Two blocks of masses `m` and `3m` are connected by an inextensible string and the string passes over a fixed pulley which is massless and frictionless. `A` bullet of mass `m` moving with a velocity `'u'` hits the hanging block of mass `'m'` and gets exbedded in it. Find the height through which block `A` rises after the collision.

Explanation

As soon as the collision occurs, the string becomes slack, tension becomes zero. Gravitational force is acting vertically downwards. But as gravitational force is a weaker force than impulse force, therefore a possible change of momentum due to gravitational force during collision can be neglected.
Hence, conserving momentum of the system during collision along vertical direction:
If v = velocity of the combined mass just after collision
Then, `mu=2mv`
`rArr v=u//2`
Block A + bullet start moving with velocity u/2.
Now, maximum height through which the combined mass rises
`h_("max")=(v^(2))/(2g)=(u^(2))/(8g)`