What is the target’s speed just after the bullet emerges?

Question

In a ballistics test, a 26g bullet traveling horizontally at 1200m/s goes through a 27-cm-thick 380kg stationary target and emerges with a speed of 870m/s . The target is free to slide on a smooth horizontal surface

Answer 1

Mass of the bullet is 26 g .

Initial velocity of the bullet is 1200 m/s .

Final velocity of the bullet is 870 m/s .

Change in velocity is 1200 - 870 = 330 m/s .

From the definition of conservation of momentum , we can found the target’s speed .

In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. 

So momentum change of bullet = mass * Change in velocity 

                                           = 0.026 kg * 330 

Momentum change in target = mass of target * Change in velocity of target 

                                     = 380 kg * Change in velocity of target 

Form the conservation of momentum 

   0.026 kg * 330 = 380 kg * Change in velocity of target 

 8.58 = 380 kg * Change in velocity of target 

Change in velocity of target = 8.58 / 380

Change in velocity of target = 0.02257 m/s .

The target is initially at rest so V_i = 0 .

Change in velocity of target =  V_f  - V_i = 0.02257 m/s 

V_f  - 0 = 0.02257 m/s 

V_f = 0.02257 m/s 

So the target’s speed just after the bullet emerges is  0.02257 m/s .