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).
Best Answer
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 $Q$.
Generally speaking, for elastic collision of two particles the ending velocity is uniquely determined by conservation of momentum and conservation of energy. For inelastic collisions, such as the one here, you must use the information given in the problem to determine the final velocity. There is no universal principle about which particle must move faster than the other.