A bullet of mass $20$ g moving with a velocity of $100$ m/s strikes a wooden block of mass $80$ g and gets embedded into it. Calculate velocity of the combined system.

The correct option is C

20 ms^-1

It is given that the bullet gets embedded into the wooden block.

So, let the velocity of the combined system(bullet and wooden block) after collision be $v$

Mass of the bullet, $m_1=20$ g $=\frac{20}{1000}=0.02$ kg

Mass of the wooden block, $m_2=80$ g $=\frac{80}{1000}=0.08$ kg

Initial velocity of the bullet, $u_1=1000$ m/s

Initial velocity of the wooden block, $u_2=0$ m/s

Final velocity of the combined system $=v$

By applying the law of conservation of momentum,

Intial momentum of the system = Final momentum of the system

$\Rightarrow m_1{u}_1+m_2{u}_2=m_{1}v+m_{2}v$

$\Rightarrow 0.02\times 100+0.08\times 0=(0.02+0.08)v$

$\Rightarrow v=\frac{2}{0.1}$

$\Rightarrow v=20$ m/s

Hence, velocity of system after impact $=20$ m/s.