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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