1 Billard Ball Into Middle of 2 Others

[agross]

Two identical billiard balls are initially at rest when they are struck symmetrically by a third identical ball moving with velocity v0 = v0i (vector)

http://session.masteringphysics.com/problemAsset/1034007/5/RW-09-66.jpg

express final velocity (vf) of third ball immediately after the elastic collision

express veloicty of the other two after elastic collision.

conservation of momentum or kinetic energy

My work
Momentum, and kinetic energy, is conserved in elastic collisions so let's use momentum -

m1v1i + m2v2i = m3v1f + m4v2f
but all masses being equal, it cancels v1i + v2i = v1f + v2f
and the other balls are at rest to begin with, so

v1i = v1f + v2f where the balls transfers momentum to two masses such that
v1i = v1f + (.5)v2f

I feel like I am on thin ice, can anyone help me here? I know that the third ball is going to
roll back wards and the other two balls will shoot out with opposite signs at angle theta off
the axis through their meeting point.
...0
.-.00

 

[LowlyPion]

You need to construct the two equations that are relevant to the elastic collision in this situation.

m1V_o = m₁V₁ +m₂V₂ + m₃V₃

1/2m₁V_o² = 1/2m₁V₁² + 1/2m₂V₂² + 1/2m₃V₃²

Because of the symmetry of the collision I think you can further say that V₃ = V₂