Two balls of masses `m_(1)=100 g` and `m_(2)=300 g` are suspended from point `A` by two equal inextensible threads, each of length `l=32/35m`. Ball of mass `m_(1)` is drawn aside and held at the same level as `A` but at a distance `(sqrt(3)/2)l` from `A`, as shown in Fig. When ball `m_(1)` is released, it collides elastically with the stationary ball of mass `m_(2)`.

Velocity `u_(1)` with which the hall of mass `m_(1)` collides with the other ball is

Options

A. `1m//s`
B. `2m//s`
C. `3m//s`
D. `4m//s`

Answer

D

Explanation

Ball of mass `m_(1)` falls freely till thread becomes taut. At that instant inclination `theta` of the thread with the vertical be given by
`sintheta=(((sqrt(3))/2l))/l=(sqrt(3))/2` or `theta=60^(@)`
Ball of mass `m_(1)` falls freely through height.

Velocity of this ball at this instant is `v=sqrt(2gxxl/2)=sqrt(gl)`
It can be resolved into two components.
i. `v cos theta`, along the thread. But thread is inextensible, hence this component decreases to zero (due to tension developed in the thread).
ii. `v sin theta`, perpendicular to the thread. Due to this component ball starts to move along a circle whose centre is at `A`.
According to law of conservation of energy,
Kinetic energy of ball `m_(1)` just before collision `=` Its kinetic energy at `AC +` Further loss of its potential energy
`1/2m_(1)u_(1)^(2)=1/2m_(1) (sqrt(gl)sintheta)^(2)+m_(1)g(l-lcostheta)`
`u_(1)=4m//s`
According to law of conservation of momentums
`m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)`
and coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))=1`
substituting `u_(1)=4 m//s, u_(2)=, m_(1)=0.1 kg` and `m_(2)=0.3kg, v_(2)=2m//s,` the height to which it rises is
`h=v_(2)^(2)/(2g)=0.20m`