Momentum of bullet fired into target

  • #1
[SOLVED] Momentum of bullet fired into target

In a ballistics test, a 23.0 g bullet traveling horizontally at 1100 m/s goes through a 40.0 cm-thick 250 kg stationary target and emerges with a speed of 800 m/s. The target is free to slide on a smooth horizontal surface. What average force does the bullet exert on the target?

Using kinematic equations, I was able to solve for the acceleration, and use that to solve for the time, which ended up being 4.21*10^-4. I then used the equation F_avg = m(v_f - v_i) / t in an attempt to solve for average force. I plugged in values for the bullet (m=0.023m v_f=800m/s v_i=1100m/s, t = 4.21*10^-4. and ended up with an answer of 17800 N.
This answer is wrong. I also came up with 16,400N, 16,340N and 16,300N, all of which tell me I am off by an "additive constant." Am I using the right equation to solve for F_avg, but just the wrong values for mass and velocity? And what is an additive constant?
 

Answers and Replies

  • #2
mgb_phys
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Have you accounted for the momentum lost to the block?
 
  • #3
Gokul43201
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In a ballistics test, a 23.0 g bullet traveling horizontally at 1100 m/s goes through a 40.0 cm-thick 250 kg stationary target and emerges with a speed of 800 m/s. The target is free to slide on a smooth horizontal surface. What average force does the bullet exert on the target?

Using kinematic equations, I was able to solve for the acceleration...
You may have assumed that the distance traveled during the deceleration phase is 40cm. This is incorrect. The target is moving while the bullet passes through it.

Hint: You can calculate the final velocity of the target independently.
 
Last edited:
  • #4
Have you accounted for the momentum lost to the block?

I hadn't. Now I tried some more things. Since P = mv, I calculated the initial momentum of the bullet to be 0.023*1100=25.3, and the final momentum of the bullet to be 0.023*800 = 18.4. Does that mean the bullet lost 6.9 whatevers (kg*m/s?) of momentum to the target? Does 6.9 = (mass of bullet + mass of target)*initial velocity of the bullet/target combo?
 
  • #5
The final velocity of the target is zero, right? Or is "final velocity" defined as the velocity of the target when the bullet leaves it?
Is the final velocity of the target 6.9/250kg = 0.276m/s?

I guess my real question is what do I need to plug in for m, v_f, and v_i in the equation I stated originally for F_avg. Is the mass the combined mass, the bullet's mass, or the target's mass, and are the velocities needed the velocities of the bullet, the target or the combined system? Is the equation the right one to use?
 
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  • #6
Gokul43201
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Look at the situation just before the bullet enters the target ("intial") and again, just after it leaves the target ("final").

Initially, the bullet has all the momentum as the target is at rest. Finally, the bullet has some of the momentum and the target has the remaining (from momentum conservation). The final velocity of the target can thus be calculated and you will get exactly what you have written down in post #5.

Next, call the contact time t, and let the average force exchanged between the target and bullet (the forces are equal by Newton's III Law) be F. From the masses of the bullet and target you can compute their average deceleration and acceleration, respectively. And finally, you can relate the contact time and the average acceleration to the change in velocity for each of them. That's 2 equations in 2 unknowns.
 
  • #7
Got it! Thank you very much for your help.
 

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