Momentum and CollisionEasy
Question
A massless inelastic thread passes over a small frictionless pulley. Two identical blocks A and B of mass m = 2 kg 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
Options
A.Velocity of centre of mass of system (block A plus bullet C and block B) just after collision is u/2
B.Velocity of centre of mass of system (block A plus bullet C and block B) just after collision is u/3
C.Time after which the string becomes taut again is t = 
D.Maximum height reached by block A plus bullet C is Hmax = 
Solution
After collision
vCM =
String become taut again when displacement of (A + C) = displacement of B ⇒ t =
When (A + C) reaches at A height of
, an impulse acts on it
which gives it a velocity in upward direction. ⇒ Hmax ≠
vCM =
String become taut again when displacement of (A + C) = displacement of B ⇒ t =
When (A + C) reaches at A height of
, an impulse acts on itwhich gives it a velocity in upward direction. ⇒ Hmax ≠
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