# How do you determine which particle has a greater speed after a collision when you are given the mass and initial velocity?

This is the specific question I refer to (exam practice):

Particle P has mass 3kg and particle Q has mass 2kg. The particles are moving in opposite directions on a smooth horizontal plane when they collide directly. Immediately before the collision, P has speed 3 ms^–1 and Q has speed 2 ms^–1. Immediately after the collision, both particles move in the same direction and the difference in their speeds is 1 ms^–1.

I did the following to (correctly) calculate the speed of each particle:

3kg * 3ms^-1 + 2kg * -2ms^-1 = 3kg * v + 2kg * (v + 1)
...
v = velocity of particle P = 0.6ms^-1
v + 1 = velocity of particle Q = 1.6ms^-1

My question is this: how do I know that the greater speed (v + 1) is for particle Q? Is it because it had the greater momentum before the collision, so it's supposed to have the greater velocity after the collision? If I assume that particle P has the greater velocity after the collision, the answer is different (and incorrect).

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The physical answer is that P can't pass through Q, so if there is a difference Q must be faster. – Ross Millikan Feb 12 '13 at 0:11
@Ross, how do you know which direction they're going, after the collision? – Gerry Myerson Feb 12 '13 at 0:13
@GerryMyerson: The net momentum is positive, so they must be going that way. – Ross Millikan Feb 12 '13 at 0:15
@RossMillikan: Brilliant, I cannot believe I hadn't thought of that. Unless there is a "proper" way to determine the faster particle with the information given, then I'd take that as the answer. – Wk_of_Angmar Feb 12 '13 at 0:17

You don't, but you can see what happens if you assume particle P has greater (more positive) velocity. Then you get that particle Q has velocity $.4\ \textrm{m/s}$ (notice that this is signed velocity, not magnitude), and particle P, $1.4\ \textrm{m/s}$. This is nonsense, though, (as Ross pointed out) since you assumed WLOG that $P$ started to the left of $Q$.