Question

A bullet of mass $20g$ and moving with $600m/s$ collides with a block of mass $4kg$ hanging with the string. What is velocity of bullet when it comes out of the block, if block rises to height $0.2m$ after collision

Answer 1

According to conservation of linear momentum,

$m_1{v}_1=m_{1}v+m_2{v}_2$

where, $v_1$ is the velocity of bullet before collision, $v$ is velocity of bullet after the collision and $v_2$ is velocity of block

Therfore,

$0.02\times 600=0.02v+4v_2$

Here, $v_2=\sqrt{2gh}=\sqrt{2\times 10\times 0.5}=2m/s$

$0.02\times 600=0.02v+4\times 2$

$0.02v=12-8$

$v=\frac{4}{0.02}=200m/s$