Question

A bullet of mass $0.01kg$ is moving with a velocity of $500m/s$ strikes a block of mass $2kg$ suspended from a $5$ metre long string (figure). The centre of gravity of the block rises vertically upwards through a height of $0.1$ metre. The velocity of the bullet after emerging out of the block will be

• A. $220m/s$
• B. $2.2m/s$
• C. $0.22m/s$
• D. $zero$

Answer 1

The correct option is A $220m/s$

According to momentum conservation law, total initial momentum $=$ total final momentum
So, $m{v}_i+M{V}_i=m{v}_f+M{V}_f....\left(1\right)$

where $v_i,v_f$ are velocities of bullet before and after collision and $V_i,V_f$ are the velocities of block before and and after collision.
Given, $m=0.01kg,M=2kg,v_i=500m/s,V_i=0$
Using $(v^2-u^2=2gs)$, we get final velocity of block $V_f=\sqrt{2g\times 0.1}=\sqrt{2}m/s$
Eqn (1) becomes, $0.01\times 500+0=0.01v_f+2\times \sqrt{2}$
$5=0.01v_f+2\sqrt{2}$
$v_f=(5-2\sqrt{2})100\sim 220m/s$