Question about cue ball collisions

[JamesL]

Heres the question:

Assume an elastic collision (ignoring friction and rotational motion).

A cue ball initially moving at 3 m/s strikes a stationary eight ball of the same size and mass. After the collision the cue ball's final speed is 1.4 m/s.

Find the eight ball's angle with respec to the original line of motion of the cue ball.

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I can't really post a diagram for you guys to see, but the cue ball is a little above the eight ball (from a birds eye view, left to right). So when the collide the eight ball is moving at a downwardish angle (this is the angle i am supposed to find) and the cue ball is moving upwards.

This question kind of threw me off bc in class we are studying work and kinetic energy etc. All of our homework problems have been about that so far. I can't seem to find a way to incorporate those concepts into this problem.

Should it just be done as a conservation of momentum problem?

Can anyone point me in the right direction?

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[Chen]

Yes, use the conservation of both energy and momentum. The only difference is that this time, you need to divide the momentum into two axes, X and Y:

m_1v_1_x + m_2v_2_x = m_1u_1_x + m_2u_2_x
m_1v_1_y + m_2v_2_y = m_1u_1_y + m_2u_2_y

And finally energy:

\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{1}{2}mu_1^2 + \frac{1}{2}mu_2^2

You can eliminate a couple of unknowns here, since you know the second ball was initally stationary. You also know that the first ball only have speed in one of the axes. Additionally, I think you also know the angle of impact between the two balls, which means you know the direction of their velocities after the impact. So:

v = \frac{v_x}{\cos \theta}

v_y = v_x\tan \theta