Two-Dimensional Elastic Collision

1. Nov 3, 2012

johnhuntsman

A particle with speed v1 = 2.64 × 106 m/s makes a glancing elastic collision with another particle that is at rest. Both particles have the same mass. After the collision, the struck particle moves off at 45º to v1. The speed of the struck particle after the collision is approximately...

I drew a diagram of the scenario. I know that the two final vectors form a right triangle with the initial v1. And I have these equations written out:

v1i = v1f + v2f (since momentum is conserved; the mass can be divided out)

v1i2 = v1f2 + v2f2 (since KE is conserved and 0.5m can be divided out)

The first equation broken up into components:

v1ix = v1fx + v2fx

0 m/s = v1fy + v2fy

What do I do? Maybe I'm the world's least intuitive man, but I don't see how this system of equations can be solved. All I need is someone to tell me where to start and I'll probably be good to go.

 Am I supposed to put the components in terms of sine and cosine and go from there? 

Last edited: Nov 3, 2012
2. Nov 3, 2012

frogjg2003

Write the energy equation in terms of the components.
You now have 3 equations and 4 unknowns (the final velocities, v1x, v1y, v2x, v2y).
There is one more equation you forgot:
v2x = v2y
which comes from the fact that v2 is at 45 degrees to vi.
Now you have 4 equations and 4 unknowns. Have at it.

3. Nov 3, 2012

johnhuntsman

Can you do that? I thought energy didn't have components since it's scalar not vector.

4. Nov 3, 2012

frogjg2003

It doesn't, it's still one equation. You just write it in terms of the individual components instead of the total velocity.
$$v_1^2=v_{1x}^2+v_{1y}^2\\ v_2^2=v_{1x}^2+v_{2y}^2$$

Another approach would have been to replace every v2y and v2x with v2/√2.

Last edited: Nov 3, 2012
5. Nov 3, 2012

johnhuntsman

Alright thanks. I got it. I appreciate it : D