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:
If `v`= velocity of the combined mass block `A` & bullet just after collision, then
`m u=2mv` or `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)`