# Newton's third law, bullet through wood question

1. Feb 13, 2013

### MightyMuddy

1. A 0.0048kg pullet traveling with a speed of 200m/s penetrates a large wooden fence post. If the average resisting force exerted on the bullet was 4500N how far did the bullet penetrate.

2. F=ma?

3.

F = ma
F/m = a
4500/0.0048 = 937500? So there's a reverse acceleration of 937500m/s?
200/937500 = 0.000213
So it takes 0.000213 seconds for the bullet to stop and it has an average speed of 100m/s.
Therefore the bullet travels 0.0213m or 2.13cm.

Well that's my wrong answer, according to the book it's 8.5cm.

Where did I go wrong, I'm sorry I'm very clueless.

2. Feb 13, 2013

### Staff: Mentor

You didn't go wrong, the book did. (Double check using energy methods.)

3. Feb 13, 2013

### MightyMuddy

Thanks for the response, fairly happy to hear that but unfortunately I'm not too sure what energy methods are? Sorry, I'm very new to learning physics.

4. Feb 13, 2013

### Staff: Mentor

By "energy methods" I just meant for you to use the concept of work and energy to solve the problem. You'll get the same answer, of course.

5. Feb 13, 2013

### haruspex

It's also a poorly worded question. What does "average force" mean? Average over distance or average over time? I feel average over time is the more natural interpretation. But in that case you only have enough info to say how long it took to come to rest. To get the distance as well you need to assume the force was constant.

6. Feb 13, 2013

### Staff: Mentor

Agreed. Sadly, such poorly worded questions are standard fare in most intro physics books.

7. Feb 13, 2013

### Andrew Mason

How would you define average force over distance? If Favg/dist = ΔKE/ΔS it would be the same as average force over time: Favg/time = Δp/Δt = mΔv/Δt = maavg/time

AM

8. Feb 13, 2013

### Staff: Mentor

Why do you think those definitions would be equal, in general?

9. Feb 13, 2013

### haruspex

Suppose the retardation force is from a spring. Accn = -k2x = -k2A sin (kt). As it goes from zero accn to max, change in momentum in time π/k = mkA, so avg accn over time is k2A/π. Change in KE is m(kA)2/2, and the distance is A, so the average accn over distance is k2A/2.

10. Feb 13, 2013

### Andrew Mason

I didn't say they were equal in general. I just meant they were equal here. Maybe I am missing something but I don't see why one would have to assume the force is constant.

The bullet comes to a stop so vf = 0. Since the stopping distance Δs = vavgΔt = (vf+vi)Δt/2 = viΔt/2

$F_{avg/dist} = m\Delta(v^2)/\Delta s = m(v_f^2 - v_i^2)/2\Delta s = mv_i^2/2\Delta s = mv_i/Δt = F_{avg/time}$

AM

Last edited: Feb 13, 2013
11. Feb 13, 2013

### haruspex

They are equal, of course, if the deceleration is constant, and you have assumed that in your equations. Try my SHM example, or perhaps simpler, just consider two different rates of deceleration over the interval.
My complaint is that the question attempted to be more general by mentioning average deceleration, instead of saying it was constant. But if it's not constant then it is necessary to state what it is averaged over.

12. Feb 13, 2013

### SammyS

Staff Emeritus
I agree with haruspex regarding this issue.

13. Feb 14, 2013

### Staff: Mentor

When you use vavg = (vf+vi)/2, you are assuming constant acceleration.

14. Feb 14, 2013

### Staff: Mentor

As do I.

15. Feb 14, 2013

### Andrew Mason

Yes, of course!

As do I!

AM