Laws of Motion

Sponsor Area

Question
CBSEENPH11020795

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

  • 200 J

  • 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
rightwards double arrow space space space space space straight W space equals space straight W subscript straight g plus straight W subscript straight f space space space space space space space space space space space... left parenthesis straight i right parenthesis
But space space space space space straight W space equals space 300 space straight J
space space space space space space straight W subscript straight g space equals space mgh space equals space 2 space cross times 10 cross times 10 space equals space 200 space straight J
putting space in space Eq. space left parenthesis straight i right parenthesis comma space we space get
space space space space space space space space space 300 space equals space 200 plus straight W subscript straight f
rightwards double arrow space space space space space space space space space space space space space straight W subscript straight f space equals space 300 space minus space 200 space equals space 100 space straight J
 

Sponsor Area

Question
CBSEENPH11020796

A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is

  • 48 N

  • 24 N

  • 12 N

  • 96 N

Solution

B.

24 N

The vector OA with rightwards arrow on top represents the momentum of the object before the collision, and the vector OB with rightwards arrow on top that after the collision. The vector AB with rightwards arrow on top represents the change in momentum of the object increment straight P with rightwards arrow on top.


As the magnitudes of  OA with rightwards arrow on top space and space OB with rightwards arrow on top are equal, the components of OA with rightwards arrow on top space and space OB with rightwards arrow on top along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is due only to the change in direction of the perpendicular components.
Hence,  increment straight p space equals space OB space sin space 30 degree space minus space left parenthesis negative OA space sin space 30 degree right parenthesis
                    equals mv space sin space 30 degree space minus space left parenthesis negative mv space sin space 30 degree right parenthesis
equals 2 space mv space sin space 30 degree
Its time rate will appear in the form of average force acting on the wall.
 therefore space space space space space space space space space straight F space cross times space straight t space equals space 2 mv space sin space 30 degree
or space space space space space space space space space space space space space space straight F space equals space fraction numerator 2 mv space sin space 30 degree over denominator straight t end fraction
   Given comma space straight m space equals space 0.5 space kg comma space space straight v space equals space 12 space straight m divided by straight s comma space space space straight t equals space 0.25 space straight s
space space space space space space space space space space straight theta space equals space 30 degree
Hence comma space space space straight F space equals space fraction numerator 2 cross times 0.5 cross times 12 space sin space 30 degree over denominator 0.25 end fraction space equals space 24 space straight N

Question
CBSEENPH11020686

A ball moving with velocity 2 ms-1 collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in ms-1) after collision will be

  • 0,1

  • 1,1

  • 1,0.5

  • 0,2

Solution

A.

0,1

If two bodies collide head on with coefficient of restitution
straight e space equals space fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
From space the space law space of space conservation space of space linear space momentum

straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 space equals space straight m subscript 1 straight v subscript 1 space plus straight m subscript 2 straight v subscript 2
rightwards double arrow space straight v subscript 1 space equals space open square brackets fraction numerator straight m subscript 1 minus em subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 plus open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
Substituting space straight u subscript 1 space equals space 2 space ms to the power of negative 1 end exponent comma space straight u subscript 2 space equals 0 comma space straight m subscript 1 equals straight m subscript 2 space and space straight m subscript 2 space equals space 2 straight m comma space straight e space equals space 0.5
we space get space space straight v subscript 1 space equals space open square brackets fraction numerator straight m minus straight m over denominator straight m plus 2 straight m end fraction close square brackets space straight x 2
similarly comma
straight v subscript 2 space equals open square brackets fraction numerator left parenthesis 1 plus straight e right parenthesis straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 1 space plus open square brackets fraction numerator straight m subscript 2 minus em subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight u subscript 2
equals space open square brackets fraction numerator 1.5 space straight x space straight m over denominator 3 straight m end fraction close square brackets space straight x space 2
space equals space 1 space ms to the power of negative 1 end exponent

Question
CBSEENPH11020539

A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is straight mu subscript straight k. When the block A is sliding on the table, the tension in the string is

  • fraction numerator left parenthesis straight m subscript 2 plus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction
  • fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 minus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

Solution

C.

fraction numerator straight m subscript 1 straight m subscript 2 left parenthesis 1 plus straight mu subscript straight k right parenthesis straight g over denominator left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction

FBD of block A,

T -m1afk ..... (i)
FBD  of block B

m2g -T = m2a ... (ii)
Adding Eqs. (i) and (ii), we get
m2g-m1a = m2a +fkrightwards double arrow space straight m subscript 2 straight g minus straight m subscript 1 straight a space equals space straight m subscript 2 straight a space plus straight f subscript straight k
rightwards double arrow space straight a equals space fraction numerator left parenthesis straight m subscript 2 minus straight u subscript straight k straight m subscript 1 right parenthesis straight g over denominator straight m subscript 1 plus straight m subscript 2 end fraction
from space equation space left parenthesis ii right parenthesis
straight T equals space straight m subscript 2 minus left parenthesis straight g minus straight a right parenthesis
equals space straight m subscript 2 open square brackets 1 minus fraction numerator left parenthesis straight m subscript 2 minus straight mu subscript straight k straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets straight g
straight T equals space fraction numerator straight m subscript 1 straight m subscript 2 space left parenthesis 1 plus straight mu subscript straight k right parenthesis over denominator straight m subscript 1 plus straight m subscript 2 end fraction straight g

Question
CBSEENPH11020772

A block B is pushed momentarily along a horizontal surface with an initial velocity v. if μ is the coefficient of sliding friction between B  and the surface, block B will come to rest after a time:

  • v/gμ

  • gμ/v

  • g/v

  • v/g

Solution

A.

v/gμ

Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, ie,
force applied = frictional force
i.e., μmg = ma
μmg = m (v/t)
t =v/μg