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. $\frac{mu}{2}$
• B. $\frac{mu}{4}$
• C. $mu$
• D. $2mu$

Answer 1

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 $+vex-$ axis direction on system of (blocks+bullet):
$p_i=p_f$
$0+0+2mu=mv+mv+2m\times 0$
(bullet comes to rest finally)
$\Rightarrow 2mu=2mv$
$\Rightarrow v=u$

Impulse on $A$ due to impulsive tension acting just after collision:
$|\overrightarrow{J}|=m|\Delta \overrightarrow{v}|$
$|\overrightarrow{J}|=m(v-0)=mv$
$\therefore |\overrightarrow{J}|=mv=mu$