Laws of Motion

Question 577
CBSEENPH11026288

A moving body of mass m and velocity 3 km/h collides with a rest body ofmass 2 m and stick to it. Now the combined mass starts to move. What will be the combined velocity?

  • 4 km/h

  • 1 km/h

  • 2 km/h

  • 3 km/h

Solution

B.

1 km/h

The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after collision is equal.

Applying law of conservation of momentum

             m1 v1 = MV

[ m2 v= 0 because v2 = 0 and M = m1 + m2 , V = final velocity 

             m × 3 = ( m + 2m ) v1

 So,          v1 = 1 km/h

Question 578
CBSEENPH11026316

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

   

  • vg μ

  • g μv

  • gv

  • vg

Solution

A.

vg μ

Block will come to rest, if force applied on it. It will vanish due to frictional force acting between block Band surface, i.e.,

    Force applied = Frictional force

⇒     μ mg  = ma

⇒      μ mg = m vt

⇒            t = vμ g

Question 579
CBSEENPH11026324

A block slides down on an incline of angle 30° with an acceleration g4, Find the kinetic 4 friction coefficient.

  • 122

  • 0.6

  • 12 3

  • 12

Solution

C.

12 3

Consider the situation

   

Let the mass of block be m. 

The equation of forces

            mg sin 30o - f = mg4

                             f  =   mg 12 - mg4              .....( since sin30o12 )

                              f = m g4 

Also, N = mg cos 30° = mg 32

As the block is slipping on the incline, friction is

               μn = FN

                    = mg4 mg 32

               μn = 123

Question 580
CBSEENPH11026339

A pendulum having a bob of mass m is hanging in a ship sailing along the equator form east to west. When the strip is stationary with respect to water, the tension in the string is T0, The difference between T0 and
earth attraction on the bob, is

  • mg + m ω2 R2

  • 2 R3

  • 2 R2

  • 2 R

Solution

D.

2 R

The attraction of earth on the bob = Weight of bob = mg

Speed of ship due to rotation of the earth

Velocity = angular velocity × radius

      v = ω R

Also due to the rotation of the earth, a centripetal force acts on the bob = mω2 R on earth 

The net force

    T0 = mg  + mω2 R

⇒  T0 - mg = mω2 R

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