There particles each of mass `m` can slide on fixed frictioless circular tracks in the same horizontal plane as shown. Particle `m_(1)(=m)` moves with velocity `v_(0)` and hits particle `m_(2)(=m)`, the cofficient of restitution being `e=0.5` . Assume that `m_(2)` and `m_(3)(=m)` are at rest initially and lie along a radial line before impact, and the spring is initially unstretched.

Velocity of `m_(2)` immediately after impact is

Options

A. `(v_(0))/4`
B. `(3v_(0))/4`
C. `(v_(0))/2`
D. `(3v_(0))/2`

Answer

B

Explanation

First we apply conservation of momentum on `m_(1)` and `m_(2): m_(1)v_(0)=m_(1)v_(1)+m_(2)v_(2)`
`v_(0)=v_(1)+v_(2).....(1)` As `m_(1)=m_(2)` Where `v_(1)` and `v_(2)` are velocities of `m_(1)` and `m_(2)` immediately after impact.
From definition of coefficient of restitution.
`e=1/2=-(v_(2)-v_(1))/(0-v_(0))` or `v_(0)=2(v_(2)-v_(1))......(2)`
On solving eqns. `(1)` and `(2)`,
We get, `v_(1)=(v_(0))/4` and `v_(2)=(3v_(0))/4`
The spring has maximum extension when angular velocity `w` of `m_(2)` about `O` is same. Now we apply conservation of angular momentum.
`m(3/4v_(0))2R=mR^(2)omega+m(2R)^(2)omega` or `r omega=3/2(v_(0))/(5R)`
velocities of `m_(2)=2 R omega=3/5 v_(0)`
Velocity of `m_(3)=R omega=3/10 v_(0)`

In order to determine maximum extension in the spring. we apply law of conservation of energy.
`1/2m(3/4 v_(0))=1/2mR^(2)(3/2(v_(0))/(5R))+1/2m(2R)^(2)(3/2(v_(0))/(5R))+1/2kDeltax_(max)^(2)`
Omn solving for `Deltax`, we get `Deltax_(max)^(2)=3/4v_(0)sqrt(m/(5R))`