# Momentum, colliding particles.

• December 10th 2012, 01:04 PM
Furyan
Momentum, colliding particles.
Hello,

I have a question about the following question:

'Two particles A and B are moving on a smooth horizontal plane. The mass of A is km where 2 < k < 3, and the mass of B is m. The particles are moving along the same straight line, but in opposite directions, and they collide directly. Immediately before they collide the speed of A is 2u and the speed of B is 4u. As a result of the collision the speed of A is halved and its direction of motion is reversed.'

There are some questions that follow this statement, which I have answered. The question I have is about knowing what happens to the direction of motion of two particles after they collide. In the last question I had about colliding particles it stated that the direction of both particles was reversed. I answered the question by assuming that the momentum of the two particles before the collision must be the same, which gave me the correct answers. But the question above has me questioning that assumption.

Taking the direction of motion of A to be positive then the momentum of A before the collision is 2kmu Ns and the momentum of B is -4mu Ns.
Since 2 < k < 3, 2kmu > 4mu. I'm probably missing something, but I would assume that since the momentum of A is larger then it is particle B that should change direction. It turns out, according to one of the other questions, that they both change direction. (Headbang)

My question is, can you tell from the relative momentum of two colliding particles what the direction of motion of each particle will be after the collision?

Thank you
• December 10th 2012, 04:18 PM
skeeter
Re: Momentum, colliding particles.
The change in momentum of an object that undergoes a collision with another depends on the impulse imparted on it ... impulse is a joint function of force on the object and the time interval that force acts between the two objects. It is often very difficult, if not impossible to determine impulse before the fact ... variables include elasticity of the objects, angle of incidence, etc. This is why physicists conduct experiments with collisions to determine the impulse between objects after the fact.

go to page 9 of the link for a more thorough explanation ...

http://ruina.tam.cornell.edu/researc...collisions.pdf
• December 10th 2012, 05:09 PM
topsquark
Re: Momentum, colliding particles.
Quote:

Originally Posted by Furyan
I answered the question by assuming that the momentum of the two particles before the collision must be the same, which gave me the correct answers.

Ahhhhh! I'm in pain! (Fubar)

The individual momenta of each object does not have to be the same either before or after the collision and I'm surprised that it gave you the correct answer.

I am assuming that there are no external forces acting on the system (of the two objects) is zero, so the total momentum of the system is constant.

I haven't read over the article skeeter posted, but I'll give a quick run-down anyway.

Call the direction "right" to be positive. Then the total momentum of the system before the collision is going to be
$2ukm - 4um$

The negative sign implies that the object is moving in the negative direction, which is to the left.

The final momentum of the system is
$-ukm + p_B$ <---We don't know the direction or momentum of B after the collision. We need to find this.

By conservation of momentum, these are equal. So....
$2ukm - 4um = -ukm + p_B$

Solving for $p_B$ gives
$p_B = 2ukm - 4um + ukm = 3ukm - 4um = um(3 - 4k)$

Now, we know that 2 < k < 3. In this domain 3 - 4k is always negative. This is a major problem because after the collision B will be moving in the negative direction but at a faster speed than A! (This is a problem since A and B would have to have switched places or moved through each other.) There is something amiss in this problem... If we look at the momentum of A and B before the collision we see that A has more momentum to the right than B does to the left.

In short, this is an impossible problem. A must be going to the right after the collision, but the problem statement says otherwise. A contradiction.

-Dan
• December 11th 2012, 02:33 AM
BobP
Re: Momentum, colliding particles.
$3ukm-4um = um(3k-4).$
• December 11th 2012, 12:20 PM
Furyan
Re: Momentum, colliding particles.
Hello skeeter, thank you for clearing that up. Thank you too for posting the link, 'Everything you wanted to know about rigid body collision, and some, but were to afraid to ask.'

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Hello topsquark, thank you very much for your efforts. I'm afraid I was unclear in my original post. It was a completely different question that I, apparently, solved by assuming the momentum of the colliding particles was the same, although I see now that I was mistaken to make that assumption.

Also for the speed of B after the collision. I got $v = u(3k - 4)$, which is positive. So B has changed direction too. You got that too, but I think I may have talked you into the problem having a problem, which I had clearly done to myself. Sorry about that and sorry it hurt your head. It was hurting mine too.

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Hello Bobp,

Thank you for pointing that out.

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Thank you all very much for your help.