Question

Two blocks $A$ and $B$ of mass $m$ each are connected by a light inextensible string going over a smooth light pulley. Initially, the system is at rest. A bullet of mass $2m$ strikes the block $A$ and sticks to it. If the bullet strikes the block with a speed $2v$, find the speed with which the block $A$ moves, immediately after the collision.

• A. $v$
  
• B. $2v$
  
• C. $\frac{v}{2}$
  
• D. $\frac{2v}{3}$
  

Answer 1

The correct option is A $v$


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 $A$ becomes $3m$ due to embedded bullet.

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

Applying impulse-momentum theorem on both blocks, considering vertically downwards as $+vey$ axis:
For block $A$:
$-J=P_f-Pi=3m\left(v^{\prime }\right)-\left[\right(m\times 0)+(2m\times 2v\left)\right]$
$\Rightarrow -J=3m{v}^{\prime }-4mv...\left(i\right)$
For block $B$:
$-J=P_f-Pi=-m\left(v^{\prime }\right)-\left[\right(m\times 0\left)\right]$
$-J=-m{v}^{\prime }...\left(ii\right)$

Equating Eq. $\left(i\right)$ and $\left(ii\right)$:
$3m{v}^{\prime }-4mv=-m{v}^{\prime }$
$\Rightarrow 4m{v}^{\prime }=4mv$
$\therefore v^{\prime }=\frac{4mv}{4m}=v$
Hence option $\left(a\right)$ is correct.

Alternative solution:
We can treat the system of blocks as a system moving together in a straight line. In that case, impulse due to tension can be neglected (since tension would be an internal force). Let the blocks move together with velocity $v^{\prime }$ after the collision.
Then, applying PCLM:
$2m\times 2v+0+0=(2m+m)v^{\prime }+m{v}^{\prime }$
(bullet sticks to the block)
$\Rightarrow 4mv=4m{v}^{\prime }$
$\Rightarrow v^{\prime }=v$