# Two spheres collide and assume that the collision is perfectly elastic

1. Aug 31, 2004

### nullspace7

Two spheres collide and assume that the collision is perfectly elastic. Also --only linear momentum.

I have the relationship:

(va' - vb') dot N = -(va - vb) dot N

Where N is the normal vector at the point of collision. va and vb are initial velocities of object A and B, respectively. And va' and vb' are the final velocities of object A and B respectively.

I want to know how this relationship is derived.

This is what I try:

Relative to object B, object A has velocity v_ab. Relative to object B,
object B has velocity 0.

m_a*v_a + m_b*v_b = m_a*v_a' + m_b*v_b'

Relative to B:

m_a*v_ab + 0= m_a*v_ab' + 0

v_ab = v_ab'

That would give me this: (va' - vb') dot N = (va - vb) dot N

But I am missing the negative sign, because they should be opposite. Please
advise. Thanks in advance.

2. Aug 31, 2004

### geometer

As I'm sure you're aware, in an elastic collision, both kinetic energy and momentum are conserved. If you write both sets of equations and solve them simultaneously, you'll get (va' - vb') = - (va - vb)

3. Aug 31, 2004

### nullspace7

Yeah (needed KE), I've solved it now. I solved it in the center of mass frame, which seemed easier.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook