Newton's third law, bullet through wood question

[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.

 

[Doc Al]

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.

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

 

[MightyMuddy]

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

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.

 

[Doc Al]

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.

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.

 

[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.

 

[Doc Al]

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.

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

 

[Andrew Mason]

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.

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

AM

 

[Doc Al]

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

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

 

[haruspex]

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

AM

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.

 

[Andrew Mason]

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

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 = v_avgΔt = (v_f+v_i)Δt/2 = v_iΔt/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

 

[haruspex]

I didn't say they were general definitions. I just said they were equal here.
The bullet comes to a stop so vf = 0. Since the stopping distance Δs = v_avgΔt = (v_f+v_i)Δt/2 = v_iΔt/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

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.

 

[SammyS]

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.

I agree with haruspex regarding this issue.

 

[Doc Al]

The bullet comes to a stop so vf = 0. Since the stopping distance Δs = v_avgΔt = (v_f+v_i)Δt/2 = v_iΔt/2

When you use v_avg = (v_f+v_i)/2, you are assuming constant acceleration.

 

[Doc Al]

I agree with haruspex regarding this issue.

As do I.

 

[Andrew Mason]

When you use v_avg = (v_f+v_i)/2, you are assuming constant acceleration.

Yes, of course!

I agree with haruspex regarding this issue.

As do I!

AM