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Define v=u+at from velocity - time graph
The variation in velocity with time for an object moving in a straight line can be represented by Velocity Time Graph. In this graph, time is represented along the x axis and the velocity is represented along the y axis.
If the object moves at uniform velocity, the height of its velocity time graph will not change with respect to time
State which of the following are examples of one, two or three dimensional motion:
(a) A kite flying in the sky.
(b) The earth revolving around the sun.
(c) A train moving along the equator of the earth in clockwise direction.
(d) Accelerated motion of a particle in a straight line.
(a) Three-dimensional motion.
(b) Two dimensional motion.
(c) Two dimensional motion.
(d) One dimensional motion.
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A particle moves from A to B along the path APB, When the distance APB will equal to magnitude of displacement ?
The magnitude of displacement will be equal to the length of path APB, if P lies in between AB on the line joining the points A and B.
We have,
Initial velocity, u = 9.8 m/s
Acceleration, a = - 9.8 m/s2
Time taken, t = 1 sec
Final velocity, v = ?
Using the formula,
Velocity after 1 sec,
v = u + at
= 9.8 + (-9.8) x 1
= 0
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Two basic equations for the kinematics of uniformly accelerated motion are,
Initial velocity of the particle = u
Rate of acceleration = a
So, dispalcement undergone by the particle is given by,
Can we find the distance travelled by particle from speed-time graph?
Yes, the area of the speed-time curve gives us the distance.
Given, a particle is moving with a constant velocity 5 m/s.
(a) Slope of position-time graph will give us the speed of the particle. Speed will be equal to 5m/s.
(b) Slope of velocity-time graph will be zero because velocity is constant.
Initial distance covered = 40 km
Distance covered after 40 mins of the trip = 76 km
Time taken = 40 min
Therefore, average speed of car =
Kinematics is the branch of classical mechanics which describes the motion of points, bodies and system of bodies without considering the masses of these objects not the forces that has caused the motion.
Dynamics is the study of motion of particle by considering the forces and their torque and their effect on motion.
Displacement and distance can be compared as follows:
(i) Both have units and dimensions same as that of length.
(ii) The change in position of particle in particular direction is known as displacement while distance is the total length of path travelled by particle.
(iii) Displacement is a vector quantity while distance is a scalar quantity.
(iv) Displacement can be negative, positive or zero while distance covered by particle is always positive.
(v) Displacement of a particle between two points is a unique path, while distance is not a unique path .
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Speed | Velocity |
Speed changes only when the magnitude of the object changes. | Velocity changes whenever there is a change in direction and speed of the particle. |
Speed is a scalar quantity. | Velocity is a vector quantity. |
Speed is either zero or positive. It cannot be negative. | Velocity can be positive, negative or zero |
Average velocity | Instantaneous velocity |
1. The velocity with which the body covers the same distance in the same time as the body actually covered with variable velocity is known as average velocity. It is measured as the ratio of total displacement to total time taken. |
1. Instantaneous velocity of a body is defined as the velocity of the body at a particular instant and it is the limiting case of average velocity as the time interval approaches zero. That is instantaneous velocity is, |
2. Average velocity between two times is equal to the slope of the line joining the two points on the position vs. time graph. | 2. Instantaneous velocity of an object at a particular time is equal to the slope of it's position vs. time graph at that time. |
Kinematics of uniform motion is given by
x = x0+ut ...(1)
Therefore x - t graph will be a straight line. Slope of graph is given by,
Hence the result.
When a particle moves with a constant velocity, displacement is given by,
S = x - x0 = ut ...(1)
For uniform motion, the velocity time graph is straight line parallel to time axis.
Now area under v-t graph from t = t1 to t = t2 is given by = Area of rectangle ABCD
= ut
= x - x0, which is the displacement of the particle.
Given,
Time taken by the particle to complete 1 revolution = 6 s
Radius of the circle = 20 cm
No. of revolutions made in 15 seconds = 2.5 revolution
After 2.5 revolutions, the particle will be at the point diagonally opposite to the starting point.
So, Displacement of particle = 2r = 40 cm
Distance travelled by the particle = Total length of the particle.
Distance = total length of the path = 2.5x2πr = 100 π cm
Displacement of partcle is,
Distance traveled by particle is,
= 57.1m
Given,
Radius of the circle = r
Distance covered = half of the circle =
Magnitude of displacement is 2r.
, which is also numerically eqal to right angle in radian.
Let the particle travel a total distance of 2S.
Distance covered in first half is S with a velocity v1 and next half with velocity v2.
Time taken to travel first half distance is,
Time taken to travel second half distance is,
Total time taken to travel 25 distance is,
Average velocity =
Let the body travel for total time to 2t.
The distance travelled by body in first t time is,
The distance travelled by body in next t time is,
Total distance travelled by the body is,
Distance travelled by light in vacum is, S1 = 100m
Distance travelled by light in media is, S2 = 100m
Velocity of light in vacum is c = 3 x 108 m/s
Velocity of media is v = 2.4 x 108 m/s
Time taken by light to cover 100m in vacum is,
Time taken by light to cover 100m in media is,
Total time is given by,
Now average velocity is given by,
Let the distance between two stations be 'd' km.
After the journey train returns to the first station.
Total distance travelled = 2d
Net displacement of the train = 0
Time taken by train to travel from one station to another,
Time taken by train to come back,
Total time taken by train to complete the round trip is given by,
Average velocity =
Given, the drunkard walks 7 steps forward and 4 steps backward.
So, the net displacement undertaken by the drunkard = 3 steps in one stroke
Also given, length of 1 step = 0.5 m
Time taken to take 1 step = 1 sec
∴ Displacement in 1 stroke = 3 x 0.5=1.5m
Therefore,
Displacement in 3 strokes = 3 x 1.5 =4.5m
Time taken to walk one step is 1 sec.
Since to complete one stroke, drunkard walks 7 steps forward and 4 steps backward i.e. total of 11 steps, therefore drunkard takes 11s to complete one stroke.
So,
Time taken to complete three strokes, t = 11 x 3 = 33s
Now, average velocity of drunkard is given by,
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Time, t (sec) | Distance, x (m) |
5 | 5 |
8 | 2 |
13 | 7 |
16 | 9 |
24 | 6 |
Speed of car, v = 30 km/hr
Distance of wall, d= 90 km
Time taken by car to reach the wall is,
Fly moves back and forth between car and the wall with constant speed 60km/hr.
Therefore, distance travelled by fly is,
S = ut
= 60 x 3
= 180 km
To find: The distance travelled by the fly in it's one complete trip.
Consider, initially when the fly is flying from the wall towards the car, it is at a distance D from the wall.
Speed at which the car is approaching the fly = 30 km/hr
Speed at which the fly is flying towards the car = 60 km/hr
Let both car and fly meet after time t1.
Thus distance travelled by car and fly respectively is,
Read each statement carefully and state if it is true or false:
A particle in one dimensional motion:
(a) with zero speed at an instant may have non-zero acceleration at that instant;
(b) with zero speed may have non-zero velocity;
(c) with constant speed must have zero acceleration;
(d) with positive value of acceleration must be speeding up.
(a) True; when the body is projected vertically upward, then the speed of the particle at the highest point is zero but has an acceleration equal to the acceleration due to gravity.
(b) False; because
|Displacement| ≤ Distance
∴|Velocity| ≥ Speed
Thus, the speed of the body zero implies that the magnitude of velocity of the particle is zero. Hence the velocity of body is zero.
(c) True; since motion is in one dimension, therefore there are only two possible directions of motion. Since speed is constant, therefore motion can be accelerated if the direction of motion changes instantaneously.
i.e. acceleration of particle is infinite, which is not possible in practice.
(d) False; because if acceleration is positive, then it need not be necessary that particle will always speed up, it may slow down. The particle will speed up if velocity is also positive.
Given, a body is thrown up with velocity = 39.2 m/s.
Acceleration due to gravity is acting vertically in a downward direction.
Acceleration, g = 9.8 m/s2
The body thrown in upward direction with velocity 39.2 m/s will take 4s to reach the maximum height.
At t = 0 and t = 3, the velocity is in upward direction.
At t = 4 sec velocity is zero and at t = 6 sec, velocity will be in downward direction.
Let, height of the cliff be 'h'.
Time taken by the stone to reach the ground = n
The distance travelled by particle in nth second is given by,
Here we have,
n = 5 s
i.e. Stone takes 5 s to reach the ground from top of cliff.
Now the height of cliff is,
Given,
Mass of the body, m = 5 kg
Height from where the mass is dropped = 100 m
Acceleartion due to gravity, g = 10 m/s2
Using the formula,
v2-u2 = 2as
Here,
u = 0,
a = g = 10 m/s2,
s = 100m
v = 44.72 m/s, is the velocity with the body will hit the ground.
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During downward motion | During upward motion | |
Position | Positive | Positive |
Velocity | Positive | Negative |
Acceleration | positive | Positive |
Given, a ball is thrown vertically upwards with a speed 29.4 m/s.
Let, ground be the reference point and vertically upward direction is taken as positive.
We have,
Initial velocity, u = 29.4 m/c
Acceleration due to gravity, a = g= -1m/s2
Time taken for the ball to reach the ground, t = 8s
Initial distance, x = 0
Let, h be the height tower.
Therefore,
Displacement, xo = h
Using the equation of motion, we have
Now
0 = h + 235.2 - 313.6
h = 78.4m, is the required height of the tower.
The velocity-time relation for uniform accelerated motion is given by,
v = u + at ...(1)
The position-time graph for uniform accelerated motion is oblique straight line.
Now area under v-t graph from t = 0 to t = t is = Area under segment AB
= Area of rectangle ACDO + Area of ∆ABC.
OA | OB | BC | |
Acceleration | positive | Zero | Zero |
AB | BC | CD | |
Velocity | Positive | Negative | Zero |
Slope of velocity-time graph gives the acceleration of the object.
Therefore,
Acceleration of object A, A = tan 30°
Acceleration of object B, B = tan 60°
Now,
Position of a particle at any instant, x = 3t+4
The x -t graph is a straight line.
The slope of the position-time graph gives us the velocity of the particle.
Here, slope of the graph is,
Therefore, velocity of the particle is 3 m/s.
One-dimensional motion of the particle is represented by graphs (b) and (d).
But, (a), (c), (e) and (f) cannot represent the motion of particle because,
(a) a particle cannot have two velocities at the same time,
(c) a particle cannot have two positions simultaneously at the same time,
(e) total length of path can never decrease with time and
(f) a particle cannot have two different accelerations at the same time.
(a) Graph (a) represents the path of a ball which is dropped from a certain height and on hitting the ground, it rebounds elastically.
(b) Graph (b) represents the path of a uniformly moving football turned back by hitting it for a very short interval of time.
(a) Who lives closer to the school?
(b) Who starts from the school earlier?
(c) Who walks faster?
(d) Who overtakes the other?
(a) Since, OP < OQ, A lives closer to the school than B.
(b) From the position-time graph, A starts at t = 0 and B at t = T. That is, A starts earlier than B.
(c) Velocity is given by the slope of x-t graph. Here, the slope of x-t graph for B is greater than A. Therefore B walks faster than A.
(d) B overtakes A.
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
(b) Since velocity is a single valued function, therefore at any instant, the particle cannot have two different velocities, hence the graph cannot represent motion.
(c) Since speed is magnitude of velocity which cannot be negative, therefore, graph is not possible.
(d) Total length of path travelled by particle cannot decrease with the time. Therefore, graph is not possible.
Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.
Position-time graph represents the position of particle at any instant, not the trajectory of particle.
For t < 0, the position-time graph is time axis which signifies that the particle is at rest.
For t > 0, the position-time graph is a parabola which signifies that the body starts moving with constant acceleration.
The given graph is a suitable context for a body is dropped from a certain height at t=0.
Which of the following equations of kinematics are of:
(i) uniform motion
(ii) uniform accelerated motion
(iii) general motion.
(a) x = 4 - 7t2+4t3
(b) v = 4 + 5t (c) x — 5
(d) a = 9t -5 (e) a = 4.
a) x = 4 - 7t1+4t3
Differentiating 'x' twice w.r.t. time, we get
a = 24 t- 14
Since acceleration depends on time, therefore this is kinematics of general motion.
(b) v = 4 + 5t
∴ a = 5
Since acceleration of particle is constant, therefore the motion is uniform motion.
(c) x = 5
∴ v = 0
i. e. particle is at rest.
(d) a = 9t-5
Since acceleration depends on time, therefore this is kinematics of general motion.
(e) a =4
Because acceleration is constant, therefore motion is uniform accelerated motion.
A car moving along a straight highway with speed of 126 km/hr is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Here,
Initial velocity, u = 126 km/hr = 35m/s
Final velocity, v = 0
Distance travelled, s = 200m
Now using the equation of motion,
v2-u2 = 2as
Substituting the value of v, u and s we get,
(0)2-(35)2 = 2a x 200 a = -3.0625 m/s2
Time taken by the car to stop is given by,
Here,
Initial velocity, u = 0
Final velocity, v = 12m/s
Time taken to acquire the velocity, t = 5s
Using the equation, we have
v = u + at
where, a is the acceleration of the moving car, and
s is the distance moved by the car.
Here,
Velocity of a particle at any instant, v = 5t + 3
So,
Velocity|t=2 is v(2) = 13 m/s
velocity|t=5 is v(5) = 28 m/s
We know displacement of particle is equal to area under v-t graph. Distance travelled = Area of trapezium ABCD
Velocity-time graph of the motion is as shown in the figure.
Describe the motion of particle. What is the significance of shaded area?
From the graph given, we can infer that the particle starts from rest at t = 0 and accelerates uniformly, acquiring a velocity of 10 m/s in 5 sec.
Thereafter, the particle moves with a constant velocity.
The shaded area represents the displacement undergone by the particle during accelerated motion.
Given that,
Initial velocity, u = 0
Acceleration, a = ?
At t = n, the velocity of particle is v is given by the first equation of motion.
Velocity of car at t = n - 2 is given by,
Now, the distance tracvelled by the car in last two second is,
Speed of car A, ua = 5 m/s
Speed of car C, uc= 25 m/s
Given, AB = BC
That is B is the mid-point of AC.
Therefore, velocity at B is given by,
v = 18 m/s is the velocity of particle at B.
A particle is moving in a straight line with constant acceleration. It travels a distance of in first n seconds,
in next n seconds and y in next n seconds.
Prove that,
α + γ = 2β.
Both reach the ground simultaneously as the acceleration due to gravity is same for both the objects.
The area under velocity time graph represents the change in velocity.
Let at time t0 the particle acquire the initial velocity.
Therefore the change in velocity of the particle in time t0 is zero. i.e. Area under a v/s t graph is zero
Thus area of ∆ AOB = Area of ∆ BCD
So, at time t=8 sec the particle acquires it's initial velocity.
A bullet fired into a very thick wall loses one fourth of its velocity after penetrating 4m. How much further will it penetrate?
What is motion? Is absolute rest or absolute motion possible?
For accelerated motion:
Initial velocity, u = 0 m/s
Acceleration = m/s2
Time taken = t1 sec
Let, v be the velocity acquired by the body at the end of accelerated motion.
By using the relation,
v = u + at,
we have
V = 0 + ...(1)
For retarded motion,
Now, using this relation, v= u + at, we have
Since it given that,
i.e.
Acceleration of the particle is directly proportional to the time, i.e. acceleration increases linearly with time.
The velocity of a particle at any instant is given by
v = 8 + 2a1(a2+3a3t)
where a1, a2 and a3 are constant. What is the acceleration of the particle?
Here,
Velocity of the particle is given by,
Given,
Position of particle at any insatnt t is,
Differentiating both sides w.r.t. tme,
Now,
The position of the particle at any instant is given by the equation
x = 3t2- 2.5t2 - 4.
what is the velocity and acceleration of particle at 3s?
The position of the particle is given by,
x=3t3 -2.5t2 -4
Velocity of the particle at any instant is, Velocity at t = 3s is,
v(3) = 9(3)2 - 5 x 3 = 81 - 15 = 66 units
Acceleration of the particle at any instant is,
Acceleration at t = 3 s is,
a(3) = 18 x 3-5 = 54 - 5 = 49 units
The acceleration of a particle at any instant is given by a = 6t-12. The particle starts from x0 = 4m with a velocity 4 m/s. What is the velocity and position of the particle at any instant?
Given,
Acceleration of the particle, a = 6t - 12
Initial distance, x0 = 4 m
Initial velocity, u = 4m/s
Let at any instant the velocity of the particle be v.
Let at any instant the position of the particle be x.
Therefore,
The position of the particle moving in one dimension, under the action of constant force is given by equation
where x is in meter and t is in second. Find the displacement of the particle when its velocity is momentarly zero.
We have,
Position of the particle moving under a constant force,
x = t2 -6t + 9
Differentiating w.r.t. r, we get
v = 2t - 6, is the velocity of the particle.
The time at which the velocity of the particle will be zero is given by,
Now,
Initial position of the particle is,
Position of the particle at t = 3 is given by,
x=(3)2 - 6(3) + 9 = 0 Displacement of the particle is,
x - x0 = 0 - 9
= - 9m
Given,
Retardation of the boat,
Therefore,
r = kv; k is the proportionality constant.
Now, we know that acceleration is given by rate of change of velocity.
From figure given, we have
Differentiating both sides w.r.t. t, we get
2xu + 2yv = 0
Here,
Initial velocity, u = 10m/s
A particle is moving along a straight line such that its displacement at any instant is given by,
s1 = αt1 + 2βt + γ
where α, β and γ are constants.
Find the acceleration of the particle.
Given here,
Differentiating (1) w.r.t. t,
Differentiating (2) w.r.t. t,
That is, acceleration of the particle is given by,
This implies,
, is the acceleration of the particle.
Acceleration of the particle, a = 12 t + 4
Acceleration is the rate of change of velocity, a = dv/dt = 12 t + 4
That is,
dv = (12 t+ 4) dt
On integrating both sides,
Here we have,
Body retards at a rate of distance travelled.
That is,
a = -s
Let the velocity of the body when it starts retarding be v0 and come to rest by travelling the distance s.
Integrating both sides of (1) between initial and final limits,
A point moves in a straight line so that its displacement x (m) at time t (s) is given by
x2 = t2+ 1.
Find its acceleration at any instant.
The position of the particle at any instant given by,
x2 = t2 + 1
x = (t2 + 1)1/2
Differentiating both sides w.r.t. to t, we get
Let at any instant the particle be at a distance x from O.
Acceleration of the particle is given by,
a =
Integrating both sides, we get
Given, the particle starts from rest.
Acceleration, a =
Ball A:
Initial position of ball, x0A = h
Initial velocity of ball, uA = 0
Acceleration, aA = -g
Therefore the position of the ball at any instant is,
Ball B:
Initial position of ball, xOB = 0
Initial velocity of ball, uB = 0
Acceleration, aB = -g
Therefore, the position of the ball at any instant is,
xB = xOB + uBt + ...(3)
When the balls meet, the position of both the balls is same.
XA = XB
h - ... (2)
Position at which the two balls meet,
x = xA = xB = h -
But, x =
Therefore,
Velocity of balls, when they meet i.e., at t = h/u
C.
The gravitational force on a particle inside a hollow isolated sphere is zero.Acceleration due to gravity =
where M is the mass of the earth and R is the radius of the Earth.
Earth rotates about its own axis as a result of which the objects on the earth experience the centrifugal force.
This centrifugal force decreases the acceleration due to gravity and magnitude of change in the value of g due to rotation of the earth depends on the value of the angle of latitude.
Acceleration due to gravity near the surface of the earth is given by,
where,
h is the distance of the place from the surface of the earth.
The effect of rotation is maximum at the equator of the Earth.
Yes, if the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).
Law of gravitation states that every body in this universe attracts the other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
i.e.,
∴
where G is constant of proportionality known as gravitational constant.
Force of gravitation,
Let,
Mass,
and
When the object falls on the ground, the earth moves up. Why the earth’s motion is not noticeable?
OR
If the force of gravity between the earth and the body is same and mutual, then does the body fall towards the earth and the earth does not move towards body?
Gravitational Force,
Here, we have
Mass of the body,
Radius, r = 1 cm
∴ Gravitational force,
A rocket is fired from the earth towards the sun. At what distance from the earth’s
centre is the gravitational force on the rocket zero ? Mass of the sun = 2× 1030 kg,
mass of the earth = 6× 1024 kg. Neglect the effect of other planets etc. (orbital radius
= 1.5 × 1011 m).
Given,
Mass of the sun, M = 2× 1030 kg
Mass of the earth, m = 6.0 × 1024 kg
Distance from the sun and the earth is,
Let the force on the rocket be zero at a distance x from the earth.
∴
where, is a mass of the rocket.
Therefore,
= 578.35
Therefore,
Gravitational force is zero on the rocket at a distance of,
, from the Earth's surface.
Given,
Mass of neutron,
Distance between the centres of two neutrons,
The gravitational force between two neutrons is,
=
Inertial mass: It is the ratio of the force applied on a body to acceleration produced in it.
Gravitational mass: It is the ratio of the gravitational force acting on a body to the acceleration due to gravity at that point.
Characteristics of inertial mass are:
(i) Inertial mass is proportional to the quantity of matter contained in a body.
(ii) It is independent of the shape of the body.
(iii) It is additive.
(iv) It is a scalar quantity.
(v) It is independent of the temperature of the body.
(vi) It is independent of the state of the body.
(vii) When two inertial masses combine physically or chemically, total inertial mass remains conserved.
According to Newton’s law of gravitation, every mass exerts a gravitational force on every other mass in the Universe.
Thus, the earth exerts a force on every mass in its surroundings. This force of attraction exerted by the earth on a body is called the force of gravity and the acceleration possessed by a body moving under the force of gravity is called acceleration due to gravity.
The value of acceleration due to gravity does not depend on the mass of a body but it depends on the mass of the earth and distance of the body from the center of the earth.
Consider, the earth to be a uniform sphere of mass M and radius R.
Let a body be placed at a distance r from the surface of the earth.
The gravitational on the body is,
The acceleration produced in the body due to force of gravity i.e. acceleration due to gravity is,
At the surface of the earth, the acceleration due to gravity is,
Acceleration due to gravity is given by,
Since Cavendish was the first scientist who measured the value of G, hence he is said to have been the first person to weigh the earth.
Given,
Mass, m = 200 kg
Weight of the body, W = 500N
Therefore,
Acceleration due to gravity,
Acceleration due to gravity,
Here,
Mass,
Radius,
Gravitational constant,
Acceleration due to gravity,
Given that,
Acceleration due to gravity,
∴
Here, we have
Acceleration,
Radius,
∴ Mass of the moon,
The value of 'g' is different at different points while G is a universal constant and has the same value everywhere in the universe.
The unit of g is m/s2.
The dimensional formula is [M0L1T2].
SI unit of G is N m2/kg2 and [M-1L3T-2] respectively.
The relation between g and G,
Acceleration due to gravity changes with the angle of latitude due to shape and rotation of the earth about its own axis.
Variation of g due to shape:
Earth is not a perfect sphere and is flat at the pole, bulges out at the equator. Therefore, the polar radius is smaller than the equatorial radius.
Acceleration due to gravity is inversely proportional to the square of the radius of the earth, therefore the value of g increases as one moves from equator to pole.
Variation of g due to rotation:
Earth rotates about its own axis as a result of which the objects on the earth experience the centrifugal force.
This centrifugal force decreases the acceleration due to gravity and magnitude of change in the value of g due to rotation of the earth depends on the value of the angle of latitude.
Consider the earth to be a homogenous sphere of mass M and radius R.
The earth rotates about the polar axis.
Let, ω be the angular velocity of rotation of the earth.
All the objects at rest on earth also revolve about its polar axis with same angular velocity ω.
Consider a body of mass M placed at a point P on the earth at an angle of latitude λ.
If the earth were at rest, then the body would have been attracted towards the center of the earth with force mg.
Due to rotation of the earth, the body describes a circle of radius OP = r = with center at O.
In rotating frame of the earth, body experiences pseudo force = along
Now the apparent weight of the body is resultant of force mg directed towards the center and force along
Applying parallelogram of vector addition, we get
As therefore
can be neglected.
Therefore,
[using binomial exp]
Substituting we get
is the g.
The acceleration due to gravity on the surface of the earth at angle of latitude is,
(i) The value of g is maximum at points where
Thus, acceleration due to gravity is maximum at poles.
∴
(ii) The value of g is minimum at points where
Thus, acceleration due to gravity is minimum at the equator.
i.e.,
The difference in the value of g at pole and equator is,
The weight of a body at depth d is equal to the weight of the body at height h above the surface of the earth. What is the ratio of h to d, for small value of h and d?
OR
Prove that the distance we have to cover into the earth below its surface is twice the distance we have to cover above the surface of the earth to get the same change in the weight of a body?
Acceleration due to gravity at a height from the Earth's surface is given by,
Since,
Therefore,
On solving we get,
, is the height at which the acceleration due to gravity is three-fourth of that at the surface of the earth.
The apparent weight of man in a lift falling with acceleration a is given by,
Wa = m(g-a)
The acceleration of free-falling lift is g.
∴ Wa = m(g-g) = 0
Hence a man feels weightlessness.
Weight of the body at any point, W = mg
where,
g is the acceleration due to gravity at that point.
The acceleration due to gravity at a point inside the earth is,
Therefore,
Weight,
At the center of the earth, d = R.
So,
Thus, the body is weightless at the centre of the earth.
If the earth stops to revolve about it's own axis then the acceleration will increase by .
Change in acceleration at equator is,
The gravitational pull of the earth on one of the mass is, F1=mg
The gravitational attraction of other mass on first mass is,
Now, the ratio of both masses is given by,
That is, the gravitational pull on one mass is 1.496 x 1010 times larger than the gravitational attraction of other mass.
We know that acceleration due to gravity at equator is,
where, is the angular velocity of the earth.
Angular velocity of Earth=
If the acceleration due to gravity at the equator becomes zero, the body at equator experiences weightlessness.
Let be the angular velocity of rotation at which the acceleration due to gravity at the equator becomes zero.
That is,
Now,
Thus, the earth should rotate 17 times faster than the present rate.
Rotation speed increases by a factor of 17.
Therefore, the duration of day decreases by factor 17.
New duration of day,
Gravitational field strength =
where,
M = mass of Earth, and
R = radius of Earth
We cannot shield a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the presence of other matter, whereas it is not so in the case of electric forces.
It means the gravitational screens are not possible.
At the surface of a planet, gravitational field strength is equal to acceleration due to gravity.
At the surface,
I =
That is, I = 1.2N/kg.
Gravitational P.E stored in mass m placed in the field of mass M at a distance r =
The space around a material body in which it's gravitational pull can be experienced is called it's gravitational field.
The gravitational field strength at any point in the gravitational field is the force experienced by unit mass placed at that point.
Gravitational field strength is a vector quantity and is directed towards the mass causing the field.
At a distance R, place the test mass m at point P to find the gravitational field strength.
Force experienced by the test mass m is,
, towards the mass M.
The gravitational field strength due to mass M at observation point P is,
, towards the mass M.
Here,
Mass of the sun,
The distance of the earth from the sun is,
Therefore, gravitational field strength due to the sun is,
Given,
Distance between the two bodies = 20 m
Let and
be the gravitational field strength due to masses 25 kg and 75 kg respectively.
Using the formula, we have
Since are in opposite direction, therefore net field strength at given point is,
Gravitational potential:
Gravitational potential at any point is defined as the amount of work done in moving a unit mass from infinity to that point.
Dimensional formula: [M0L2T2].
SI unit: J/ kg
CGS unit: erg/g
Gravitational potential energy:
Gravitational potential energy is the amount of work done against the gravitational force to bring the body from infinity to that point.
Dimensional formula: [M1L2T2].
SI unit: J
CGS unit: erg.
Given here,
Mass of the earth,
Radius of earth,
Height,
Gravitational potential at the site of geostationary satellite is,
V is the required potential due to Earth;s gravity at the site of this satellite.
Derive an expression for the potential energy stored in the mass m placed in the gravitational field of mass M.
Given,
Mass of first body, m1 = 25 kg
Mass of second body, m2 = 75 kg
r1 = 5 m and r2 = 15 m
Let be the gravitational potential due to masses 25 kg and 75 kg respectively.
Using the formula for gravitational potential,
and
The total potential at the given point is,, is the gravitational potential at a point at a distance of 5m from the 25 kg mass.
A rocket is fired vertically with a speed of 5 km/s from the earth’s surface. How far from the earth’s surface does the rocket go before returning to the earth? Mass of the earth = 6.0 x 1024 kg; mean radius of the earth = 6.4 x 106 m; G =6.67 x 10-11 N m2 kg-2.
Acceleration due to gravity at the surface of the earth = 9.8 m/s2.
The gravitational potential due to some distribution of mass is given by Find the gravitational field due to that distribution of mass.
Gravitational field is given by,
Gravitational P.E is given by,
We know that field the negative gradient of gravitational potential energy.
∴
This is the required expression for gravitational field due to the distribution of mass.
A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant:
(a) linear speed
(b) angular speed
(c) angular momentum
(d) kinetic energy
(e) potential energy
(f) total energy throughout its orbit?
Neglect any mass loss of the comet when it comes very close to the sun.
The torque exerted by the sun on the comet is zero.
Therefore, angular momentum of comet remains constant.
i.e., mvr = constant vr = constant [because m is constant].
In an elliptical orbit distance of the comet from the sun changes, therefore, the speed of the comet also changes.
(a) Linear speed of the comet is not constant.
(b) Angular speed ω of the comet is not constant.
(c) Angular momentum of the comet is constant.
(d) Kinetic energy of comet 1/2 mv2 changes because the speed of the comet changes.
(e) Potential energy of the comet is
(f) Total energy of the comet remains constant.
(b)
(i) Gravitational forces
(ii) Electromagnetic forces
(iii) Nuclear forces
A non-homogeneous sphere of radius R has the following density relation:
What is the total mass of sphere and gravitational field at the surface of sphere?
Let be the velocities of two stars when they collide.
According to the law of conservation of momentum,
According to the law of conservation of energy,
We have
Mass of the star,
Distance between the stars,
Radius of star,
The distance between two stars when they collide is,
2r = 2 x 107m
∴ Speed with which the stars collide is given by,
The gravitational field due to two masses are equal and opposite, therefore net gravitational field at P is zero.
The gravitational potential at P is,
The net gravitational field at P is zero, therefore the object placed at P will be in equilibrium.
The equilibrium is an unstable equilibrium.
Escape velocity is independent of the mass of the body which is projected from the surface of the earth.
Define orbital velocity.
Orbital velocity of a satellite revolving in a circular orbit at a height h from the surface of the planet is,
The total energy of satellite =
where
m is the mass of satellite,
M is the mass of the earth and
r is the radius of orbit.
Escape velocity is the minimum velocity with which a body must be projected vertically upward so that it may just escape the surface of the Earth.
Expression for escape velocity:
Let a body of mass m be escaped from the gravitational field of the earth.
During the course of motion, let at any instant, body be at a distance r from the centre of the earth.
The gravitational force between body and the earth is,
Work done to raise the body by distance dr is,
Total work done, W in raising the body from the surface of the earth to infinity is,
If we throw the body upward with a velocity ve, then work done to raise the body from surface of the earth to infinity is done by kinetic energy.
Therefore,
... (1)
Substituting the values in equation (1), we have
g = 9.81m/sec and R = 6.4 x 106m
Escape velocity is given by,
Does the escape speed of a body from the earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Let v∞ be the velocity of body at infinity.
By using law of conservation of energy, we have
Here, we have
Velocity with which the body is projected,
Escape velocity,
∴ Velocity of body at infinity,
Two bodies of masses m1 and m2 (m1 > m2) are to be projected out of the field of the earth.
(i) Which of the two must be thrown with greater velocity?
(ii) Which of the two must be thrown with greater momentum?
(iii) Which of the two must be thrown with greater kinetic energy?
(i) Escape velocity is independent of the mass of body projected and is same for all the bodies.
i.e., 11.2 km/s, therefore, both must be projected with equal velocity.
(ii) Momentum of body of mass m1 is,
Momentum of the body of mass is,
Since
∴
That is, a heavy body must be thrown with greater momentum.
(iii) Kinetic energy of mass is,
Kinetic energy of mass is,
Since m1 > m2, therefore K1 > K2.
That is, i.e., heavy body must be thrown with greater kinetic energy.
Expression for orbital velocity:
Consider, a satellite of mass 'm' is revolving around the earth in a circular orbit of radius 'r', at a height 'h' from the surface of the earth.
Let 'M' be the mass of the earth and 'R' be the radius of the earth.
Therefore,
r = R + h
The centripetal force that is required to revolve the satellite =
where,
is orbital velocity.
Orbital velocity is produced by the gravitational force between the earth and the satellite =
Therefore,
But,
GM=gR2
where,
g is the acceleration due to gravity.
So,
Let g' be the acceleration due to gravity in the orbit i.e. at a height h from the surface.
Therefore,
∴ , is the required orbital velocity.
First, the rocket should be fired with such a velocity that it attains the height at which it is required to revolve.
Orbital velocity rovided to the rocket = , in horizontal direction to put into orbit.
The orbit of the satellite should be above the earth's atomosphere.
2. Second law: Ariel velocity of the planet is constant.
3. Third law: Square of the time period of one complete revolution is directly proportional to the cube of the semi-major axis of the elliptical orbit.
Kepler’s third law states that the square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the elliptic orbit.
If orbit of the planet is circular, then square of the time period will be proportional to the cube of radius of the orbit.
Let a planet revolve around the sun in a circular orbit of radius r.
The necessary centripetal force required is provided by the gravitational pull,
i.e.,
i.e.,
When a satellite suddenly jumps to higher orbit then how does the following change:
i) angular velocity,
ii) linear momentum,
iii) kinetic energy, and
iv) potential energy?
We have, ... (1)
So, when a satellite suddenly jumps to the higher orbit, r increases.
Therefore, from equation (1), T increases.
i) Angular velocity is given by,
Therefore, as T increases, angular velocity decreases.
ii) We know,
Orbital velocity,
Therefore, linear momentum of satellite is,
Since r increases, therefore, linear momentum decreases.
iii) The kinetic energy of satellite is,
The kinetic energy of satellite decreases with increase in r.
iv) The potential energy of satellite is,
The potential energy of satellite increases with increase in r.
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to the satellite to rocket out of the earth’s gravitational influence?
Mass of the satellite = 200 kg, Mass of the earth = 6.0 x 1024 kg, Radius of the earth = 6.4 x l06 m; G = 6.67 x 10-11 Nm2/kg2.
A satellite orbits the earth at a height of 500 km from its surface. Compute its:
(a) kinetic energy
(b) potential energy
(c) total energy.
Mass of the satellite = 300 kg,
Mass of the earth = 60 x 1024kg,
Radius of the earth = 6.4 x 106m
G = 6.67 x 10-11 Nm2/kg2.
Will your answers alter if the earth shrinks suddenly to half of its present size?
We have,
Mass of the satellit, m = 300 kg
Mass of the Earth, M= 6.0 x 1024kg
Radius of Earth, R = 6.4 x 106m
Gravitational constant, G = 6.67 x 10-11 Nm2/kg2
(a) Kinetic energy of satellite is,
(b) Potential energy of satellite is,
(c)Total energy of satellite is,
E = K.E. + P.E.
= 8.7 x 109 -17.4 x 109
= -8.7 x 109J
(d) If the earth shrinks to half its size without changing its mass, the answers will remain the same.
A satellite which appears to be stationary to an observer standing on the earth is known as a geostationary satellite.
The conditions for satellite to appear stationary are:
(i) The time-period should be 24 hours.
(ii) Its orbit should be in the equatorial plane of the earth.
(iii) Its direction of motion should be the same as that of the earth about its polar axis.
Uses of satellites are:
(i) They are used for telecommunication purposes,
(ii) They are used to weather forecasting,
(iii) They are used to study the shape of the earth,
(iv) They are used in military and country’s safety purposes,
(v) They are used in studying the environmental conditions of the earth.
We have,
Mass, m1 = 40 kg ; m2 = 60 kg
Also, F = 4 mgf = 4 x 10-6 x 10
= 4 x 10-5 N
G = 6 x 10-11 N m2 kg-2
Distance between the masses, r = ?
Gravitational force is given by, F =
This implies,
r =
Let, the acceleration due to gravity at height h be the same as that at a depth, d, deep into the earth.
We have, d = 100 km
Therefore,
g' = g
Gravitational field is given by,
... (1)
The mass lies at the periphery of the shell. And at the centre mass is equal to 0.
Therefore, according to equation (1),
Gravitational field at a point inside the shell = 0
The value of escape velocity does not depend on the mass of the body and angle of projection from the surface of the earth. Therefore, both the objects will escape from the surface at the same velocity.
Gravitational P.E is given by,
U =
Negative sign implies that the P.E. is due to the attractive gravitational force exerted by the Earth on the body. In order to bring a body from infinity to a particular point, work has been performed by the gravitational field of the earth. That is, there is an attractive force.
Given that the force equal to the body's weight is applied in the horizontal direction. Therefore, the body will not accelearte.
A body is projected with velocity less than escape velocity. What is the sign of total energy at:
(i) surface of the earth,
(ii) the highest point?
(i) Total energy at the surface of the Earth is negative.
(ii) Total energy at the highest point is negative.
Energy is negative because it is due to the attractive gravitational potential force exerted by earth on the body.
Why does the moon has no atmosphere?
Weight of the body = mg = 64 N
So, mass of the body, m = 64 / g = 64/ 10 = 6.4 kg
At height h, the value of g' is given by,
If the Earth stops rotating about its axis, by what value the acceleration due to gravity at equator will change?
The value of g will increase by R
where,
R is the radius of the earth, and is the angular velocity of the rotation of the Earth.
How much below the surface does the acceleration due to gravity become 70% of it's value on the surface of the earth. Radius of the Earth = 6.4 x 106 m.
Here,
g'/g = 70 / 100 = 7/10 ; d = ?
Now, g' = g(1-d/R);
Therefore,
The value of acceleration due to gravity varies with altitude, depth, shape and rotation of earth about it's axis.
i) On the surface of the earth,
g = ... (1)
ii) At a height h above the surface of the earth,
g' = ... (2)
iii) At a height below the surface of the earth,
g ' = ... (3)
From (1), (2) and (3), we can say that the value of 'g' decreases with height and is maximum on the surface of the earth.
That is why, the value of 'g' is lesser at mountains than in plains.
The mass of the body remains the same. But, the weight of the body changes.
At the centre of the earth, acceleration due to gravity is zero.
Weight, W = mg
Therefore,
Weight of the body = 0
All bodies have a definite mass.
Acceleration due to gravity on the moon is one-sixth of acceleration due to gravity on earth.
Therefore, the body will possess some weight, as weight is the product of mass and acceleration due to gravity.
The value of g decreases, due to rotation of the Earth.
S. No. | Orbital Velocity | Escape velocity |
1. | The minimum velocity required to put the satellite into the Earth's orbit is called orbital velocity. | The minimum velocity required to throw a body out of the gravitational field of earth is called escape velocity. |
2. | Orbital velocity of a satellite at height, ![]() |
Escape velocity of a satellite near the Earth, ve = ![]() |
3. | When satellite is orbiting close to earth, vo = ![]() |
Escape velocity f a satellite near the earth, ve = ![]() |
If the diameter of the earth becomes three times its present value weight of the body on the surface of the earth decreases to one-ninth.
Two bodies of masses 100 kg and 10,000 kg are at a distance 1 m apart. At which point on the line joining them will be the resultant gravitational field intensity zero?
Let, x be the distance of the point P from 100 kg body where the resultant gravitational intensity is zero.
Therefore, gravitational intensity at P due to body of mass 100 kg is equal and opposite to that due to body of mass 10, 000 kg.
Hence,
At this point, the gravitational field intensity will be 0.
Gravitational Binding Energy is the minimum energy that is required to free a body from the gravitational attraction. Binding Energy is the negative of the total energy.
Mass of the satellite = 300 kg
Height = 500 km, from the surface of the Earth
Here, we have
r = R + h
= 6.4 x 106 + 500 x 103
= 7.0 x 106 m
m = 300 kg;
M = 6 x 1024 kg;
G = 6.67 x 10-11 N m2 kg-2
i) K.E of satellite =
We have,
h = 1600 km
R = 6400 km = 6.4 x 106 m
Therefore,
r = R + h = 6400 +1600
= 8000 m = 8.0 x 106 m
Hence, escape velocity is given by,
Escape velocity is given by,
The mass and size of different planets is different. Therefore, escape velocity is different for different planets.
The relation for orbital velocity , v =
where,
h is the height of the satellite above the Earth's surface.
It is evident from the above formula that, smaller the value of h, greater is the value of v and vice versa.
Therefore, the satellite which is revolving close to the earth has larger speed.
Work done inorder to take the body from the surface of the earth (x-R) to aheight equal to R (x = 2R) is given by,
We have,
Tm is the time period of revolution of mars.
Orbital velocity is given by,
In which of the following examples of motion, can the body be considered approximately a point object:
(b) a monkey sitting on top of a man cycling smoothly on a circular track
In which of the following examples of motion, can the body be considered approximately a point object:
c) a spinning cricket ball that turns sharply on hitting the ground.
In which of the following examples of motion, can the body be considered approximately a point object:
(d) a tumbling beaker that has slipped off the edge of a table.
In case of train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Since the train is moving with uniform velocity,
Acceleration, aI = 0
Now, using second equation of motion,
Distance covered by train A is given by,
For train B,
Initial velocity, uB = 72 km/h = 20 m/s
Acceleration, a = 1 m/s2
Time, t = 50 s
Now, using second equation of motion,
Distance travelled by second train B is,
Length of both trains = 2
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.
Let V be the speed of the bus running between towns A and B.
Given,
Speed of the cyclist = 20 km/hr
Relative speed of bus moving in the direction of the cyclist = V - v = (V-20) km/hr
Every 18 mins, the bus went past the cyclist. moving in the direction of the bus.
That is,
Distance covered by the bus =
One bus leaves every t minutes.
Therefore,
Distance travelled by the bus = V ... (ii)
Equations (i) and (ii) are equal.
Relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h
Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h
Therefore,
Now, from equations (3) and (4), we get
Acceleration due to gravity always acts in the downward direction, irrespective of the direction of motion.
At t= 0.3 sec,
Negative, Negative, Positive.
At t= 1.2 sec,
Positive, Positive, Negative
At t= -1.2 sec
Negative, Positive, Positive
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by,
a = – ω2x ; ω → angular frequency … (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the position-time graph will also be negative. Therefore, both position and velocity are negative.
However, using equation (i), acceleration of the particle will be positive.
t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x-t graph will also be positive. Therefore, both position and velocity are positive.
However, using equation (i), acceleration of the particle comes to be negative.
t = – 1.2 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive.
From equation (i), it can be inferred that the acceleration of the particle will be positive.
n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Sn | 0.5 | 1.5 | 2.5 | 3.5 | 4.5 | 5.5 | 6.5 | 7.5 | 8.5 | 9.5 |
Does the escape speed of a body from the earth depend on:
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?
Choose the correct answer from among the given ones:
For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)
D.
Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R
Orbit velocity of a satellite
therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.
A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms−2 )
2s
C.
A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support.
Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 )
C.
The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.
Consider the stones thrown up simultaneously as shown in the diagram below.
Considering the motion of the second particle with respect to the first we have relative acceleration
|a21| = |a2-a1| = g-g = 0
Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
Thus, distance covered by the second particle with respect to the first particle in 8s is
S12 = (v21) t = (40-10)(8s)
= 30 x 8 = 240m
Similarly, time taken by the second particle to strike the ground is given by
Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.
From the tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is
2gH = n2u2
gH = (n-2)2u2
2gH = nu2(n-2)2u2
gH = (n-2)2u2
C.
2gH = nu2(n-2)2u2
Time is taken to reach the maximum height t1 = u/g
If t2 is the time taken to hit the ground,
i.e,
What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?
5GmM/6R
2GmM/3R
GmM/2R
GmM/3R
A.
5GmM/6R
From conservation of energy,
Total energy at the planet = Total energy at the altitude
In its orbit the necessary centripetal force provided by gravitational force.
∴
The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g'and 'R'(radius of earth) are 10 m/s2 and 6400km respectively. The required energy for this work will be;
6.4 x1011 J
6.4 x108 J
6.4 x109 J
6.4 x109 J
D.
6.4 x109 J
Potential energy on the earth's surface is -mgR while in free space, it is zero. So, to free the spaceship minimum required energy is
E =mgR
= 103 x 10 x 6400 x 103 J
= 6.4 x 1010 J
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)
B.
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to
x2
ex
x
logex
A.
x2
In this problem acceleration (a) is given in terms of displacement (x) to determine the velocity with respect to position or displacement we have to apply integration method.
From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant.
Let for any displacement from 0 to x , the velocity changes from vo to v
A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed ν0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?
yes, 60°
yes, 30°
no
yes, 45°
A.
yes, 60°
Man will catch the ball if the horizontal component of velocity becomes equal to the constant speed of man i.e.
At what height from the surface of earth the gravitation potential and the value of g are -5.4 x 107 J kg-2 and 6.0 m/s2 respectively? (Take the radius of the earth as 6400 km)
1600 km
1400 km
2000 km
2600 km
D.
2600 km
Gravitational potential at some height h from the surface of the earth is given by,
V = ... (i)
Acceleration due to gravity at some height h from the earth surface can be given as, ... (ii)
From equation (i) and (ii), we get ... (iii)
V = -54 x 107 J kg-2
g' = 6.0 m/s2
Radius of Earth, R = 6400 km
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,
90o
45o
180o
0o
A.
90o
There are two vectors P and Q.
It is given that,
Let, angle between P and Q is ,
Two identical charged spheres suspended froma common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as
vx
vx-1
B.
Two identical charged spheres suspended from a common point by two massless strings of length L.
In ,
The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.
Therefore, equation (i) can be rewritten as,
A ship A is moving Westwards with a speed of 10 kmh-1 and a ship B 100km south of A, is moving Northwards with a speed of 10 kmh-1. The time after which the distance between them becomes shortest is
0 h
5 h
B.
5 h
It is clear from the diagram that the shortest distance between the ship A and B is PQ.
Two spherical bodies of masses M and 5M and radii R and 2R 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 before collision is
2.5 R
4.5 R
7.5 R
1.5 R
C.
7.5 R
Suppose, the smaller body cover a distance x before collision, then
Mx = 5M (9R-x)
Or,
x= 45R- 5x
Or,
x = 45 R\6 = 7.5 R
Which one of the following plots represents the variation of the gravitational field on a particle with distance r due to a thin spherical shell of radius R? (r is measured from the centre of the spherical shell).
B.
Inside the shell, E =0
and outside the shell,
Hence, graph (b) shows the exact result.
The force F acting on a particle of mass m is indicated by the force-time graph as shown below. The climate change in momentum of the particle over the time interval from zero to 8 s is,
24 Ns
20 Ns
12 Ns
6 Ns
C.
12 Ns
The area under F-t graph gives change in momentum.
For 0 to 2 s,
For 2 to 4 sec,
For 4 to 8 sec,
So, total change in momentum for 0 to 8 sec is,
A stone falls freely under gravity. It covers distances h1 , h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relations between h1, h2 and h3 is,
h1 = 2 h2 = 3 h3
h1 =
h2 = 3 h1 and h3 = 3 h2
h1 = h2 = h3
B.
h1 =
Relations between h1, h2 and h3 is,
Three blocks will masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)
zero
2 mg
3 mg
6 mg
A.
zero
Since all the blocks are moving with constant velocity, then the net force on all blocks will be zero.
A uniform force of (3 i + j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2 i + k) m to position ( 4 i + 3 j - k) m. The work done by the force on the particle is,
9 J
6 J
13 J
15 J
A.
9 J
Given,
Force, F = 3 i + j
r1 = ( 2 i + k) m
r2 = (4 i + 3 j - k) m
Therefore,
S = r2 - r1
Therefore,
W = F. s
= (3i + j). (2 i +3j - 2k)
= 3 x2 + 3x0
= 6 + 3 = 9 J
Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1m, 2 m, 4 m, 8 m, respectively from the origin. The resulting gravitational potential due to this system at the origin will be
-G
-4G
D.
-4G
The resulting gravitational potential is,
A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest
at the instant just before the body hits the earth
it remains constant all through
at the instant just after the body is projected
at the highest position of the body
A.
at the instant just before the body hits the earth
We know
P = F. v = Fv cos θ
so just before hitting θ is zero and both F and v are maximum
A boy standing at the t op of a tower of 20 m height drops a stone. Assuming g = 10 ms-2, the velocity with which it hits the ground is
20 m/s
40 m/s
5 m/s
10 m/s
A.
20 m/s
Given, g = 10 ms-2 and h = 20 m
We have
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Then,
the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
the total mechanical energy of S varies periodically with time
the linear momentum of S remains constant in magnitude
the acceleration of S is always directed towards the centre of the earth
D.
the acceleration of S is always directed towards the centre of the earth
As we know that, the force on satellite is an only gravitational force which will always be towards the centre of the earth. Thus, the acceleration is S is always directed towards the centre of the earth
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30o, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be respectively
0.6 and 0.6
06 and 0.5
0.5 and 0.6
0.4 and 0.3
B.
06 and 0.5
Given a plank with a box on it one end is gradually raised about the end having angle of inclination is 300, the box starts to slip and slides down 4 m the plank in 4 s as shown in a figure.
The coefficient of static friction.
So, distance covered by a plank,
Thus, coefficient of kinetic friction between the box and the plank is 0.51
A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25 x 106 m above the surface of the earth. if earth 's radius is 6.38 x 106 m and g = 9.8 ms-2 then the orbital speed of the satellite is
7.76 kms-1
8.56 kms-1
9.13 kms-1
6.67 kms-1
A.
7.76 kms-1
Given, height of a satellite
h = 0.25 x 106 m
Earth's radius, Re = 6.38 x 106m
For the satellite revolving around the earth, orbital velocity of the satellite
A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is
v1v2 / v1 + v2
2v1v2/v1 + v2
2v12v22 / v12 + v22
v1 + v2 / 2
B.
2v1v2/v1 + v2
Velocity v = s / t
s = vt
The average speed of particle
A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box - will move relative to belt before coming to rest on it taking
g= 10 ms-2 is
1.2 m
0.6 m
zero
0.4 m
D.
0.4 m
The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2(R2> R1) is
D.
The kinetic energy changing the orbit of satellite
(1) Centre a gravity (CG) of a body is the point at which the weight of the body acts.
(2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
(3) To evaluate the gravitational field intensity due to anybody can be considered to be concentrated at its CG.
(4) The radius of gyration of anybody rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.
Which one of the following pairs of statements is correct?
(4) and (1)
(1) and (2)
(2) and (3)
(3) and (4)
A.
(4) and (1)
Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.
A body mass 1 kg is thrown upwards with velocity 20 ms-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 ms-2)
20 J
30 J
40 J
10 J
A.
20 J
The energy lost due to air friction is equal to the difference of initial kinetic energy and final potential energy.
Initially, body possess only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy = KE - PE
=
The figure shows elliptical orbit of a planet m about the sun S. The shaded are SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then,
t1 > t2
t1 =4 t2
t1 = 2t2
t1 = t2
C.
t1 = 2t2
Apply Kepler's law of area fo planetary motion.
The line joining the sun to the planet sweeps out equal areas time interval ie, areal velocity is constant.
A roller coaster is designed such that riders experience " weightlessness" as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between
14 m/s and 15 m/s
15m/s and 16 m/s
16 m/s and 17 m/s
13m/s and 17 m/s
A.
14 m/s and 15 m/s
Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at a temperature to C, the power received by a unit surface (normal to the incident rays) at a distance R from the centre of the sun is :
D.
From Stefan's law, the rate at which energy is radiated by sun at its surface si
P = σ x 4 πr2T4
Two satellites of earth S1 and S2, are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statement is true?
The time period of S1 is four times that of S2
The potential energies of earth and satellite in the two cases are equal
S1 and S2 are moving with the same speed
The kinetic energies of the two satellites are equal
C.
S1 and S2 are moving with the same speed
In same orbit, orbtial speed of satellites remains same. When two satellites of earth are moving in same orbit, then time period of both are equal.
From Kepler's third law
Time period is independent of mass, hence their t ime periods will be equal.
The potential energy and kinetic energy are mass dependent, hence the PE and KE of satellites are not equal.
But, if they are orbiting in a same orbit, then they have equal orbital speed.
A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation , where t is in s. Work done by the force in 2 s is:
C.
If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
Work done by the force = Force x Displacement
or W = F x s ...(i)
But from Newton's 2nd law, we have
Force = Mass x Acceleration
i.e., F = ma ...(ii)
Hence, from Eqs. (i) and (ii), we get
Hence, Eq (ii) becomes
We have given
A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by
How long would the particle travel before coming to rest?
24m
40m
56m
16m
C.
56m
Velocity is rate of change of distance or displacement.
Distance travelled by the particle is
We know that, velocity is rate of change of distance i.e.,
but final velocity v = 0
Hence, distance travelled by the particle before coming to rest is given by
The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fve, where ve is its escape velocity from the surface of the earth. The value of f is
B.
If energy in the form of kinetic energy which is equal to binding energy, is supplied to the sphere, it leaves the gravitational field of earth.
At a platform at a height h,
escape energy = binding energy of sphere
or
But at surface of earth,
As given,
Hence,
Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is
5/4
12/5
5/12
4/5
D.
4/5
As bodies are dropped from a certain height, their initial velocities are zero i.e., u = 0.
For free fall from a height u = 0 (initial velocity).
From second equation of motion
or
The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is
0
5 m/s2
-4 m/s2
–8 m/s2
C.
-4 m/s2
x = 5t – 2t2 y = 10t
Acceleration of particle at t = 2 s is = –4 m/s2
What do you mean by mean free path and write its formula.
In kinetic theory the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles. The formula still holds for a particle with a high velocity relative to the velocities of an ensemble of identical particles with random locations.
A police jeep is chasing with velocity of 45 km/h. A thief in another jeep moving with velocity 153 km/h. Police fires a bullet with muzzle velocity of 180 m/s. The velocity it will strike the car of the thief is
150 m/s
27 m/s
450 m/s
250 m/s
A.
150 m/s
Relative velocity is the vector difference between the velocities of two bodies, the velocity of a body with respect to another regarded as being at rest.
Here, we calculate the relative velocity of the jeep:
153 -45 =108 km/h
Therefore, striking speed = relative speed of a bullet with respect to the thief's car:
180 - 30 = 150 m/s
Two similar charges having mass m and 2m are placed in an electric field. The ratio of their kinetic energy is :
4:1
1:1
2:1
1:2
C.
2:1
Force on charged particle is electric field F = q E
A car starts from rest, moves with an acceleration a and then decelerates at a constant rate b for sometimes to come to rest. If the total time taken is t. The maximum velocity of car is given by :
A.
Let car accelerates for time t1 and decelerates for time t2 then
t1 + t2 = t
From ν = u + at
v = u + at1
at1 = bt2
A river is flowing from west to east with a speed of 5 m/min. A man can swim in still water with a velocity 10 m/min. In which direction should the man swim so, as to take the shortest possible path to go to the south?
30o east of south
60o east of south
60o west of south
30o west of north
A.
30o east of south
Let the swimmer swims at an angle θ with the vertical.
The swimmer should swim 30o east of south to take the shortest possible path to go to the south.
Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1 The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is
B.
Distance covered by boy A in time t:- AC= vt
Distance covered by boy B in time t
BC = ν1 t
Using Pythagoras theorem
Ac2 = AB2 + BC2
⇒ ( νt )2 = a2 +( ν1t )2
⇒ a2 = ν2 t2 - (ν1t)2
A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one-half of its maximum height. How high does the ball rise? ( Take, g = 10 m/s2 )
15 m
10 m
20 m
5 m
B.
10 m
The problem can be solved using third equation of motion at A and O'.
Let maximum height attained by the ball be H. Third equation of motion gives
v2 = u2 - 2gh
v2 - final velocity ; u2 = initial velocity
At A,
⇒ u2 = 100 + 10 H ....(i)
At 'O'
( 0)2 = u2 - 2 × 10 × H ....(ii)
Thus, from Eqs. (i) and (ii), we get
20 H = 100 + 10 H
⇒ 10 H = 100
∴ H = 10 m
A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd The average speed for this round trip is
A.
Average velocity is defined as the change in position or displacement divided by the time intervals ( Δt ) in which displacement occurs.
Let t1 and t2 times taken by the car to go from X to Y and then from Y to X respectively.
Total distance travelled = XY + XY
= 2XY
Therefore, average speed of the car for this round trip is
A particle moves along a straight line OX. At a time t (in second) the distance x ( in metre ) of the particle from O is given by x = 40 + 12 t - t3 . How long would the particle travel before coming to rest?
24 m
40 m
56 m
16 m
C.
56 m
Speed is rate of change of distance.
Distance travelled by the particle is
x = 40 + 12 t - t3
We know the speed is rate of change of distance
i.e
= 0 + 12 - 3t2
but final velocity ν = 0
∴ 12 - 3t2 = 0
⇒
⇒ t = 2s
Hence, distance travelled by the particle before coming to rest is given by
x = 40 + 12(2)-(2)3
x = 64 - 8
x = 56 m
A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc?
0.4 cm
2.4 cm
1.8 cm
1.2 cm
A.
0.4 cm
The situation can be shown as: Let radius of complete disc is a and that of small disc is b. Also let centre of mass now shifts to 02 at a distance X2 from original centre
The position of new centre of mass is given by
Here, a = 6 cm, b = 2 cm, x1 = 3.2 cm
⇒ XCM = 0.4 cm
The displacement of particle is given by .What is its acceleration?
Zero
B.
Acceleration is the rate of change of velocity and velocity is the rate of change of displacement.
The displacement equation is given by
Veiocity = rate of change of displacement
Acceleration = rate of change of velocity
The area of the acceleration-displacement curve of a body gives
impulse
change in momentum per unit mass
change in KE per unit mass
total change in energy
C.
change in KE per unit mass
Area of the acceleration-displacement curve gives change in KE per unit mass.
Velocity-time curve for a body projected vertically upwards is
parabola
ellipse
hyperbola
straight line
D.
straight line
When the ball goes upwards the direction of velocity of the ball and the direction of acceleration of gravity are in the opposite directions of acceleration of gravity are in the opposite directions so the velocity of the ball decrease until it reaches the top and stops momentarily.
v = u - gt ... (i)
Then from the peak position, the ball will start to descend downwards due to the acceleration due to gravity that is g.
v = -gt ....(ii)
( negative sign is due to motion in downward direction)
From both equations, the velocity and time are linearly proportional to each other which is the same as the equation of a straight line. So this graph will be a straight line starting from initial velocity (u) on the y-axis and slope g with time as the x-axis.
The numerical ratio of displacement to the distance covered is always
less than one
equal to one
equal to or less than one
equal to or greater than one
C.
equal to or less than one
Distance is scalar, it is never negative. Therefore distance can be equal or greater than displacement. That implies the ratio of displacemrnt to distance is always equal to or less than one.
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
east-north direction
west-north direction
south-east direction
None of the above
B.
west-north direction
Let O be the origin, then
passenger in the train at P observes the car at Q along direction PQ, where direction PQ is west north direction.
A metro train starts from rest and in 5 s achieves 108 km/h. After that it moves with constant velocity and comes to rest after travelling 45 m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.
12.2 s
15.3 s
9 s
17.2 s
D.
17.2 s
Given:- v= 108 km/h = 30 m/s
For first equation of motion
v = u + t
30 = 0 + × 5 ( u = 0)
⇒ a = 6 m/s2
So, distance travelled by metro train in 5s
=
Distance travelled before coming to rest = 45 m
So, from third equation of motion
⇒
⇒
Time taken in travelling 45 m is
= 3s
Now, total distance = 395 m
i.e 75 + s' + 45 = 395 m
i.e s' = 395 - (75 + 45)
⇒ s' = 275 m
∴ t2 =
t2 = 9.2 s
Hence, total time taken in whole journey
= t1 + t2 + t3
= 5 + 9.2 + 3
= 17.2 s
A police party is moving in a jeep at a constant speed v. They saw a thief at a distance x on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration a. Which of the following relations is true if the police is able to catch the thief?
C.
Let the police party catch the thief after t
Distance travelled by police party in t = v t
and distance travelled by thief =
⇒
⇒
Since by using the Quadratic equation formula
⇒
⇒
The engine of a jet aircraft applies a thrown force of 105 N during take off and causes the plane to attain a velocity of 1 km/s in 10 s. The mass of the plane is
102 kg
103 kg
104 kg
105 kg
B.
103 kg
Given:- Force F = 105 N
Velocity during take off = 1 km/s = 1000 m
i.e initial velocity = 0
Hence Change in velocity = (103 - 0)
And time t = 1
The acceleration produced in jet
= 100 m/s2
∴ Mass =
=
⇒ Mass = 103 kg
A stone is weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 s. The force of friction (assuming it to be constant) will be
- 20 N
-0.2 N
0.2 N
20 N
B.
-0.2 N
u = 2 m/s
v = 0
f = 0s
The kinematic eqution is
Where u is the initial velocity
v - final velocity
α - acceleration
t - time
⇒
=
⇒
Friction force =
= 1 × (-0.2)
= -0.2 N
When a ceiling fan is switched on, it makes 10 revolution in the first 3 s. Assuming a uniform angular acceleration, how many rotation it will make in the next 3 s
10
20
30
40
C.
30
Angle turned in three second and θ1 s = 2 n x 10
= 20 π rad
According to the kinematic equation of Rotational motion
Now, angle turned in 6 s from the starting
= 0 +
θ2 = 80 π rad
Since angle turned between t = 3 s to t = 6 s
= 80π - 20 π
Number of revolution =
= 30
Hence Number of revolution in next 3s is 30
Two boys are standing at the ends A and B of a ground, where AB= a. The boy at B starts running in a direction perpendicular to AB with velocity The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is
B.
Distance covered by boy A in time t
AC = vt ...(i)
Distance covered by boy B in time t
BC = v1 t ... (ii)
Using Pythagourus theorem
AC2 = AB2 + BC2
( vt )2 = a2 + (v1 t)2
⇒
⇒
⇒
A force F is given F = at + bt2 where, t is time. What are the dimensions of a and b?
[ M L T-1 ] and [M L To ]
[ M L T-3 ] and [ M L2 T-4 ]
[ M L T-4 ] and [ M L T1 ]
[ M L T-3 ] and [ M L T-4 ]
B.
[ M L T-3 ] and [ M L2 T-4 ]
D.
[ M L T-3 ] and [ M L T-4 ]
Force F = at + bt2
From principle of homogeneity
Dimension of at =
=
Dimension of at = [ M L T-3 ]
Similarly,
dimensions of bt2 =
=
dimension of bt2 = [ M L T-4 ]
A particle moves towards east for 2 s with velocity 15 m/s and move towards north for 8 s with velocity 5 m/s. Then, average velocity of the particle is
1 m/s
5 m/s
7 m/s
10 m/s
B.
5 m/s
Distance covered towards east
= speed × time
= 2 × 15 = 30 m
Distance covered towards north
= speed × time
= 8 × 5 = 40 m
Total displacement
=
= 50 m
Average velocity =
=
= 5 m/s
The distance x (in µm) covered by a molecule starting from point A at time t = 0 and stopping at another point B in given by the equation The distance between A and B ( in μm) is closed to
10.7
20.7
40.7
50.7
A.
10.7
Given that distance
On differentiating
For x to be maximum,
⇒ 4t - t2 = 0
⇒ t = 4
∴ Maximum distance
xmax =
=
= 16
xmax = 10.7
The density of a rod having length l varies as ρ = c + dx, where x is the distance from the left end. The centre of mass is
A.
Let the cross-sectional area is . The mass of an element of length dx located at a distance x away from the left end is (C + Dx) dx. The x-coordinate of the centre of mass is given by
=
Where limit is from 0 to l.
=
Xcm =
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. (Take, g = 10 m/s2)
30 cm
zero
20 cm
25 cm
C.
20 cm
The acceleration (downward) of the elevator is 12 m/s2 which is greater than gravitational acceleration g. So, the block will act as a freely falling body.
From the motion of the block
s = ut +
= 0 × t + × 10 × (0.2)2
= 10 × 0.2 × 0.2
s = 20 cm
A rod PQ of length l is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end Q is 2 m/s towards negative direction of x. The speed of end P will be
B.
From the diagram in question
x = l cosθ
y = l sinθ
∴ = -l sinθ .....
vy = = l cosθ
∴ .....
vy = - vx cot θ
Since vx = -2 m/s
vy = -2 cot 60o ....( Given:- θ = 60o )
=
vy = ms-1
The negative sign indicates the motion towards negative direction of Y.
The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, and β are positive constants. The velocity of the particle will
decrease with time
be independent of
drop to zero when
increase with time
D.
increase with time
Given:-
x = a e-αt + beβt
We know that
velocity v =
v =
= -a e-αt + bβ eβt
v =
v = A + B
where A = - a , B = bβ e-βt
The value ot term A = -a increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time.
If A + B = C and that C is perpendicular to A. What is the angle between A and B, if
rad
rad
rad
rad
C.
rad
Given:-
A + B = C
Since, B = C A
Also
B2 = 2 A2
⇒ B = A
Now
A2 + B2 + 2A B cosθ = C2 = A2
∴
This gives
rad
A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?
2 m/s
3 m/s
3.25 m/s
4.25 m/s
C.
3.25 m/s
Given:- r = 25 cm = m
ω = 13 rad/s
Linear speed = Radius × angular speed
V = rω
= 0.25 × 13
3.25 m/s
Assertion: If = 2t for a particle moving in a straight line starting with initial velocity 4 m/s from the origin, then distance travelled by it in 2 s is same as displacement.
Reason: Velocity changes direction after 2 s only.
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.
B.
If both assertion and reason are true but reason is not the correct explanation of assertion.
Acceleration is given by
= 2t
2t
⇒ dv = t dt
⇒ =
⇒ v 4 = t2
⇒ v = 4 t2
⇒ v > 0 for t < 2
v = 0 for t = 2
v < 0 for t > 2
The body reverses its direction of motion after 2 s.
A man is at a distance of 6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is
2 m s-1
4 m s-1
6 m s-1
8 m s-1
C.
6 m s-1
If the man did not run, the bus would be at a distance s1 at time t given by
s1 = 6 + t2
= 6 + × 3 × t2
s1 = 6 + t2
If v is the speed of man, he would cover a distance s2 = vt in time t.
To catch the bus,
s1 = s2
6 + t2 = v t
⇒ t2 t + 4 = 0
which gives
t =
Now t will be real if is positive or zero.
Minimum v corresponds to = 0 which gives v = 6 m s-1
A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after
3 s
2 s
2.5 s
5 s
C.
2.5 s
Let balls meet after t s. The distance travelled by the ball coming down is
s1 = gt2
Distance travelled by the other ball
s2 = 40 t gt2
s1 + s2 = 100 m
∴ gt2 + 40 t gt2 = 100m
t =
∴ t = 2.5 s
The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is proportional to
t
t2
t3/2
B.
t
As given the displacement of a body is given to be proportional to the cube of time elapsed.
s ∝ t3
s = k t3
velocity:- Velocity of an object is the rate of change of its position with respect to time.
v =
=
v = 3 kt2
Acceleration:- Acceleration is the rate of change of velocity with respect to time.
=
= 6 kt
∴
Assertion: The speedometer of an automobile measures the average speed of the automobile.
Reason: Average velocity is equal to total distance divided by total time taken.
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.
D.
If both assertion and reason are false.
The speedometer of an automobile measures the instantaneous speed of the automobile.
Average velocity =
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