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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