# Accounting for momentum in a collision

1. Feb 12, 2014

### mstachowsky

1. The problem statement, all variables and given/known data
Consider a bar of length L, mass M, and uniform density floating in free space (ie: neglect gravity and assume that it is not contacting anything). The bar is initially stationary. Imagine a bullet of mass m travelling at a speed v hits the bar at one end. What is the linear momentum of the bar, the angular momentum of the bar, and the linear momentum of the bullet after the collision?

2. Relevant equations
p_bullet = mv before collision
Momenta of the bar are zero, for both angular and linear.

3. The attempt at a solution
The best I can come up with is that in the collision, we have:

mv = p_bar + Jw_bar + mv2

Where v2 is the speed of the bullet after collision and Jw_bar is the angular momentum of the bar. The problem I'm having is I can't figure out how to find the expressions for p_bar and Jw_bar in terms of the known variables M, m, and v.

2. Feb 12, 2014

### BvU

Hello M, and welcome to PF. I can almost picture this violent event. But the aftermath I can not. If the bar is made of glass (I know, it is not), the bullet just shatters the thing and contimues. If it's made of wood the bullet stays stuck and you can do quite a lot of calculating. If it's made of stainless steel the bullet bounces off and you might do some calculating as well,but different. (in fact bullet itself is a comparable story).

You make good use of the template, for which I am grateful. So I want to help. Is it reasonable to hypothesize that the bullet stays stuck in the bar and the lot starts to move and rotate ?

Did you already learn that the momentum vector of the center of gravity of colliding objects is a constant in these collisions, whether elastic or inelastic ?

3. Feb 12, 2014

### mstachowsky

Thanks for the reply. The way the prof asked the question didn't specify whether the collision was elastic or inelastic. I just assumed that there was some elasticity (like your stainless steel example) so the bullet keeps travelling in a possibly different direction.

We haven't learned about the momentum vector yet, but that sounds like it will help a lot in answering the question. I think the prof asked this question to get us thinking about the vector form of momentum...brilliant So somehow I have to think of the momentum as a vector acting on what exactly? Both objects together?

4. Feb 12, 2014

### BvU

Momentum is movement. And you know that more mass means more momentum, just as well as more speed means more momentum. And because it has a direction it is a vector.

You have me puzzled by asking a good question: what is the momentum acting on ? Well, the thing is that that only comes into play when there is inter-action. It sounds corny, but the action is on (or from) whatever it is interacting with. Can be a gravity field for instance, or another object (collision), whatever.

In situations where there is no external force messing things up, you have momentum conservation:
$$\vec F = {d\vec p \over dt} = 0 \enspace \Rightarrow \enspace\vec p = \vec {constant}$$

And the good news about this momentum conservation is that you can do some nice things with it.

In your relevant equations you only state the mv of the bullet before the collsion, not some conservation of mv.

If I follow your "we haven't learned about the momentum vector yet" and combine it with the original problem formulation, that directly asks for linear and angular momentum, I am a bit puzzled again.

I also have some reservation about mv = p_bar + Jw_bar + mv2. For one, the dimensions don't fit and second: the bar appears twice, which I don't believe. Leave out the Jw and I am a lot happier.

I respect your idea that there is very little wood floating around in space, but then stainless steel is so heavy that my idea is that it's hard to find there too. We might compromise on aluminum. And that the bullet gets stuck in it. Makes things less complicated, but still complicated enough. We can always track back afterwards and talk about bouncing off. And if we forget: you'll run into that kind of exercises soon enough.

So now we can concentrate on

mv = p_bar + mv2 along some x-axis, which I lured you into looking upon as mv = (M+m)v2.

I am extremely lazy, so I even think about making it still simpler by setting m = M. That would leav you with the work of restoring the results for the more general case.

Thas was already quite something. There musts be questions (and I probably also have made errors--so check me) Ask away on this story so far.

I will prepare a next post about the angular momentum.

5. Feb 12, 2014

### mstachowsky

So far so good! Thanks for your help, it's really appreciated. Your reply made me look back at my notes and read "Momentum is a vector but for now let's just consider it in one direction". That's probably why I was ignoring the vector stuff so far and why I thought we hadn't been taught that yet Sorry.

Now, we have:

mv = (M + m)v2 -> this is the total momentum before (left side) and after (right side) the collision. Logically, if m = M, then v2 = 1/2v is that right?

Of course, my big question now becomes: when we add angular momentum, where does it "come from" if momentum is conserved? Maybe I'm mixing up some concepts but this is where I'm getting lost:

The bullet starts by moving with velocity v. Yes there is an external force that causes that to happen but once it's done acting the system is just the already moving bullet and the currently stationary bar, and it's a closed system at that point, correct?

If momentum is conserved, then we have a total amount of momentum equal to mv (I'm again falling back on the whole "it's all in a single direction" to avoid dealing with the vector stuff explicitly). Upon collision, the bar starts moving in a straight line and, because we hit it on the end and not right at the center of mass, it starts to rotate. If we've already accounted for all the momentum in the system: mv = (M+m)v2, then how can there be "left over" momentum to start it rotating? does that make any sense?

Thinking about it from the perspective of conservation of energy makes it even more strange to me, and this way we don't need to think about vectors anyway:

total energy in the system is 1/2 mv^2 before the collision. After the collision, it is still supposed to be 1/2mv^2 since energy is conserved. But, if m = M and v2 = 1/2*v, then the energy of the bullet/bar after the collision is 1/2*(2M*1/4*v^2) = 1/4Mv^2. At this point, we've apparently lost energy (if I've done everything right...) and we haven't even considered rotation, so it's not like we've accounted for the lost energy by dumping it into the rotation. If the system was closed, why did we lose energy? Where did it go?

6. Feb 12, 2014

### BvU

Boy, you type fast. I type slow and read slow.

Yes. The dumb astronaut that fired the thing is flying backwards (and rotating like crazy if he didn't align his center of gravity and the line of firing!)

But I am very sensitive to small mistakes. In particular:
Energy is conserved, but kinetic energy doesn't have to be conserved. In completely elastic collisions it is (that is the definition of an elastic collision...;-)
In the kind I propose here, it is not. Some kinetic energy is transformed into deformation and ultimately into heat.

I had drawn a picture, but you already describe it so nice that it's almost superfluous.
At least I can explain why I liked m=M: I can draw the center of gravity halfway.

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7. Feb 12, 2014

### mstachowsky

Haha my typing speed is directly related to how much coffee I've had and how much I'm trying to avoid studying

I agree that some energy will be lost, for sure, to heat and deformation, sound etc. But didn't we just ignore all of those things in the equations I put forward? Somehow I've failed to account for half of the total energy, unless I've missed the point and the fact that the collision isn't elastic therefore makes us lose the energy?

8. Feb 12, 2014

### BvU

Now we come to the rotation stuff. In the picture I had drawn the total momentum before and after and placed it at the center of mass C.

So on the left there is only one red arrow, which I translated from the bullet to the center of mass, i.e. over a certain distance perpendicular to the vector itself.

That translation can only be done if I add angular momentum. Completely analogous to translating a force and adding torque. Torque and angular momentum are pseudo-vectors, but never mind at the moment.

(And now I come to a point where I have to admit my little plan of picking the center of mass to work on the rotaton stuff isn't useful, because we will have to go to the center of gyration. Quantitatively quite some work. My turn to shout help! It's been so long ago I had to do this...). Qualitatively there is no problem, because for angular stuff you can use any origin.

Angular momentum $\vec L \equiv \vec r \times \vec p$ and it is also conserved (in the absence of external torque being applied, which is OK here). The x is a vector cross product; you know about those?

Pick a point and sum the angular momenta. Lazy me: the top of the bar - at rest. Before the collision: sum is zero. Therefore after the collision: zero also. But now the center of gravity has momentum mv at L/4 distance. So there must be something else moving the other way! Namely the center of gravity of the Bar. (Oh boy, uncertainty whether I should search for the center of gravity, gyration, or still otherwise -- help ! again...)

Ok, go to go for training. Pick up later...

9. Feb 12, 2014

### mstachowsky

Hm, and here we get to my confusion as well: where *does* that extra momentum "come from"?

10. Feb 12, 2014

### haruspex

Here's the rule for picking the reference point for conservation of angular momentum. You can use either:
- any point that's fixed in the (inertial) reference frame; or
- the centre of mass of the complete system.
In problems like this, finding the centre of mas can be a little messy, so I tend to go for the first option. Sometimes the simplest is to use the instantaneous centre of rotation, but here I'd go for that point in space where the centre of the bar was before impact.
The bullet will have angular momentum about that point both before and after impact.
The bar will have angular momentum about that point due to its rotation about its centr, but its overall linear motion will not have any moment about it.
Note that if you were to select the midpoint of the bar (even as it moves) as the reference point you would get all the same numbers except for one: you would underestimate the angular momentum of the bullet after impact.

11. Feb 12, 2014

### BvU

Spent a while chewing on this. Can't think of something decent with m and M, so continue with m=M. Give it a long shot with angular momentum around center of gravity.

Have to correct the picture too: before collision center of gravity is halfway between center of bar and center of bullet, so more to the left than where is was drawn in the picture. Linear momentum still mv ( =(m+M)v/2 ).

W.r.t. center of gravity:

Before collision: bullet has speed v/2 vertical distance L/4, mass m and the bar v/2, L/4, mass M so (m+M)vL/8 .

After collision center of gravity is on the bar. Now my guess/long shot: look at the (conserved) angular momentum with this point as rotation axis and write it as £ = I $\omega$ to at least get some angular velocity $\omega$ .

(I use £ -- a pound sign -- for the angular momentum because the L is already in use for the length of the rod)

I is the moment of inertia and now it boils down to finding he moment of inertia of our bar+bullet around axis through C. A bar mass M length L around axis at end has I = M L2/3.

Bullet at distance L/4: m (L/4)2
bar length L/4: M/4 (L/4)2/3
Remainder of bar: 3M/4 (3L/4)2/3

So something like (m + 7/48 M) (L/4)2

which would yield $\omega = { (m+M) 2 v \over (m+7M/48) L}$

but this was for the case m = M.
So say a 1 kg "bullet" hits a 1 kg bar of 1 m at 1 m/s, then $\omega =$ 3.5, so 1.1 rotations/sec. Who knows, but:

I wouldn't bet on this. Bedtime.

12. Feb 13, 2014

### mstachowsky

Hm, i'm starting to recognize that this may have been a more challenging question that I'd anticipated before I'll have to think a lot more on this, or just ask the prof tomorrow...

13. Feb 13, 2014

### BvU

Like to know what was intended, so if you'd post that I'd be grateful!

14. Feb 13, 2014

### haruspex

I don't understand where this thread has wandered off to. Why are we taking the two masses to be the same? Just use the reference point I suggested in post #10. The problem is not particularly difficult.

15. Feb 13, 2014

### BvU

Mstach: Haru is probably quite right. Give it a shot. Better than my blabbing, and who knows you can write down something a lot more general! Good luck.

16. Feb 14, 2014

### mstachowsky

Oh wow. Here's what he wanted us to think about: consider the fact that collisions don't happen instantly. Can you perform the calculations based on a collision that occurs over some small time period t_o?

So...yeah. Turns out we CAN consider the time it takes, and therefore the impulse required and off to the races we go. Wow. That's what I get for not asking first. Neat thread and great insights, though!

17. Feb 14, 2014

### haruspex

Is 'he' me? I wasn't suggesting considering time. I'm just saying to take the point of reference for conservation of angular momentum to be that point in space where the centre of the bar is at the start.
What is the angular momentum of the bullet about that point before impact?
If the rod moves off at linear speed v' and angular speed ω, what are the angular momenta of the bullet and bar about that point after impact?

18. Feb 14, 2014

### mstachowsky

Oh, no. Sorry. I was talking about the prof. I mentioned that I'd explain what the question was asking when I found out.

19. Feb 14, 2014

### haruspex

Well, I don't see any need for your Prof's approach. It really is not a difficult question. Can you now see how to do it my way?