Question

In a ballistics test, a 23.0g bullet traveling horizontally at 1300m/s goes through a 40.0cm -thick 450kg stationary target and emerges with a speed of 900m/s . The target is free to slide on a smooth horizontal surface. How long is the bullet in the target? What average force does the bullet exert on the target? What is the target's speed just after the bullet emerges?

Answer 1

Is the target mass 450 or 45.0 kg? I will use 450 as stated in the post, 45.0 would be a more realistic scenario, though.

Let's first calculate the speed of the target when the bullet emerges using conservation of momentum

.021*1300=450*v+.021*900

solve for v, this is the speed of the target

v=0.056/3 m/s

Now consider the bullet w/r/t the target from the moment of impact to the moment the bullet emerges from the target. This is an accelerating reference frame, so I will assume the acceleration is constant.

In this reference frame, the starting speed of the bullet is 1300 m/s, and the emerging speed of the bullet is 900-0.056/3 m/s or 899.98 m/s

the equations of motion of the bullet are

x(t)=v0*t-.5*a*t^2

and

v(t)=v0-a*t

where t is in seconds while the bullet is in the target, and a is the acceleration magnitude.

solve for t, the time the bullet is the target. With the values, the two equations are:

30=1300*t-.5*a*t^2

899.98=1300-a*t

30=1300*t-.5*400.02*t

t=0.027 seconds

now that we have t, consider the target in the Earth's frame of reference

v=a*t

a=0.684 m/s^2

And F=m*a, where F is the average force the bullet exerts on the target

308 N

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