A 10 g bullet traveling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off?

 As given, 

Mass of the bullet $m_1=10\ g=0.01\ kg$

Velocity of the bullet $v_1=200\ m/s$

After the collision with the target:

Mass $m_2=2\ kg+0.01\ kg=2.01\ kg$

Let $v_2$ be the velocity of the target, then according to the law of conservation of the momentum:

$m_1\times v_1=m_2\times v_2$

Or $0.01\times200=2.01\times v_2$

or $v_2=\frac{2}{2.01}\ m/s$

Or $v_2=1\ m/s$

Tutorialspoint

Updated on: 10-Oct-2022

25 Views

Kickstart Your Career

Get certified by completing the course

Get Started