Question

A system of two blocks $A$ and $B$ are connected by an inextensiible massless string as shown in figure. The pulley is massless and frictionless. Initially, the system is at rest. A bullet of mass $m$ moving with a velocity $u$ as shown in the figure hits block $B$ and gets embedded into it. Then find out the impulse imparted by tension force to the block of mass $3m$, immediately after the collision takes place.

• A. $\frac{5}{4}mu$
• B. $\frac{4}{5}mu$
• C. $\frac{2}{5}mu$
• D. $\frac{3}{5}mu$

Answer 1

The correct option is D $\frac{3}{5}mu$

Tension in both strings will be equal, hence impulse due to tension on both blocks will be the same.
$J_1=J_2=-J$ (assuming $-vey$ direction upwards)

After collision, mass of block $B$ becomes $2m$ due to embedded bullet.

$\Rightarrow$The system of (blocks+bullet) will move with common speed $v$, to satisfy string constraint. Block $A$ moves upwards with speed $v$ and $B$ downwards with speed $v$

Applying impulse-momentum theorem on both blocks, considering vertically downwards as $+vey$ axis:
For block $A$:
$-J=P_f-P_i=3m(v_f-v_i)$
$=3m(-v-0)=-3mv...\left(i\right)$
For block $B$:
$-J=P_f-Pi=2m\left(v\right)-\left[\right(m\times 0)+(m\times u\left)\right]$
$=2mv-mu...\left(ii\right)$

Equating Eq. $\left(i\right)$ and $\left(ii\right)$:
$-3mv=2mv-mu$
$\Rightarrow 5mv=mu$
$\therefore v=\frac{u}{5}$

Hence, impulse on block $A$:
$=-J=2mv-mu=\frac{2}{5}mu-mu$
$=-\frac{3}{5}mu$
$\therefore |J|=\frac{3}{5}mu$