Question

A white billiard ball with mass mw = 1.38 kg is moving directly to the right with a speed of v = 3.19 m/s and collides elastically with a black billiard ball with the same mass mb = 1.38 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 60.0° and the black ball ends up moving at an angle below the horizontal of θb = 30.0°.

1) What is the final speed of the white ball?

Answer 1

What does it mean when there's an "elastic" collision?

Answer 2

It means all energy remains as kinetic energy.

Answer 3

Conservation of energy equation due to "elastic" collision:
\[\frac{ 1 }{ 2 }m _{w}v _{w}+\frac{ 1 }{ 2 }m _{b}v _{b} = \frac{ 1 }{ 2 }m _{w} v'_{w} + \frac{ 1 }{ 2}m _{b}v'_{b}\] Where v'_w and v'_b equaly the velocity of the two billiard balls after collision. Remember to that angle of their velocities into account and their respective "degrees" as example of the final velocity of the black ball in the y component is

\[\sin(-30.0°)\]

Solve for their vector components of the velocities, for example, in the "x" and "y" direction and then you'll end up with two equations and two unknowns. Solve them simultaneously and you'll get your final velocity of the white ball.

Answer 4

The square of the velocities

Answer 5

^Very true. Add on squares to the velocities of each one. Thank you @wio =)

Answer 6

I nice hint I'd like to add is that there was originally no vertical momentum, so the vertical momentum of the balls is equal in magnitude and opposite in direction.