Elastic Collisions: some special cases
Let’s examine the elastic collision of two bodies, mass m1 and m2 with initial velocities v1i and v2i, final velocities v1f and v2f. Since the collision is elastic, kinetic energy (Key Point 3.16) and momentum (Key Point 4.10) will be conserved.
- Kinetic energy conservation gives:
- Momentum conservation gives:
m1v1i+m2v2i=m1v1f+m2v2f
To solve these equations for v1f and v2f, it is convenient to rewrite the kinetic energy equation as
m1(v1i-v1f)(v1i+v1f)=-m2(v2i-v2f)(v2i+v2f)and the momentum equation as
m1(v1i-v1f)=-m2(v2i-v2f)After dividing these two equations and doing some more algebra we obtain
and
Special cases
Equal masses; stationary target
So m1=m2=m, and v2=0. The equations then give
v1f=0; v2f=v1iSo particle 1 transfers all its momentum to particle 2.
Massive stationary target
Here m2>m1, and v2=0. The equations then give
The small mass bounces back; the big mass moves off slowly.
Massive projectile: stationary target
This time m1>m2, and v2=0. The equations then give
v1f∼v1i; v2f=2v1iThis is not entirely unexpected. It is the same situation as above, but viewed in a different reference frame; the momentum change of the light particle is ∼2v in both cases.
Center of mass motion
Since no external force acts on the center of mass, it should have a constant velocity. Therefore the center of mass velocity (Key Point 4.2) calculated before and after the collision should be the same.
Before the collision:
After the collision:
Substituting for v1f and v2f (try it) one obtains
as before.