Physics Help?

Question

A 22-g bullet traveling 290 m/s penetrates a 2.0 kg block of wood and emerges going 140 m/s .

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Answer 1

Mass of the tagret is 2 kg.

Mass of the bullet is 22 g=0.022 kg.

Initial velocity of the bullet is 290 m/s.

Final velocity of the bullet is 140 m/s.

Change in velocity is m/s.

Momentum change of bullet.

                                          

Momentum change of target.

                                          

Find the target’s speed just after the bullet emerges.

Conservation of momentum:

In a collision, the momentum change of bullet is equal to and opposite of the momentum change of target.

Change in velocity of target = 1.65 m/s .

The target is initially at rest so V_i = 0.

Change in velocity of target V_f  - V_i = 1.65 m/s.

V_f  - 0 = 1.65

V_f = 1.65.

The target’s speed just after the bullet emerges is 1.65 m/s.