Laws of Motion

• Question 581

## 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 bev' = v, θ = θ' v' = $\frac{\mathrm{v}}{2}$, θ = 2θ' v' = 2v, θ = 2θ' v' =

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' =

and       tanθ' = $\frac{\mathrm{tan\theta }}{\mathrm{e}}$

For elastic collision, e = 1, so that

θ' = θ  and  v' = v

Question 582

## 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

Solution

B.

See the diagram

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

⇒                    v2 = R g cosθ

Again from work-energy  principle

change in KE = work done

⇒

v2 = 2gR ( 1 - cosθ)

From Eqs. (i) and (ii)

Rg cosθ = 2gR ( 1 - cosθ )

⇒           3 gR cosθ = 2gR

⇒

Question 583

## 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

## 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 is13 g 9 g 22 g 12 g

Solution

D.

12 g

Consider the diagram in balance

8x = wy

⇒                               ...(i)

Also,  wx = 18 y

⇒                              ....(ii)

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

⇒         1 =

⇒          w =

w = 12 g