Question

A massless inelastic thread passes over a small frictionless
pulley. Two identical blocks A and B of mass m = 2kg are
attached with the thread. The system is initially at rest. A
bullet C also of mass m strikes the block A from below with
initial speed u as shown in figure and gets embedded in
block A at t = 0 second. Then
(A) Velocity of centre of mass of system (block A plus bullet C
and block B) just after collision u/2
(B) Velocity of centre of mass of system (block A plus bullet C u C
and block B) just after collision is u/3
(C) Time after which the string becomes taut again is t  u
2g
(D) Maximum height reached by block A plus bullet C is
u

Answer 1

Answer:

(B) v(COM) = u/3

(C) t = u/2g

Solution:

mu = 2mv , or v =u/2

So, v(COM) = {(u/2)(2m) + m(0)}/{3m} = u/3

Also, ½ut – ½gt² = 0 + ½gt²

So, t = u/2g

H(max) = (u/2)²/(2g) = u²/8g

Hope it helps!