Laws of Motion

  • Question 581
    CBSEENPH11026342

    A ball of mass m hits the floor with a speed v making an angle of incidence θ with the normal. The coefficient of restitution is e. The speed of reflected ball and the angle of reflection of the ball will be

         

    • v' = v, θ = θ'

    • v' = v2, θ = 2θ'

    • v' = 2v, θ = 2θ'

    • v' = 3v2, θ = 2θ'3

    Solution

    A.

    v' = v, θ = θ'

    The coefficient of restitution is a number which indicates how much kinetic energy ( energy of motion ) remains after collision of two object.

    The parallel component of velocity of the ball remains unchanged. This gives

               v' sinθ = v sinθ                   .... (i)

    For the components normal to the floor, the velocity of separation = v' cosθ

    Hence  v' cosθ' = e v cosθ               .....(ii)

    From equations (i) and (ii)

                v' = sin2θ + e2 cos2θ

    and       tanθ' = tanθe

    For elastic collision, e = 1, so that 

              θ' = θ  and  v' = v

    Question 582
    CBSEENPH11026343

    A particle slides on surface of a fixed smooth sphere starting from top most point. The angle rotated by the radius through the particle, when it leaves contact with the sphere, is

    • θ =cos-1 13

    • θ = cos-1 23

    • θ = tan-1 13

    • θ = sin-1 43

    Solution

    B.

    θ = cos-1 23

    See the diagram

       

    Let the velocity be v when the body leaves the surface. From free body diagram,

                         mv2 R = mg cosθ

    ⇒                    v2 = R g cosθ

    Again from work-energy  principle

              change in KE = work done

    ⇒             12 mv2 - 0 =  mg ( R -R cosθ )  

                    v2 = 2gR ( 1 - cosθ)

    From Eqs. (i) and (ii)

              Rg cosθ = 2gR ( 1 - cosθ )

    ⇒           3 gR cosθ = 2gR

    ⇒              θ = cos-123

    Question 583
    CBSEENPH11026347

    A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration 1 m/s, how much force should it apply to the rope if rope is 5 m long and the monkey starts from rest?

    • 150 N

    • > 160 N

    • 165 N

    • 150 < T ≤ 160 N

    Solution

    C.

    165 N

    The mass of monkey = 15 kg

       Acceleration     a = 1 m/s2

    For the motion of the monkey T - [ 15 g + 15 (1) ]

    Hence, T is tension in the string T = 15 g + 15

                  T = 15 ( 10 + 1)

                      = 15 × 11

                 T = 165 N

    The monkey should apply 165 N force to the rope.

    Question 584
    CBSEENPH11026350

    A body weighing 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

    • 13 g

    • 9 g

    • 22 g

    • 12 g

    Solution

    D.

    12 g

    Consider the diagram in balance

           

             8x = wy

    ⇒        xy = w8                       ...(i)

    Also,  wx = 18 y

    ⇒         xy = 18w                     ....(ii)

    Dividing Eq. (i) and Eq. (ii) we get

                xyxy = w818w

    ⇒         1 = w218 × 8

    ⇒          w = 18 × 8

                 w = 12 g

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