Three particles each of mass `m` can slide on fixed friction less 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 coefficient 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. `1/4 v_( 0)sqrt((m)/(5K))`
B. `3/4 v_(0)sqrt((m)/(5K))`
C. `1/3 v_(0)sqrt((m)/(5K))`
D. `1/8 v_(0)sqrt((m)/(5K))`

Answer

B

Explanation

By the conservation of linear momentum between `m_(1)` and `m_(2)` and by using coefficient of restitution velocity `m_(2)` just after collision `(3v_(0))/(4)`.
After the collision, the maximum extension in the spring occurs when angular velocity of `m_(2)` and `m_(3)` about O become same. Now by conservation of angular momentum about .O. and energy conservation
Velocity of `m_(2)=2 R omega=3/5 v_(0)`
Velocity of `m_(3)=R omega = (3)/(10)v_(0)` and `Deltax` maximum `=3/4 v_(0) sqrt((m)/(5K))`