opposite direction. After collision the two bodies stick together. Find the loss of energy in the process.

    Solution

    Given,
    Mass of first body,  straight m subscript 1 space equals space 2 kg
    Initial velocity of the body, straight u subscript 1 equals 3 straight m divided by straight s  
    Mass of the second body, space straight m subscript 2 equals 3 kg
        
     Initial velocity of the body, straight u subscript 2 equals negative 4 straight m divided by straight s      

    Let, v be the velocity of composite mass after collision.
    Therefore by conservation of momentum, 
                  straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v 
    rightwards double arrow       2 cross times 3 plus 3 left parenthesis negative 4 right parenthesis space equals space left parenthesis 5 right parenthesis straight v
    rightwards double arrow                           straight v equals negative 1.2 space straight m divided by straight s
    Now kinetic energy of system before collision is, 
    1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 equals 1 half cross times 2 left parenthesis 3 right parenthesis squared plus 1 half cross times 3 left parenthesis negative 4 right parenthesis squared
                               equals 9 plus 24 equals 33 straight J 
    Kinetic energy of the sytem after collision is,
     1 half left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v squared equals 1 half cross times 5 cross times left parenthesis 1.2 right parenthesis squared equals 3.6 straight J      

    Loss of energy = 33 - 3.6 = 29.4 J

    Question 197
    CBSEENPH11017323
    Wired Faculty App

    A body is dropped from height h1 and it rebounds to height h2.Find the coefficient of restitution.

    Solution
    Given, a body is dropped from a height h1. 
    Let v1 be the velocity of the ball just before it hits the plane. 
    ∴         straight v subscript 1 superscript 2 minus 0 squared equals 2 gh subscript 1   
    rightwards double arrow             straight v subscript 1 equals square root of 2 gh subscript 1 end root 
    Let the ball rebound with velocity straight v subscript 2. 
    As it attains the height straight h apostrophe comma 
    ∴        0 squared minus straight v subscript 2 superscript 2 equals 2 gh subscript 2   
    rightwards double arrow            straight v subscript 2 equals square root of 2 gh subscript 2 end root 
    Now the coefficient of restitution is given by, 
                  straight e equals straight v subscript 2 over straight v subscript 1 
    ∴            straight e equals square root of straight h subscript 2 over straight h subscript 1 end root, is the coefficient of restitution. 
    Question 198
    CBSEENPH11017324
    Wired Faculty App

    A body is dropped from a height h on an inelastic floor. If the coefficient of restitution is e, then find the height of the body after nth collision.

    Solution
    Let v0 and v1 be the velocities of body just before and after collision when it hits first time.
    By definition, coefficient of restitution is given by
                        space space straight e equals straight v subscript 1 over straight v subscript 0 
    rightwards double arrow                 straight v subscript 1 equals ev subscript 0                     ...(1) 
    The velocity of rebound after first collision is v1, therefore second time it will hit the ground with velocity v1. 
    Let after collision it rebound with velocity v2. 
    i.e.,                    straight e equals straight v subscript 2 over straight v subscript 1 
    rightwards double arrow                    straight v subscript 2 equals ev subscript 1 equals straight e squared straight v subscript 0          ...(2) 
    Similarly, the velocity of body after third fourth.....nth collision is, 
    straight v subscript 3 equals straight e cubed straight v subscript 0 , straight v subscript 4 equals straight e to the power of 4 straight v subscript 0, ....<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>         ...(3) 
    As the body falls from height h, therefore it will hit first time with velocity v0 given by 
    straight v subscript 0 equals square root of 2 gh                                          ...(4) 
    After nth collision, it rebounds with velocity vn and let it attain the height h0.
    Thus, 
                          straight v subscript straight n equals square root of 2 gh subscript straight n end root                   ...(5) 
    Now from (3), 
                       straight v subscript straight n squared over straight v subscript 0 squared equals straight e to the power of 2 straight n end exponent                        ...(6) 
    From (4) and (5), 
                     straight v subscript straight n squared over straight v subscript 0 squared equals straight h subscript straight n over straight h                         ...(7) 
    ∴                 <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>, is the height of the body after the nth collision. 
    Question 199
    CBSEENPH11017325
    Wired Faculty App

    A body is dropped from a height h on an inelastic floor of the coefficient of restitution e. Find the total distance travelled by the body before it comes to rest.

    Solution
    The body is dropped from height h.
    Let h1, h2, h3..........be the heights attained by the body after first, second, .... collision.
    Total distance travelled by the body just before coming to rest is,
    straight S equals straight h subscript 0 plus 2 straight h subscript 1 plus 2 straight h subscript 2 plus 2 straight h subscript 3.......... 
     equals straight h subscript 0 plus 2 straight e squared straight h subscript 0 plus 2 straight e to the power of 4 straight h subscript 0 plus 2 straight e to the power of 6 straight h subscript 0........ 
     equals straight h subscript 0 plus 2 straight h subscript 0 straight e squared left parenthesis 1 plus straight e squared plus straight e to the power of 4........ right parenthesis 
     equals straight h subscript 0 plus 2 straight h subscript 0 fraction numerator straight e squared over denominator 1 minus straight e squared end fraction equals straight h subscript 0 fraction numerator 1 plus straight e squared over denominator 1 minus straight e squared end fraction
    Question 200
    CBSEENPH11017331
    Wired Faculty App

    What is the amount of work done by a man in holding a weight of 50N in his hands for 2.6s?

    Solution

    The amount of work done by the man = 0 Joules. 
    This is because, work is a path dependent function. And, here the path has not changed for the weight held by man in his hands. 

    Question 201
    CBSEENPH11017333
    Wired Faculty App

    A body of mass 1 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table. Calculate the work done by the force in 10 seconds. 

    Solution

    Given, 
    Mass, m = 1.0 kg 
    Initial velocity, u = 0 m/s
    Force = 0.5 N 
    Time, t = 10 sec
    Now,
    Acceleration, a = F/m = 0.5 / 1.0 = 0.5 m/s2 
    Now, using the second equation of motion, we have 
                    s = ut + 1/2 at2 
    i.e.,           s = 0 + 1/2 X (0.5) X (10)2 
                       = 25 m
    Therefore, 
    Work done, W = F X s
                          = 0.5 X 25
                          = 12.5 Joule
     

    Question 202
    CBSEENPH11017334
    Wired Faculty App

    A body falls freely from height ‘h’. Plot the graph of kinetic energy and potential energy of body versus height from the ground.

    Solution

    When a body is falling from a height h, the kinetic energy of the body increases.
    Gain in K.E is at the expense of gravitational P.E. So, the P.E of the body decreases as it falls from a height h.
    But, the total mechanical energy remains conserved. 
    Therefore, the graph of K.E and P.E of the body vs. height from the ground is given by, 

    Question 203
    CBSEENPH11017335
    Wired Faculty App

    How many ergs/s are there in 1HP?

    Solution

    One of the popular unit of power is horse power.
    1 horse power = 746 W
    1 W = 107 erg s-1
    That is, 
    1 h.p = 7.46 X 109 erg s-1. 

    Question 204
    CBSEENPH11017336
    Wired Faculty App

    Convert 1 KWt.hr. into joules.

    Solution

    Watt is defined as Joule per second. 
    Therefore, 
    1 KW. hr = 1000 X 3600
                  = 3600000 Joules 
                  = 3.6 X 106 J 

    Question 205
    CBSEENPH11017337
    Wired Faculty App

    A body of mass m1 moving with velocity straight u with rightwards arrow on top collides with target of mass straight m subscript 2 at rest. Can the mass straight m subscript 2 move with velocity opposite to the direction of straight u with rightwards arrow on top after collision? Explain. (Let collision be elastic collision).

    Solution

    In an elastic collision, 
    If the mass m1 of the moving ball is greater than the mass m2 of the ball at rest, then after collision both the velocities v1 and v2 will be positive. And two balls will move in the same direction. 
    And, if the moving ball is lighter than the ball at rest, the two balls will move in the opposite direction. 

    Question 206
    CBSEENPH11017352
    Wired Faculty App

    A lorry and a car moving with same kinetic energy are brought to rest by application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance?

    Solution

    Given that the two bodies are moving with the same K.E. And also, brakes are applied with equal retarding forces. Therefore, both the bodies will come to rest at the same time. 

    Question 207
    CBSEENPH11017358
    Wired Faculty App

    A spring of force constant K obeys Hook's law. It requires 4J of work to stretch it through 10 cm beyond its unstretched length.

    Calculate: 

    (i) The value of K. 

    (ii) The extra work required to stretch it through additional 10 cm.

    (Take g = 10 m/s2)
    Solution

    Work done to stretch the spring, W = 4 J 
    Distance through which the spring is stretched, x= 10 cm 
    Work done is given by, 
     straight W space equals space 1 half space kx squared space Therefore comma space straight K space equals space fraction numerator 2 straight W over denominator straight x squared end fraction space equals space fraction numerator 2 space straight X space 4 over denominator left parenthesis 10 to the power of negative 1 end exponent right parenthesis squared end fraction space equals space 800 space straight N divided by straight m space
    Total work done in stretching the spring through 
    x' = 20 cm = 1 half space k space x apostrophe squared
        = 1 half cross times space 800 space cross times space open parentheses 20 over 100 close parentheses squared space equals space 16 space J
    Therefore, 
    Extra work required = 16 J - 4 J = 12 J 

    Question 208
    CBSEENPH11017359
    Wired Faculty App

    A truck draws a tractor of mass 1000 kg at a steady rate of 20 m/s on a level road. The tension in the coupling is 2000 N. Calculate the power spent on the tractor? 


    A truck of weight 2x 10    5N can move up a road

    having a slope of 1 in 50 with a speed of 8 m/s. The resistive force of road is one twenty-fifth of the weight of truck. What is power of truck?

    Solution

    Given, 
    Mass of the tractor, m = 1000 kg 
    Speed of the tractor, v = 20 m/s
    Tension in coupling = Force applied
    Force, F = 2000 N
    Using the formula for power, we have
    Power spent on the tractor, P = 2000 x 200 = 40000 Watt = 40 kW

    Question 209
    CBSEENPH11017360
    Wired Faculty App

    Find the power of a person who can chew 30 g of ice in one minute. Latent heat of ice = 80 cal/gm

    Solution

    Mass of ice chewed, m = 30 g 
    Time taken, t = 1 min  = 60 sec
    Latent heat of ice, L = 80 cal/g
    Heat produced = mL = 30 x 80
                                    = 2400 cal 
                                    = 2400 x 4.2 J
    Now, using the formula for power
    Power = Work / Time taken 
    power, P = fraction numerator 2400 space straight x space 4.2 over denominator 60 space end fraction space Watt space equals space 168 space Watt

    Question 210
    CBSEENPH11017361
    Wired Faculty App

    The nucleus 57Fe emits a γ-ray of energy 14.4 KeV. If the mass of the nucleus is 56.935 amu, calculate the recoil energy of the nucleus.

    Solution

    Here,
    Energy of γ-ray is, E(γ) = 14.4 KeV 
                                        = 14.4 x 1.6 x 10–16 J
                                        = 2.24 x10–15J 
    Mass of nucleus, m(Fe) = 56.935 amu 
    We know the momentum of proton is given by, 
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    According to conservation of momentum, 
                    straight p left parenthesis straight alpha right parenthesis space equals space straight p left parenthesis Fe right parenthesis 
    Thus the recoil energy of nucleus is, 
    space space space straight E left parenthesis Fe right parenthesis space equals space fraction numerator straight p left parenthesis Fe right parenthesis squared over denominator 2 straight m left parenthesis Fe right parenthesis end fraction equals fraction numerator straight p left parenthesis straight gamma right parenthesis squared over denominator 2 straight m left parenthesis straight gamma right parenthesis space end fraction equals fraction numerator straight E left parenthesis straight gamma right parenthesis squared over denominator 2 straight m left parenthesis straight gamma right parenthesis straight c squared end fraction  

              equals fraction numerator left parenthesis 2.24 cross times 10 to the power of negative 15 end exponent right parenthesis squared over denominator 2 cross times 56.935 cross times 1.661 cross times 10 to the power of negative 27 end exponent cross times left parenthesis 3 cross times 10 to the power of 8 right parenthesis squared end fraction

              equals 3.12 cross times 10 to the power of negative 22 end exponent straight J equals 1.95 cross times 10 to the power of negative 6 end exponent KeV 

    Question 211
    CBSEENPH11017363
    Wired Faculty App

    A body of mass 4 kg moving with velocity 10 m/sec collides inelastically with another body of mass 10 kg moving with a velocity of 4m/sec. If coefficient of restitution is 0 5, then find the velocities of two after collision and loss of energy.

    Solution
    After collision,
    Let v1 be the velocity of mass 4 kg. 
    Let v2 be the velocity of mass 10 kg. 
    According to the law of conservation of momentum, 
                straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals straight m subscript 1 straight v subscript 1 plus straight m subscript 2 straight v subscript 2 
    Here we have,

    Mass, straight m subscript 1 equals 4 kg space 
    Mass,straight m subscript 2 equals 10 space kg
    Initial velocity, straight u subscript 1 equals 10 space straight m divided by sec comma
    Velocity after collision, straight u subscript 2 equals 4 space straight m divided by sec 
    Therefore, 
               space space space 4 cross times 10 plus 10 cross times 4 equals 4 straight v subscript 1 plus 10 straight v subscript 2 
    rightwards double arrow               4 straight v subscript 1 plus 10 straight v subscript 2 equals 80               ...(1) 
    Coefficient of restitution is, straight e equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction 
               straight v subscript 2 minus straight v subscript 1 equals straight e left parenthesis straight u subscript 1 minus straight u subscript 2 right parenthesis 
                        equals 0.5 left parenthesis 10 minus 4 right parenthesis  
    rightwards double arrow       straight v subscript 2 minus straight v subscript 1 equals 3                                 ...(2) 
    Solving (1) and (2), we get 
    Velocities of bodies after collision is, 
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    Total kinetic energy before collision is, 
           space space space straight K subscript straight i equals 1 half straight m subscript 1 straight u subscript 1 superscript 2 plus 1 half straight m subscript 2 straight u subscript 2 superscript 2 
                equals 1 half cross times 4 cross times left parenthesis 10 right parenthesis squared plus 1 half cross times 10 cross times left parenthesis 4 right parenthesis squared 
                equals 200 plus 80 equals 280 space straight J 
    Total kinetic energy after collision is, 
             straight K subscript straight f equals 1 half straight m subscript 1 straight v subscript 1 superscript 2 plus 1 half straight m subscript 2 straight v subscript 2 superscript 2 
                equals 1 half cross times 4 cross times left parenthesis 25 over 7 right parenthesis squared plus 1 half cross times 10 cross times left parenthesis 46 over 7 right parenthesis squared 
                 equals 241.43 space straight J 
    ∴   Loss of energy  = 280 - 241.43 = 38.57 J 
     
    Question 212
    CBSEENPH11017364
    Wired Faculty App

    A nucleus of radium (226Ra88) decays to radon (222Rn86) by the emission of an α-particle of energy 4.8 MeV. Calculate the recoil energy of the daughter nucleus.
    (m(α) = 4.003 amu, m(Rn) = 222.0 amu).
    Solution

    Decay of radium is given by, 
          Ra presuperscript 226 subscript 88 rightwards arrow Rn presuperscript 222 subscript 86 plus He presuperscript 4 subscript 2 plus Energy 
    We have,
      straight m left parenthesis straight alpha right parenthesis space equals space 4.003 space amu comma space space space straight m left parenthesis Rn right parenthesis space equals space 222.0 space amu 
        space space straight E subscript straight alpha space equals space 4.8 space MeV 
    Let stack straight p subscript straight alpha with rightwards arrow on top space and space stack straight p subscript Rn with rightwards arrow on top be the momentum of straight alpha minusparticle and radon.
    Let, straight E subscript straight alpha space and space straight E subscript Rn be the energies of alpha-particle and radon respectively.
    According to the conservation of momentum, we have 
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    rightwards double arrow                   space space straight p subscript straight alpha equals straight p subscript Rn 
    We know that,
    Energy, straight E equals fraction numerator straight p squared over denominator 2 straight m end fraction 
    ∴   straight E subscript Rn over straight E subscript straight alpha equals fraction numerator straight p subscript Rn squared over denominator 2 straight m left parenthesis Rn right parenthesis end fraction cross times fraction numerator 2 straight m left parenthesis straight alpha right parenthesis over denominator straight p subscript straight alpha squared end fraction equals fraction numerator straight m left parenthesis straight alpha right parenthesis over denominator straight m left parenthesis Rn right parenthesis end fraction 
    rightwards double arrow  straight E subscript Rn equals fraction numerator straight m left parenthesis straight alpha right parenthesis over denominator straight m left parenthesis Rn right parenthesis end fraction straight E subscript straight alpha equals fraction numerator 4.003 over denominator 222 end fraction 4.8 space MeV 
               equals 0.87 space MeV, is the recoil energy of the daughter nucleus. 

    Question 213
    CBSEENPH11017365
    Wired Faculty App

    The absolute value of potential energy and therefore total energy has no physical significance. It is the difference of potential energies that matters. One can, therefore, add or subtract a constant to the potential energy provided we do it to potential energy at every position for a given force without any change in the physical situation. By convention, for forces which fall off to zero at large distances, the potential energy at infinity is taken to be zero. With this choice, is the potential energy positive or negative for, 

    (a) electron-position bound state,

    (b) planet-satellite system,

    (c) electron-electron system?
    Solution

    (a) Because e–, e+ pair attracts each other due to electrostatic force, and if potential energy at infinity is taken to be zero, then an electron-positron pair in the bound state will have negative potential energy.
    (b) The planet-satellite system attracts each other due to gravitational force. Therefore, the system has negative energy. 
    (c) Electron-electron repels each other, therefore to bring them close to each other, energy is required which is stored in the form of potential energy. Therefore, the potential energy of system becomes positive.

    Question 214
    CBSEENPH11020093
    Wired Faculty App

    The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: 

    (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
    Solution
    Positive
    In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket. 
    Question 215
    CBSEENPH11020094
    Wired Faculty App

    The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

    b) Work done by gravitational force when a man in lifting a bucket out of a well by means of a rope tied to the bucket.
    Solution
    Negative
    In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative. 
    Question 216
    CBSEENPH11020095
    Wired Faculty App

    The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:


    (c) work done by friction on a body sliding down an inclined plane
    Solution
    Negative quantity
    The direction of the frictional force is opposite to the direction of motion. Therefore, the work done by the frictional force is negative in this case.
    Question 217
    CBSEENPH11020096
    Wired Faculty App

    The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:


    (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity
    Solution
    Positive quantity.
    In this case, the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.
    Question 218
    CBSEENPH11020097
    Wired Faculty App

    The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative,


    (e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
    Solution
    Negative Quantity.
    The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, negative work is done in this case.
    Question 219
    CBSEENPH11020098
    Wired Faculty App

    Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
    Solution
    Total energy of a system is, 
                           Energy = P.E. + K. E. 
    ∴                          K.E. = E – P.E
    The Kinetic energy of a body is a positive quantity.
    It cannot be negative.
    Therefore, the particle will not exist in a region where K.E. becomes negative.
     
    (i) For x > a, P.E. (V0) > E
    Therefore, 
    K.E. becomes negative.
    Hence, the object cannot exist in the region x > a.
     
    (ii) For x < a and x > b, P.E. (V0) > E.
    Therefore,
    Kinetic Energy becomes negative.
    Hence the object cannot be present in the region x < a and x > b.
     
    iii) x > a and x < b < –V1
    In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between x > a and x < b.

    The minimum P.E in this case is –V1.
    Therfore, K.E. = E – (–V1) = E + V1.
    Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than –V1.
    So, the minimum total energy the particle must have is –V1.
    iv)  -b/2 <  x <  a/2 ; a/2 < x < b/2 ; -V1

    In the given case, the potential energy (V0) of the particle becomes greater than the total energy (E) for -b/2 < x < b/2 and -a/2 <  x < a/2.
    Therefore, the particle will not exist in these regions. 
    The minimum potential energy in this case is –V1.
    Therefore, K.E.  = E – (–V1) = E + V1.
    Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than –V1.
    So, the minimum total energy the particle must have is –V1.
    Question 220
    CBSEENPH11020099
    Wired Faculty App

    Underline the correct alternative:

    (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
    Solution
    Decreases
    Positive work is done on a body by a conservative force when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.
    Question 221
    CBSEENPH11020100
    Wired Faculty App

    Underline the correct alternative:

    (b) Work done by a body against friction always results in a loss of its kinetic/potential energy.

    Solution
    Kinetic energy
    The work done against the direction of friction reduces the velocity of a body. Hence, the body loses an amount of Kinetic energy. 
    Question 222
    CBSEENPH11020101
    Wired Faculty App

    Underline the correct alternative:

    (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

    Solution
    External force
    Internal forces cannot produce any change in the total momentum of a body, irrespective of the direction. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system. 
    Question 223
    CBSEENPH11020102
    Wired Faculty App

    Underline the correct alternative:

    (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 

    Solution
    Total linear momentum
    The total linear momentum always remains conserved, whether it is an elastic collision or an inelastic collision. 
    Question 224
    CBSEENPH11020103
    Wired Faculty App

    A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to,    

    • (i) t1/2 

    • (ii) t  

    • (iii) t 3/2

    • (iv) t2
    Solution

    B.

    (ii) t  

    Tips: -

    From,

    v = u + at
    v = 0 + at = at 

    As power, P = F × v

    ∴            P = (ma) × at = ma2t

    As m and a are constants, therefore, P ∝ t
    Question 225
    CBSEENPH11020104
    Wired Faculty App

    A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to: 
    • (i) t1/2
    •  (ii) t 
    • (iii) t3/2
    • (iv) t2
    Solution

    C.

    (iii) t3/2
    As
    Power, P = force × velocity 
             P = [MLT-2] [LT-1] = [ML2T-3] 
    As,     P = [ML2T-3] 
               = constant  
    ∴  L2T-3 = constant 
    rightwards double arrow L2/T3 = constant
    ∴        L2 ∝ T3 
    rightwards double arrow      L ∝ T3/2

    Tips: -

    As 
    Power, P = force × velocity 
             P = [MLT-2] [LT-1] = [ML2T-3] 
    As,     P = [ML2T-3] 
               = constant  
    ∴  L2T-3 = constant 
    rightwards double arrow L2/T3 = constant
    ∴        L2 ∝ T3 
    rightwards double arrow      L ∝ T3/2
    Question 226
    CBSEENPH11020105
    Wired Faculty App

    A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1?
     
    Solution
    Given, 
    Radius of the rain drop, r = 2 mm = 2 × 10–3 m
    Volume of the rain drop, V = 4 over 3πr3

                                       = 4 over 3 × 3.14 × (2 × 10-3)3 m-3
    Density of water, ρ = 103 kg m–3
     
    Mass of the rain drop, m = ρV 

                                     = 4 over 3 × 3.14 × (2 × 10-3)3 × 103 kg
    Gravitational force, F = mg
                                =  4 over 3× 3.14 × (2 × 10-3)3 × 103 × 9.8  N
    The work done by the gravitational force on the drop in the first half of its journey, 
                            WI = F . s 

                                 =4 over 3 × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 250
                                 =  0.082 J 
    This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey.

    i.e., WII = 0.082 J 
    According to the law of conservation of energy,
    If no resistive force is present, then the total energy of the rain drop will remain the same. 
    ∴ Total energy at the top, 
         ET = mgh + 0 
             = (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 500 × 10-5 

              = 0.164 J 
    Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.
    ∴Total energy at the ground is, 
          EG = (1/2) mv2 + 0 
               = (1/2) × (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × (10)2 

               = 1.675 × 10-3 J 
    ∴ Resistive force = EG – ET = –0.162 J
    Question 227
    CBSEENPH11020106
    Wired Faculty App

    A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

    Solution
    Given,
    Volume of the tank, V = 30 m3 
    Time of operation, t = 15 min
                                  = 15 × 60
                                  = 900 s 
    Height of the tank, h = 40 m 
    Efficiency of the pump, η = 30 % 
    Density of water, ρ = 103 kg/m3 

    Mass of water, m = ρV = 30 × 103 kg 
    Output power can be obtained as, 
    P0 = fraction numerator Work space done over denominator Time end fraction = mgh over straight t
        = 30 × 103 × 9.8 × 40 over 900 
        =  13.067 × 103 W 
    For input power Pi,, efficiency η, is given by the relation, 
     
    η =straight p subscript straight o over straight P subscript straight i=  30% 
    Pi = fraction numerator 13.067 space straight x space 100 space straight x space 10 cubed over denominator 30 end fraction 
       = 0.436 × 105 W 
       =  43.6 kW, is the electric power consumed by the pump. 
    Question 228
    CBSEENPH11020107
    Wired Faculty App

    Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision? 

    Solution
    The total momentum before and after a collision in each case is constant.
    For an elastic collision,
    Total kinetic energy of a system remains conserved before and after the collision. 
    For mass of each ball bearing m, we can write, 
    Total kinetic energy of the system before collision, 
    = 1 halfmV2 +1 half(2m) × 02
    =1 halfmV2

    Case (i)
    Total kinetic energy of the system after collision, 
    = 1 half m × 0 +1 half (2m) (straight V over 2)2
    =open parentheses 1 fourth close parentheses mV2
    Hence, the kinetic energy of the system is not conserved in case (i).
    Case (ii)
    Total kinetic energy of the system after collision, 
    =open parentheses 1 half close parenthesesx(2m) × 0 + 1 halfmV2 
    = 1 half mV2
    Hence, the kinetic energy of the system is conserved in case (ii). 

    Case (iii)
    Total kinetic energy of the system after collision, 
    = (1 half)(3m)(straight V over 3)2
    = (open parentheses 1 over 6 close parentheses)mV2 

    Hence, the kinetic energy of the system is not conserved in case (iii). 
    Hence, Case II is the only possibility.
    Question 229
    CBSEENPH11020108
    Wired Faculty App

    A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m -1/2 s-1. What is the work done by the net force during its displacement from x = 0 to x= 2 m?

    Solution
    Given,
    Mass of the body, m = 0.5 kg
    Velocity of the body is governed by the equation,
                                      v = ax3/2
    Acceleration, a = 5 m -1/2 s-1

    Initial velocity, u (at x = 0) = 0 
    Final velocity v (at x = 2 m) = 10√2 m/s 
    Work done, W = Change in kinetic energy 
                       = (1 half) m (v2 - u2) 
                       = (1 half) × 0.5 [ (10√2)2 - 02]
                       = (1 half) × 0.5 × 10 × 10 × 2
                       = 50 J
    Question 230
    CBSEENPH11020109
    Wired Faculty App

    The blades of a windmill sweep out a circle of area A. 
     
    (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
     
    (b) What is the kinetic energy of the air? 
     
    (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. 
    What is the electrical power produced?
    Solution
    Given,
    Area of the circle swept by the windmill = A
    Velocity of the wind = v

    Density of air = ρ 

    (a) Volume of the wind flowing through the windmill per sec = Av 

    Mass of the wind flowing through the windmill per sec = ρAv 

    Mass m, of the wind flowing through the windmill in time t = ρAvt

    (b) Kinetic energy of air = open parentheses 1 half close parentheses mv2 

                                    = open parentheses 1 half close parentheses (ρAvt)v2 
                                    = (1/2)ρAv3t
    (c) Area of the circle swept by the windmill, A = 30 m2 

    Velocity of the wind = v = 36 km/h
    Density of air, ρ = 1.2 kg m–3 

    Electric energy produced = 25% of the wind energy
                                   = (25 over 100) × Kinetic energy of air
                                   = open parentheses 1 over 8 close parentheses ρ A v3t 
    Electrical power = fraction numerator Electrical space Energy over denominator Time end fraction
                         = open parentheses 1 over 8 close parentheses ρ A v3t / t
                         = open parentheses 1 over 8 close parentheses ρ A v3
                         =open parentheses 1 over 8 close parentheses × 1.2 × 30 × (10)3
                         = 4.5 kW
    Question 231
    CBSEENPH11020110
    Wired Faculty App

    A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

    (a) How much work does she do against the gravitational force?

    (b) Fat supplies 3.8 × 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate.
    How much fat will the dieter use up?

    Solution
    Given, 
    (a) Mass of the weight, m = 10 kg
    Height to which the person lifts the weight, h = 0.5 m
    Number of times the weight is lifted, n = 1000
    Therefore,
    Work done against gravitational force, 
    W = n(mgh)
       = 1000 × 10 × 9.8 × 0.5 
       =  49 kJ
    (b) Energy equivalent of 1 kg of fat = 3.8 × 107 J 
    Efficiency rate = 20% 
    Mechanical energy supplied by the person’s body, 
    E = (20/100) × 3.8 × 107 J 
      = (1/5) × 3.8 × 107 J 
    Equivalent mass of fat lost by the dieter, 
    m = [ 1 / (1/5) × 3.8 × 107 ]× 49 × 103 

       = (245 / 3.8) × 10-4 
       = 6.45 × 10-3 kg
    Question 232
    CBSEENPH11020111
    Wired Faculty App

    A family uses 8 kW of power.

    (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

    (b) Compare this area to that of the roof of a typical house.

    Solution
    (a) 
    Power used by the family, P = 8 kW
                                              = 8 × 103 W 
    Solar energy received per square metre = 200 W 
    Efficiency of conversion from solar to electricity energy = 20 %
    Area required to generate the desired electricity = A 

    We have
    8 × 103 = 20% × (A × 200)
                 = (20 /100) × A × 200 
     
    ∴        A = 8 × 103 / 40  =  200 m2
    (b) 
    The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m. (≈ √200).
    Question 233
    CBSEENPH11020112
    Wired Faculty App

    A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

    Solution
    Given,
    Mass of the bullet, m = 0.012 kg 
    Initial speed of the bullet, ub = 70 m/s 
    Mass of the wooden block, M = 0.4 kg 
    Initial speed of the wooden block, uB = 0 
    Final speed of the system of the bullet and the block = ν 
    Using the law of conservation of momentum, 
                    mub + MuB = (m + M) v

    0.012 × 70 + 0.4 × 0 = (0.012 + 0.4) v 
    ∴      v = 0.84 / 0.412 = 2.04 m/s 
    For the system of the bullet and the wooden block, 
    Mass of the system, m' = 0.412 kg 
    Velocity of the system = 2.04 m/s 
    Height up to which the system rises = h 

    Applying the law of conservation of energy to this system, 
    Potential energy at the highest point = Kinetic energy at the lowest point 
                                            m'gh = open parentheses 1 half close parenthesesm'v2

    ∴                                           h = open parentheses 1 half close parenthesesstraight v squared over straight g
                                                   = open parentheses 1 half close parentheses × fraction numerator left parenthesis 2.04 right parenthesis squared over denominator 9.8 end fraction 
                                                   = 0.2123 m  
    The wooden block will rise to a height of 0.2123 m. 
    Heat produced = Kinetic energy of the bullet – Kinetic energy of the system 
                       =open parentheses 1 half close parentheses mu2 - open parentheses 1 half close parentheses m'v2 
                       =open parentheses 1 half close parentheses × 0.012 × (70)2 - open parentheses 1 half close parentheses × 0.412 × (2.04)2
                       = 29.4 - 0.857
                       = 28.54 J
    Question 234
    CBSEENPH11020113
    Wired Faculty App

    Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

    Solution

    The given question can be illustrated using the figure below: 
     

    AB and AC are two smooth planes inclined to the horizontal at ∠θ1 and ∠θ2 respectively.
    The height of both the planes is the same, therefore, both the stones will reach the bottom with same speed.
    As P.E. at O = K.E. at A = K.E. at B 
    Therefore, 
         mgh = 1/2 mv12 = 1/2 mv22 

    ∴                          v1 = v2 

    As it is clear from fig. above, acceleration of the two blocks are
     a1 = g sin θ1 ,
     a2 = g sin θ2 

    As θ2 > θ1 

    ∴       a2 > a1 

    From v = u + at 
                = 0 + at 

    rightwards double arrow      t = v/a 

    As t ∝ 1/a, and a2 > a1 

    ∴    t2 < t1 

    That is, the second stone will take lesser time and reach the bottom earlier than the first stone.

    Question 235
    CBSEENPH11020114
    Wired Faculty App

    A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

    Solution
    Given,
    Mass of the block, m = 1 kg 
    Spring constant, k = 100 N m–1 

    Displacement in the block, x = 10 cm = 0.1 m
    The given situation can be illustrated in the following figure, 

    At equilibrium, 
    Normal reaction, R = mg cos 37° 
    Frictional force, f = μ R
                              = mg Sin 370 

    where, μ is the coefficient of friction
    Net force acting on the block = mg sin 37° – f 

                                                = mgsin 37° – μmgcos 37° 
                                                = mg(sin 37° – μcos 37°) 
    At equilibrium, the work done by the block is equal to the potential energy of the spring.
    i.e., mg(sin 37° – μcos 37°)x = open parentheses 1 half close parentheses kx2
     1 × 9.8 (Sin 370 - μcos 37°) =open parentheses 1 half close parentheses × 100 × (0.1) 
                    0.602 - μ × 0.799 = 0.510 
    ∴                        μ = fraction numerator 0.092 over denominator 0.799 end fraction  =  0.115 


    Question 236
    CBSEENPH11020115
    Wired Faculty App

    A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

    Solution
    Given,
    Mass of the bolt, m = 0.3 kg 
    Speed of the elevator = 7 m/s 
    Height, h = 3 m 
    Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. 
    Heat produced = Loss of potential energy  
                          = mgh 
                          = 0.3 × 9.8 × 3 
                          = 8.82 J 
    The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
    Question 237
    CBSEENPH11020116
    Wired Faculty App

    A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

    Solution
    Given,
    Mass of the trolley, M = 200 kg 
    Speed of the trolley, v = 36 km/h = 10 m/s 
    Mass of the boy, m = 20 kg 
    Initial momentum of the system of the boy and the trolley, 
    = (M + m)v 
    = (200 + 20) × 10 
    = 2200 kg m/s 
    Let v' be the final velocity of the trolley with respect to the ground. 
    Final velocity of the boy with respect to the ground = v' - 4 
    Final momentum = Mv' + m(v' - 4) 
     
                               = 200v' + 20v' - 80 
                               = 220v' – 80 
    As per the law of conservation of momentum,
     
    Initial momentum = Final momentum 
                       2200 = 220v' – 80 
    ∴                       v' =2280 over 220 space equals space 10.36 space m divided by s
    Length of the trolley, l = 10 m 
    Speed of the boy, v'' = 4 m/s 
    Time taken by the boy to run, t = 10 over 4 = 2.5 s
    ∴ Distance moved by the trolley = v'' × t 
                                                    = 10.36 × 2.5
                                                    = 25.9 m
    Question 240
    CBSEENPH11020271
    Wired Faculty App

    When a rubber band is strecthed by a distance x, it exerts a restoring force of magnitude F = ax +bx2, where a and b are constants. The work done in stretching are unstretched rubber-band by L is

    • aL2 +bL2

    • 1 half left parenthesis aL squared plus bL cubed right parenthesis
    • aL squared over 2 space plus bL cubed over 3
    • 1 half space open parentheses aL squared over 2 plus bL cubed over 3 close parentheses

    Solution

    C.

    aL squared over 2 space plus bL cubed over 3 straight U subscript straight f minus straight U subscript straight i space equals space minus straight W space equals space minus space integral subscript straight i superscript straight f straight F. dr
    Given, F = ax +bx2
    We know that work done in stretching the rubber band by L is 
    |dW|= |Fdx|
    vertical line straight W vertical line space equals space integral subscript 0 superscript straight L left parenthesis ax space plus bx squared right parenthesis dx space equals space open square brackets ax squared over 2 close square brackets subscript straight O superscript straight L space plus space open square brackets bx cubed over 3 close square brackets subscript straight O superscript straight L equals open square brackets aL squared over 2 minus fraction numerator ax space left parenthesis 0 right parenthesis squared over denominator 2 end fraction close square brackets space plus space open square brackets fraction numerator straight b space straight x space straight L cubed over denominator 3 end fraction minus fraction numerator straight b space straight x space left parenthesis 0 right parenthesis cubed over denominator 3 end fraction close square brackets vertical line straight W vertical line space equals space aL squared over 2 plus bL cubed over 3
    Question 242
    CBSEENPH11020377
    Wired Faculty App

    An athlete in the Olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    • 200 J − 500 J

    • 2 × 105J − 3 × 105J

    •  20,000 J − 50,000 J

    • 2,000 J − 5,000 J
    Solution

    C.

     20,000 J − 50,000 J

    Approximate mass = 60 kg Approximate
    velocity = 10 m/s Approximate
    KE = (1/2)x60 x1000 = 3000 J
    = KE range ⇒ 2000 to 5000 joule

    Question 246
    CBSEENPH11020469
    Wired Faculty App

    A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

    • 7.2 J

    • 3.6 J

    • 120 J

    • 1200 J 
    Solution

    B.

    3.6 J

    Mass per length
    = M/L
    = 4/2 = 2 kg/m
    The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
    The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
    Hence, work done in pulling the chain on the table
    W =mgh
    = 1.2 x 10 x 0.3
    = 3.6 J

    Question 252
    CBSEENPH11020569
    Wired Faculty App

    A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed (v). The total kinetic energy generated due to explosion is,

    • mv2

    • 3 over 2 m v squared

    • 2mv2

    • 4 mv2
    Solution

    B.

    3 over 2 m v squared
    As per the question, the third part of mass 2m will move because the total momentum of the system after explosion must remain zero.
    Let the velocity of the third part be v'.
    According to the law of conservation of momentum,
    space space space space square root of 2 space left parenthesis m v right parenthesis space equals space left parenthesis 2 m right parenthesis space x space v apostrophe rightwards double arrow space space space space v apostrophe space equals space fraction numerator v over denominator square root of 2 end fraction
    Total kinetic energy generated by the explosion,
    equals space 1 half mv squared space plus space 1 half mv squared space plus space 1 half space left parenthesis 2 straight m right parenthesis space straight v apostrophe squared
    equals space mv squared space plus space mv squared over 2 equals space 3 over 2 mv squared
    Question 254
    CBSEENPH11020593
    Wired Faculty App

    Two cities are 150 km apart. electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 V and the average resistance per km is 8 V and the average resistance per km is 0.5 space straight capital omega. The power loss in the wire is,

    • 19.2 W

    • 19.2 kW

    • 19.2 J

    • 12.2 kW
    Solution

    B.

    19.2 kW

    Potential difference (drop) between two cities = 150 x 8 = 1200 V
    Average resistance of total wire = 0.5 x 150 = 75 space straight capital omega
    Power loss, P = straight V squared over straight R equals fraction numerator 1200 space x space 1200 over denominator 75 end fraction equals 19200 space W 
                      = 19.2 kW

    Question 255
    CBSEENPH11020638
    Wired Faculty App

    The potential energy of a system increase if work is done

    • By the system against a conservative force

    • By the system against a nonconservative

    • upon the system by a conservative force

    • upon the system by a nonconservative
    Solution

    A.

    By the system against a conservative force

    The potential energy of a system increases if work is done by the system against a conservative force. 
    -ΔU = Wconservative face

    Question 260
    CBSEENPH11020744
    Wired Faculty App

    Water falls from a height of 60 m at the rate of 15 kg /s to operate a turbine. The losses due to frictional 

    • 8.1 kW

    • 10.2 kW

    • 12.3 kW

    • 7.0 kW
    Solution

    A.

    8.1 kW

    Power generated by the turbine is,
    Pgenerated = Pinput  x 90/100

    equals Mgh over straight t space straight x 90 over 100 Putting space the space given space values straight M over straight t space straight x space 15 space kg divided by straight s comma space straight g space equals space 10 space straight m divided by straight s squared space comma space straight h space equals 60 space straight m straight P subscript generated space end subscript space space equals space left parenthesis 15 space straight x space 10 space straight x 60 right parenthesis space straight x 90 over 100 space equals space 8.1 space kW

    Question 261
    CBSEENPH11020805
    Wired Faculty App

    A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is

    • MLω squared over 2
    • fraction numerator ML squared straight omega over denominator 2 end fraction
    • MLω squared
    • fraction numerator ML squared straight omega squared over denominator 2 end fraction

    Solution

    A.

    MLω squared over 2

    Let the length of a small element of tube be dx. 
    Mass of this element
              dm space equals space straight M over straight L dx

    where M is mass of filled liquid and L is length of tube. 
    Force on this element
          dF space equals space dm space cross times space xω squared space space space space space integral subscript 0 superscript straight F dF space equals space straight M over straight L straight omega squared integral subscript 0 superscript straight L xdx or space space space space space space straight F space equals space straight M over straight L straight omega squared open square brackets straight L squared over 2 close square brackets space equals space MLω squared over 2 or space space space space space space space straight F space equals space 1 half MLω squared

    Question 264
    CBSEENPH11024903
    Wired Faculty App

    Define the term elasticity
    Solution

    Elasticity is a measure of a variable's sensitivity to a change in another variable. In business and economics, elasticity refers to the degree to which individuals, consumers or producers change their demand or the amount supplied in response to price or income changes. It is predominantly used to assess the change in consumer demand as a result of a change in a good or service's price.

    Question 265
    CBSEENPH11026146
    Wired Faculty App

    Which of the following is not an inelastic collision ?

    • A man jump on a cart

    • A bullet embedded in a block

    • Collision of two glass ball

    • none of the above
    Solution

    C.

    Collision of two glass ball

    Inelastic collision:- The deformation may not be relieved and the two bodies could move together after the collision. A collision in which two particles move together after the collision is called completely inelastic collision. Collision in which momentum of the system is conserved but not the kinetic energy are called inelastic collisions. In the above cases, collision of two glass ball is an inelastic collision.

    Question 266
    CBSEENPH11026163
    Wired Faculty App

    A force F acting on an object varies with distance x as shown here. The force is in newton and distance is in metre. The work done by the force in moving the object from x = 0 to x = 6 m is

    • 4.5 J

    • 13.5 J

    • 9.0

    • 18.0 J
    Solution

    B.

    13.5 J

    The work done will be the area of the F-x graph. Work done in moving the object from x = O to x = 6 m is given by

     

    W = area of rectangle + area of triangle

        = 3 × 3 +12 × 3 × 3

    W   = 9 + 4.5 = 13.5 J

    Question 267
    CBSEENPH11026166
    Wired Faculty App

    A 5 A fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is

    • 0.2 Ω

    • 5 Ω

    • 0.4 Ω

    • 0.04 Ω
    Solution

    D.

    0.04 Ω

    Power P = i V = i × iR

            P = i2 R

    ⇒ R = Pi2P = 1 W, i = 5 AR = 152R = 0.04 Ω

    Question 268
    CBSEENPH11026179
    Wired Faculty App

    A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is

    • -0.5 J

    • -1.25 J

    • 1.25 J

    • 0.5 J
    Solution

    B.

    -1.25 J

    The height (h) traversed by particle while going up is

    h = u22gh  = 252 × 9.8

    Work done by gravity force = m g→. h→= 0.1 × g × 252×9.8cos180o

    [Angle between force and displacement is 180°)

    W = -0.1 × 252

    ⇒ W  = -1.25 J

     

    Question 269
    CBSEENPH11026180
    Wired Faculty App

    A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

    • Mg 2 + 1

    • Mg 2

    • Mg2

    • Mg 2 -1
    Solution

    D.

    Mg 2 -1

    Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work-energy theorem. Let us assume that body is taken slowly so that its speed doesn't change, then  ΔK = 0

         = WF + WMg + Wtension

    [symbols have their usual meanings]

    WF = F × l sin 45o

    WMg = Mg ( l - l cos 45o )

    Wtension = 0

    F = Mg 2 - 1

    Question 270
    CBSEENPH11026185
    Wired Faculty App

    An alpha nucleus of energy 12 mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

    • v2

    • 1/ m

    • 1/ν4

    • 1/ Ze
    Solution

    B.

    1/ m

    Since, here nuclear target is heavy it can be assumed safely that it will remain stationary and will not move due to the Coulombic interaction force.

    At distance of closest approach relative velocity of two particles is v. Here target is considered as stationary, so a-particle comes to rest instantaneously at distance of closest approach. Let the required distance is 'r'  then from work energy-theorem.

    The change in kinetic energy of a particle is equal to the work done on it by the net force.

    0 - mv22 = -14πεo Ze × 2e r⇒       r    ∝ 1m ⇒      r   ∝ 1ν2 ⇒      r  ∝  Ze2

    Question 271
    CBSEENPH11026191
    Wired Faculty App

    For inelastic collision between two spherical rigid bodies

    • the total kinetic energy is conserved

    • the total mechanical energy is not conserved

    • the linear momentum is not conserved

    • the linear momentum is conserved
    Solution

    D.

    the linear momentum is conserved

    In an inelastic collision, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of particles no longer remains conserved. However in the absence of external forces, law of conservation of linear momentum still holds good.

    Question 272
    CBSEENPH11026197
    Wired Faculty App

    300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g =10 m/s2, work done against friction is

    • 200J

    • 100 J

    • zero

    • 1000 J
    Solution

    B.

    100 J

    Net work done in sliding a body up to a height h on inclined plane

            = Work done against gravitational force + Work done against frictional force

    ⇒             W = Wg  +  Wf

                    W = 300 J 

       but ,      Wg = mgh = 2 × 10 × 10

                          = 200 J

      Putting in Eq. (i), we get
                    300 = 200 + Wf

    ⇒               Wf = 300 - 200

                      Wf = 100 J

    Question 273
    CBSEENPH11026209
    Wired Faculty App

    A motor cycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motor cycle is ascending on the overbridge, the normal force on it

    • increases

    • decreases

    • remains the same

    • fluctuates erratically
    Solution

    A.

    increases

    Reaction of normal force on the motorcyclist

                 mg cosθ - N = m v2R 

               

    As the motorcyclist moves upward on the bridge, θ decreases and therefore, N increases and becomes maximum at the highest point.

    Question 274
    CBSEENPH11026210
    Wired Faculty App

    If we throw a body upwards with velocity of 4 m/s, at what height does its kinetic energy reduce to half of the initial value? ( Take g = 10 ms-1 )

    • 4 m

    • 2 m

    • 1m

    • 0.4 m
    Solution

    D.

    0.4 m

    At a given height the half of the kinetic energy of the body is equal to the potential energy.

    Initial kinetic energy of the body

                               = 12mv2

                              = 12 m 42

                                = 8 m

    Let at height h, the kinetic energy reduces to half, i.e, it becomes 4 m. It is also equal to potential energy. Hence

                      mgh = 4 m

                       ⇒ h = 4g        = 410

                              h = 0.4 m 

    Question 275
    CBSEENPH11026214
    Wired Faculty App

    Two spherical bodies of masses M and SM and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision,

    • 1.5 R

    • 2.5 R

    • 4.5 R

    • 7.5 R
    Solution

    D.

    7.5 R

    Let at O there will be a collision. If smaller sphere moves x distance to reach at O, then bigger sphere will move a distance of ( 9R - x )

              

              F = GM × 5M12 R - x2

              asmall  = FM

              asmall = G × 5M12 R - x2

               abig = F5M

               abig = GM12R -x 2

                x = 12asmall t2                     

                 x =12G × 5M12 R - x2 t2             .....(i)

        ( 9R - x ) = 12abig t2

         (9R - x)  = 12GM12 R - x2t2           ....(ii)

    Thus dividing equation (i) by equation (ii), we get

                    x9R- x = 5

    ⇒            x = 45 R - 5x

    ⇒             6x = 45 R

    ∴               x = 7.5 R    

    Question 276
    CBSEENPH11026241
    Wired Faculty App

    A ball of mass 2 kg moving with velocity 3 m/s collides with spring of natural length 2 m and force constant 144 N/m. What will be length of compressed spring?

    • 2 m

    • 1.5 m

    • 1 m

    • 0.5 m
    Solution

    B.

    1.5 m

    Let spring is compressed by a length x. 

    Kinetic energy of ball = Potential energy of spring

    i.e,        12m v2 = 12 k x2

    Given m = 2 kg, v = 3 m/s , k = 144 N/m

    ∴           12 × 2 × 32 = 12 × 144 × x2

    ⇒                      9 = 72 x2

    ∴                      x = 972

    ⇒                    x = 122m

    Hence, length of compressed spring

                                 = 2  -  122

                                  = 4 2 - 12 2

    length of compressed spring = 1.5 m                

    Question 277
    CBSEENPH11026255
    Wired Faculty App

    The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

    • 225 J

    • 200 J

    • 400 J

    • 175 J
    Solution

    B.

    200 J

    Work done W = Area ABCEFDA

                         = Area ABCD + Area CEFD

     

                        = Area ABCD + Area CEFD

                        =  12 × 15 + 10 × 10 + 12 + 10 + 20  × 5

    Work done W = 125 + 75

                         = 200 J

    Question 278
    CBSEENPH11026271
    Wired Faculty App

    A body of mass m1 = 4 kg moves at 5 i  m/s and another body of mass m2 = 2 kg moves at 10 i  m/s. The kinetic energy of centre of mass is

    • 2003J

    • 5003 J

    • 4003J

    • 8003J
    Solution

    C.

    4003J

    The total kinetic energy of a multiparticle system is equal to the sum of particles in the systems.      

    The centre of mass is a point that represents the average location for the total mass of system.

    vCM =m1 dr1dt + m2 dr2dtm1 + m2  

          vCM = 4 × 5i + 2 ×10i4 + 2

           VCM = 40 i6

                 = 203i

    The kinetic energy

           K = 12 m v2

           K =12 × 4 + 2 ×20 × 203 × 3

           K = 12× 6 × 20 × 203 × 3

           K = 4003 J

    Question 279
    CBSEENPH11026278
    Wired Faculty App

    The potential energy of a body is given by U = A - Bx2 (where x is the displacement ) The magnitude of force acting on the particle is

    • constant

    • proportional to x

    • proportional to x2

    • inversely proportional to x
    Solution

    B.

    proportional to x

    The potential energy is given by 

    U = A - B x2

    x - displacement

          F = - dUdv

              = -ddxA - Bx2

            = 2 Bx

    ⇒    F ∝ x

    Question 280
    CBSEENPH11026285
    Wired Faculty App

    A gas expands 0.25 m3 at constant pressure 103 N/m2, the work done is

    • 250 N

    • 250 W

    • 250 J

    • 2.5 erg
    Solution

    C.

    250 J

    From the formula of work done

           W = P ΔV

    Where W is the work done, P is the pressure and ΔV is the change in volume

        W = 103 × 0.25

         W = 250 J

    Question 281
    CBSEENPH11026299
    Wired Faculty App

    A ball is released from certain height which losses 50% of its kinetic energy on striking the ground it will contain a height again

    • 14th of initial height

    • 12th of initial height

    • 34th of initial height

    • None of the above
    Solution

    B.

    12th of initial height

    Since it loses 50% of its kinetic energy on striking ground, its PE is also reduced by half.   

    R     Ratio of heights

                    h1h2 = 100100 - 50

                     h2 = h12

                      h2 = 12 of the initial height

    Question 282
    CBSEENPH11026306
    Wired Faculty App

    If we increase kinetic energy of a body 300%, then per cent increase in its momentum is

    • 50%

    • 300%

    • 100%

    • 150%
    Solution

    C.

    100%

     Kinetic energy           

                     E = 12 mv2 

                     E = 12m2v2m

                     E = 12p2m

                      p2 = 2ME

    ∴                 p ∝ E

    ∴                 p1p2 = E1E2

    ∴                        = EE + 3E

                        p1p2 = 12

    ⇒                  p2 = 2 p1

                     % increase = p2 - p1p1 × 100

    ⇒                 2p1 - p1p1 × 100 = 100%

    Question 283
    CBSEENPH11026317
    Wired Faculty App

    A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field, when the particle completes one full circle, is

    • M v2R 2πR

    • zero

    • BQ 2πR

    • BQv 2πR
    Solution

    B.

    zero

    When particle describes circular path in a magnetic field, its velocity is always
    perpendicular to the magnetic force

        Power P = F . v

                 P = Fv cosθ
     Hence     

              θ = 90o

               P = 0

    But     P = Wt

             W = P .t

    Hence, work done W = 0 (everywhere)

    Question 284
    CBSEENPH11026318
    Wired Faculty App

    A particle of mass 100 g is thrown vertically upward with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is 

    • -0.5 J

    • -1.25 J

    • 1.25 J

    • 0.5 J
    Solution

    B.

    -1.25 J

    The height (h) transversed by particle while going up is 

           n = u22 g

           n = 252 × 9.8

    Work done by gravity force = mg.h

                   = 0.1 × g × 252 × 9.8cos 180o

    (angle between force and displacement is 180)

    ∴          W = - 0.1 × 252

               W = -1.25 J

    Question 285
    CBSEENPH11026323
    Wired Faculty App

    One end of a massless spring of constant 100 N /m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of 2 rad/s, then elongation of spring is

    • 0.1 m

    • 10 cm

    • 1 cm

    • 0.01 cm
    Solution

    C.

    1 cm

    For the circular motion acceleration towards the centre is v2r.

    The horizontal force on the particle is I due to the spring and kl,

     where, l is the elongation 

               k is spring constant of spring.

         kl = mv2r

             = mω2 r

         kl = mω2 ( lo + l )

    ⇒  kl - mω2 ( lo + l ) = 0

    ⇒ ( k - mω2) l = mω2 lo

    ⇒             l = mω2 lok - mω2 

    Putting tle values l = 0.5 × 4 × 0.5100 - 0.5 × 4

                               = 198m

                                = 0.010 m

                             l = 1 cm

    Hence elongation(l) in spring is 1 cm.

    Question 286
    CBSEENPH11026345
    Wired Faculty App

    A block of mass 2 kg is pushed against a  rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied
    on the block in a direction parallel to the wall. If the block will move, then its direction would be

    • 15o with 15 N force

    • 53o with 15 N force

    • 45o with 15 N force 

    • 75o with 15N force
    Solution

    B.

    53o with 15 N force

    Now, When a force of 40N is applied, it will be balanced by the normal reaction from the wall

    The net force = 202 + 152 

                         = 625

                   F = 25 N

    This is greater than the 20N (friction), therefore block will move

    ∴ Direction 

              tanθ = 2015

                     =  43

    ∴           θ = tan-1 43

    So, the block will move at an angle 53° with the 15 N force. 

    Question 287
    CBSEENPH11026363
    Wired Faculty App

    Two wires are stretched through same distance. The force constant ofsecond wire is half as that of the first wire. The ratio of work done to stretch first wire and second wire will be

    • 2 : 1

    • 1 : 2

    • 3 : 1

    • 1 : 3
    Solution

    A.

    2 : 1

    As W = 12 kx2

    If both wires are stretched through same distance, then   

         W ∝ k

        W1W2 = k1k2

                = k1k12

                = 21

         W1W2 = 2 : 1

    Question 288
    CBSEENPH11026368
    Wired Faculty App

    A light string passes over a frictionless pulley. To one of its ends a mass of 8 kg is attached. To its other end two masses of 7 kg each are attached. The acceleration of the system will be

      

    • 10.2 g

    • 5.10 g

    • 20.36 g

    • 0.27 g
    Solution

    D.

    0.27 g

    free body diagram,

         

    For body A, having mass m1 forces are given by

              m1a = T1 - m1 g                   ....(i)

    For body B

                m2a = m2 g + T2 - T1            .....(ii)

    For body C

                m3 a = m3g - T2                  ......(iii)

    On solving Eqs. (i), (ii) and (iii), we get

                a = m2 + m3 - m1 gm1 + m2 + m3

     As      m1 = 8 kg, m2 = m3 = 7 kg

               a = 7 + 7  - 8 g22

                a = 622g

                 a = 0.27 g

    Question 289
    CBSEENPH11026373
    Wired Faculty App

    A sphere of mass m moving with velocity v hits inelastically with another stationary sphere of same mass. The ratio of their final velocities will be (in terms of e)

    • v1v2 = 1 + e1 - e

    • v1v2 = 1 - e1 + e

    • v1v2 = 1 + e2

    • v1v2 = 1 - e2
    Solution

    B.

    v1v2 = 1 - e1 + e

    As coefficient of restitution,

              e = v2 - v1u1 - u2

                = v2 - v1u - 0

    ⇒    v2 - v1 = eu                   ...... (i)

    By conservation of momentum,

    Momentum before collision =  Momentum after collision

         m u = mv1 + mv2

    ⇒   v1 + v2 = u                   ......(ii)

    On solving Eqs. (i) and (ii), we get

             v1 = u2 ( 1 - e )

    and    v2 = u2 ( 1 + e )

    ∴     v1v2 = 1 - e1 + e

    Question 290
    CBSEENPH11026387
    Wired Faculty App

    A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution is

    • 122

    • 13

    • 12

    • 32
    Solution

    C.

    12

    Let u1 and v1 be the initial and final velocities of ball 1 and u2 and v2 be the similar
    quantities for ball 2.

    Here, u2 = 0 and v1 = 0

    ∴    initial K.E 

              Ki =  12 mu12 + 12 mu22

              Ki = 12 mv12

    and  final K.E

              Kf = 12 mv12 + 12 mv22

              Kf = 12 mv22

    Loss of KE

             ΔK = Ki - Kf

                  = 12 m12 - 12 mv22

    According to question

            12 12 mv12 = 12 mu12 - 12 mu22

    ( since half KE is lost by impact )

            u12 = 2 v22

     ⇒     v2 = u12

    The coefficient of restitution is defined as the ratio of the final velocity to the initial velocity after their collision.

    ∴  Coefficient of restitution,

        e = v2 - v1u1 - u2 

            = v2v1

       e = 12

    Question 291
    CBSEENPH11026388
    Wired Faculty App

    A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents are respectively

    • 40 A and 16 A

    • 16 A and 40 A

    • 20 A and 40 A

    • 40 A and 20 A
    Solution

    A.

    40 A and 16 A

    Given:-

       η = 80%

        Pi = 4 kW

    ⇒  Pi = 4000 W

         Vp = 100 V

         Vs = 200 V

          Ip = 4000100

          Ip = 40 A

    The efficiency of the transformer is defined as the ratio of useful power output the input power, the two being measured in the same unit

    Transformer efficiency given by,

            η = Vs IsVp Ip

    Where Vs → Secondary voltage

               Vp → Primary voltage

              Is → secondary current

              Ip → primary current

    ⇒    80100 = 200 Is4000

                   = Is20

    ⇒    Is = 20 × 80 100

    ⇒   Is = 16 A

    Question 292
    CBSEENPH11026397
    Wired Faculty App

    Same spring is attached with  2 kg,  3 kg  and  1 kg blocks in three different cases as shown in figure. If  x1 , x2, x3 be the extensions in the spring in the three cases, then

                   

    • x1 = 0,  x3 > x2

    • x1 > x2 > x3

    • x3 > x2 > x1

    • x2 > x1 > x3
    Solution

    D.

    x2 > x1 > x3

    If T1, T2, T3 are the tensions in the strings in the three cases.

    We have

             T1 = 2 m1m2 gm1 + m2

                   = 2 × 2 × 2g2 + 2

                T1 = 2g

         T2 = 2 × 3 × 2g3 +2 

          T2 = 2.4 g

         T3 = 2 × 1 × 2g1 + 2

           T3 = 1.33 g

    As

               x ∝ T

    and    T2 > T1 > T3

    ∴   x2 > x1 > x3

    Question 293
    CBSEENPH11026402
    Wired Faculty App

    Assertion: A particle strikes head-on with another stationary particle such that the first particle comes to rest after collision. The collision should necessarily be elastic.

    Reason: In elastic collision, there is a loss of momentum ofthe system of the particles.

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.
    Solution

    D.

    If both assertion and reason are false.

    In elastic collision there is no loss in kinetic energy in the system as a result of the collision. Both momentum and kinetic energy are consrved quantities in elastic collisions.

    In an Inelastic collision, there is a loss of kinetic energy. While the momentum of the system is conserved in an elastic collision, kinetic energy is not.

    Momentum of a system remains conserved in any kind of collision.

    Question 294
    CBSEENPH11026406
    Wired Faculty App

    Assertion: Work done in moving a body over a closed loop is zero for every force in nature.

    Reason: Work done does not depend on nature of force.

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.
    Solution

    D.

    If both assertion and reason are false.

    Work done in the motion of a body over a closed loop is zero only when the body is moving under the action of conservative forces ( like gravitational or electrostatic forces ). It is not zero when the forces are non-conservative e.g. frictional forces etc.

    Question 296
    CBSEENPH11026418
    Wired Faculty App

    A gun of mass 10 kg fires 4 bullets per second. The mass of each bullet is 20 g and the velocity of the bullet when it leaves the gun is 300 m s-1. The force required to hold the gun when firing is

    • 6 N

    • 8 N

    • 24 N

    • 240 N
    Solution

    C.

    24 N

    Momentum of one bullet

                 p = mv

                     = 20 × 10-3 × 300 

                 p = 6 kg m s-1

    N = number of bullet per sec = 4

    ∴   dpdt  = change of momentum per sec or force

                = N ( p - 0 )

         dpdt = 4 × 6

           dpdt = 24 N

    Question 297
    CBSEENPH11026420
    Wired Faculty App

    A stone of mass 0.3 kg attached to a 1.5 m long string is whirled around in a horizontal circle at a speed of 6 m s-1 The tension in the string is

    • 10 N

    • 20 N

    • 7.2 N

    • 30 N
    Solution

    C.

    7.2 N

    Here,    mass of the stone m = 0.3 kg

               Length of a string = 1.5 m 

                            Speed v = 6 m/s

                   T = mv2R

    Where R is the length of the string.

                       = 0.3 621.5

                 T = 7.2 N

    Question 298
    CBSEENPH11026422
    Wired Faculty App

    If the linear momentum is increased by 50%, then kinetic energy will increase by

    • 0 %

    • 100 %

    • 125 %

    • 25 %
    Solution

    C.

    125 %

    Kinetic energy of a body

                   K = p22m                                  K = 12 mv2

    where p is the momentum of the body and m is the mass of the body

                p' = p + 50100p

                p' = 32 p

    Assuming m remains constant

    ∴        K'K = p'p

               K'K= 94

    % increase in kinetic energy

                          = K' - KK × 100%

                           = 94 - 1 × 100%

                           = 125 %

    % increase in kinetic energy = 125 %

    Question 299
    CBSEENPH11026426
    Wired Faculty App

    A body moves from rest with a constant acceleration. Which one of the following graphs represents the variation of its kinetic energy K with the distance travelled (x)?

     
    Solution

    C.

    Kinetic energy is given by,     

          K = 12 mv2

    where     v = 0 + αt

    ∴          K = 12 mα t2

                   = 12 m αt2

                K = 12m α2t2

    ∴   As α = [ L T-2 ]

    ∴         L ∝ T2

    ∴         K = 12 m α2.L

    ∴           K ∝ L

    ∴           K ∝ x

    The graph is straight line.

    Question 300
    CBSEENPH11026429
    Wired Faculty App

    Consider the system shown in figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (take g = 10 ms-2)

            

    • 0 N

    • 1 N

    • 2 N

    • 5 N
    Solution

    D.

    5 N

    If α is acceleration of m2, then from fig

           

         m2 g - T = m2 α                            .....(i)

    and   T = m1α = 1α                             ....(ii)

    From (i)

          m2 g - T = m2 α

              1 g - m1α = m2 α

               1 g = (m1 + m2 ) α

                      = ( 1 + 1 ) α

                    α =  g2

                       = 102

                 α  = 5 ms-2

    ∴        T = m1 α

              T = 1 × 5

              T = 5 N

    Question 301
    CBSEENPH11026431
    Wired Faculty App

    Assertion:  Heavy water is used as moderator in nuclear reactor.

    Reason:  Water cool down the fast neutron

    • If both assertion and reason are true and reason is the correct explanation of assertion.

    • If both assertion and reason are true but reason is not the correct explanation of assertion.

    • If assertion is true but reason is false.

    • If both assertion and reason are false.
    Solution

    C.

    If assertion is true but reason is false.

    When two bodies of same mass undergo an elastic collision, their velocities are interchanged after collision. Water and heavy water are hydrogenic materials containing protons having approximately the same mass as that of a neutron. When fast moving neutrons collide with protons, the neutrons come to rest and protons move with the velocity of that of neutrons.

    Question 303
    CBSEENPH11026438
    Wired Faculty App

    A mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at height h above the ground at equilibrium. If the distance between poles is L the tension in each string will be

    • Mg L22 +  H - h 22 H - h

    • Mg L2 + H22 H - h

    • Mg L2 + H22 H

    • Mg h2 + L22 L
    Solution

    A.

    Mg L22 +  H - h 22 H - h

    At equilibrium 

              Mg = 2 T sinθ    

         

                    T = M g2 sinθ                                  .......(i)

    Here,       sinθ = H - hL22 + H - h2

    From equaton (i)

                   T = Mg L2 + H - h 22 H - h

    Question 304
    CBSEENPH11026444
    Wired Faculty App

    A body is moved in straight line by machine with constant power. The distance travelled by it in time duration t is proportional to

    • t3/2

    • t1/2

    • t2

    • t
    Solution

    A.

    t3/2

          P = constant

           ∆W∆t = P

         ΔW = P Δt

    By using work-energy theorem

        12 mv2 = Pt

         v2 = 2Pm t1/2 dt

    Integrating both sides

        ∫0xdx = 2Pm ∫0tt1/2 dt

             x = 2Pm 23 t3/2

    ∴      x  ∝  t3/2

         displacement ∝ (time)3/2

    Question 305
    CBSEENPH11026445
    Wired Faculty App

    Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

    • Ring

    • Solid cylinder

    • Solid sphere

    • All reach the ground with same velocity
    Solution

    C.

    Solid sphere

              

    The total mechanical energy of a system is conserved if the forces doing work on it. 

    Law of conservation of mechanical energy: the total amount of mechanical energy, in a closed system in the absence of dissipative forces ( eg. friction, air resistance), remains constant. This means that potential energy can become kinetic energy or vice versa, but energy cannot disappear. 

    According to conservation of mechanical energy 

           mgh = 12 m v2 1 + k2R2

    ⇒       v2 = 2 gh1 + k2R2

    It is independent of the mass of the rolling body.

    For a ring

              k2 = R2

              vring = 2 gh1 + 1

    For solid cylinder

            k2 = 2 gh1 + 1

            k2 = g h

    For solid cylinder

                k2 = R22

         vcylinder = 2 gh1 + 12

                     = g h

    For a solid sphere

          vsphere = 2 gh1 + 25

            vsphere = 10 gh7

    Among the given three bodies the solid sphere has the greatest and the ring has the least velocity at the bottom of the inclined plane.     

    Question 306
    CBSEENPH11026453
    Wired Faculty App

    An air compressor is powered by a 200 rad/s electric motor using a V-belt drive. The motor pulley is 8 cm in radius, and the tension in the V-belt is 135 N on one side and 45 N on the other. The power of the motor will be

    • 1.44 kW

    • 14.4 kW

    • 2.88 kW

    • 28.8 kW
    Solution

    A.

    1.44 kW

    Since the net force on the belt is the difference between the two tensions, the torque exerted by the motor on the belt is 

              Τ = F r

                 = ( 135 N - 45 N ) ( 0.08 m )

            Τ = 7.2 N m

    The power output of the motor is therefore

         P = Τ ω

            = ( 72 N.m ) ( 200 rad s-1 )

         P = 1440 W

        P = 1.44 kW

    Question 307
    CBSEENPH11026455
    Wired Faculty App

    A string 0.5 m long is used to whirl a 1 kg stone in a vertical circle at a uniform velocity of 5 m s-1.  What is the tension in the string when the stone is at the top of the circle?

    • 9.8 N

    • 30.4 N

    • 40.2 N

    • 59.8 N
    Solution

    C.

    40.2 N

    The centripetal force needed to keep the stone moving at 5 m s-1 is

            Fc = m v2r

                 = 1 kg 5 m s-10.5 m

          Fc = 50 N

    The weight of the stone is

       W = m g

            = ( 1 kg ) ( 9.8 m s-2 )

        W = 9·8 N

    At the top of the circle

       T = Fc - 9·8 N

        T = 40.2 N

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