A particle of mass m1 and velocity v1 collides elastically (in one dimension) with a particle of mass m2 and velocity v2. What are the velocities of m1 and m2 after the collision?
A particle of mass m1 and velocity v1 collides elastically with a particle of mass m2, with initial velocity v2.
After the collision, m1 has velocity v1', and m2 has velocity v2'. What are v1' and v2'?
To find the velocities of the particles after the collision, you can:
Gee, this seems awfully easy - one simple calculation, a couple of additions and subtractions - does it really work, always? Well, yes, it does. Here is a proof.
Actually, the solution for a one-dimensional elastic collision is even easier than that! It can be shown (quite easily) that
v1_after = 2vcm - v1, and v2_after = 2vcm - v2
A 2 kg meatball with a velocity of 5 m/s collides head-on, elastically, with a 3 kg meatball with a velocity -5 m/s. If the collision is confined to one dimension, what will be their velocities after the collision?
The center of mass of the two-meatball system is easy to find:
The velocity of the 2 kg. meatball is 2vcm - v1 = 2(-1 m/s) - 5 m/s = -7 m/s
The velocity of the 3 kg. meatball is 2vcm - v2 = 2(-1 m/s) - (-5 m/s) = 3 m/s
Can it really be that easy? Yes, it can.
If m1 is much more massive that m2, then:
v1_after = 2vcm - v1 = 2v1 - v1 = v1 (approximately)
The equations predict that the velocity of m1 will be essentially unchanged, which corresponds to experience, and:
v2_after = 2vcm - v2 = 2v1 - v2 (approximately)
If m2 started at rest (v2 = 0), its velocity after the collision will be approximately 2v1.
If the two masses are equal,
In other words, if the masses are equal, the two objects simply exchange velocities in an elastic collision.
If the masses are equal and m2 is initially at rest (a special case of part B above):
The velocities after the collision are:
So the two objects exchange velocities - m1 stops and m2 takes its velocity. Try it on an air track!