Question

In a ballistics test, a 20.0 g bullet traveling horizontally at 1300 m/s goes through a 30.0 cm -thick 450 kg stationary target and emerges with a speed of 900 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 the bullet exert on the target?

C) What is the target's speed just after the bullet emerges?

Answer 1

Part a)
0.3m = s = target length
u = 1300m/s
v = 900m/s
v^2 = u^2 + (2*a*s) rearranging for acceleration gives a = (v^2 - u^2)/(2*s) =-1.46*10^6

so acceleration = - 1*46*10^6 m/s^2 (notice it is negative acceleration)

now v = u + a*t we now know a, so rearrange for t, time t = (v-u)/a

time inside the target = 0.000273 seconds

Part b)

Average force F = ma

m = 0.02kg and we know acceleration so:

F = 29200N, 29.200kN

Part c)

Again F = ma but mass of target = 450kg and we now know F so:

rearrange for a to give a=F/m

a = 29200/450 = 64.888m/s^2

Again v = u + a*t

u = 0, a = 64..888, t = 0.000273

so velocity of target = 64.888 * 0.000273 = 0.0177m/s