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Question

Two identical blocks A and B, connected through a massless string are placed on a frictionless horizontal surface. A bullet having mass 2 times that of a single block, moving with velocity u strikes block B from behind as shown in the figure. If the bullet comes to rest after striking the block, find the impulse on block A due to tension in the string.


A
mu2
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B
mu4
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C
mu
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D
2mu
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Solution

The correct option is C mu
Mass of each block is m, hence mass of bullet is 2m.
From string constraint, both blocks will be moving with the same velocity after the collision of bullet.


Let velocity of both blocks after the bullet strikes be v.
Applying momentum conservation along +ve x axis direction on system of (blocks+bullet):
pi=pf
0+0+2mu=mv+mv+2m×0
(bullet comes to rest finally)
2mu=2mv
v=u

Impulse on A due to impulsive tension acting just after collision:
|J|=m|Δv|
|J|=m(v0)=mv
|J|=mv=mu

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