Momentum, force, speed: bullet and ballistics?

Question

In a ballistics test, a 25 g bullet traveling horizontally at 1200 m/s goes through a 30 cm thick 350 kg stationary target and emerges with a speed of 850 m/s. The target is free to slide on a smooth horizontal surface.


(a) How long is the bullet in the target?

(b) What average force does it exert on the target?
N (magnitude only)

(c) What is the target's speed just after the bullet emerges?
m/s

Answer 1

a) Initial velocity, u , of the bullet as it hits target=1200m/s
final velocity, v, as it emerges =850m/s
distance traversed,s = thickness of target
=30cm
=0.30m

v^2-u^2=2as

850^2-1200^2=2a*0.30
722500-1440000=0.6a
-717500=0.6a
a=-717500/.6
=-1,195,833m/s^2

v=u+at

850=1200+(-1,195,833)t
1,195,833t=1200-850
t=350/1,195,833
=2.93*10^-4 s


b) F=ma
mass of bullet, m=25grams or 0.025kg

F=0.025*1,195,833
=29,896N

c)Initial momentum of bullet=25*1200
=30,000
Initial momentum of target=350*0
=0

Final momentum of bullet=25*850
=21,250
Final momentum of target=350v

Based on conservation of momentum:

Initial momentum of the system=Final momentum of the system

Substitute known values:

30,000+0=21,250+350v
350v=30,000-21,250
v=8750/350
=25m/s

Answer 2

It is easiest to calculate c) first. The momentum gained by the target will be the momentum lost by the bullet. The momentum lost by the bullet is m(v_i - v_f), where v_i and v_f are the initial and final speeds of the bullet, respectively. The momentum gained by the target is (350 kg)*v_t, where v_t is the speed of the target when the bullet emerges. So (0.025 kg)*(1200 m/s - 850 m/s) = (350 kg)*v_t, and you can solve alebraically for v_t.

Now you can do a). The bullet enters the near side of the target at a position 0 and time 0. It exits from the far side of the target at position x and time t, and note that the far side has an initial position of 35 cm. We have to assume that both objects have constant acceleration during the encounter. If the bullet has acceleration a_b and the target has acceleration a_t, the position of the bullet at time t is v_i*t + 0.5*a_b*t^2 = (1200 m/s)t + 0.5*a_b*t^2 and the position of the far side of the target at time t is x_0 + 0.5*a_t*t^2 = (.35 m) + 0.5*a_t*t^2. Now, the acceleration of the bullet would be (800 m/s - 1200 m/s)/t and the acceleration of the target would be v_t/t, and the bullet will have the same position as the far side of the target when it exits. So subbing in the a values and equating the two displacements, we get (1200 m/s)t + 0.5*(-400 m/s)*t = (.35 m) + 0.5*v_t*t, where you can solve algebraically for t, because you already know v_t from above. That gives you the desired time.

b) is pretty easy. Average force is change in momentum divided by change in time. I gave the formula for change in momentum in part c), and you calculated the time in part a), so you can do this simple calculation easily.