Angle of Queue Ball after elastic collision

1. Oct 24, 2007

luna02525

1. The problem statement, all variables and given/known data

Assume an elastic collision (ignoring friction and rotational motion).
A queue ball initially moving at 2.2 m/s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball's final speed is 0.61 m/s.

Find the queue ball's angle $$\theta$$ with respect to its original line of motion. Answer in units of degrees.

2. Relevant equations

$$\frac{1}{2}mv_1_i^2+\frac{1}{2}mv_2_i^2=\frac{1}{2}mv_1_f^2+\frac{1}{2}mv_2_f^2$$

3. The attempt at a solution

$$\frac{1}{2}mv_1_i^2+\frac{1}{2}mv_2_i^2=\frac{1}{2}mv_1_f^2+\frac{1}{2}mv_2_f^2$$
$$v_1_i^2+v_2_i^2=v_1_f^2+v_2_f^2$$
$$v_2_f=2.114 m/s$$

From here I am unsure of how to come to the angle $$\theta$$ the question is asking for.

I thought it might be:

$$tan\theta=\frac{v_2_f}{v_1_i}$$

This is incorrect, though.

Any guidance would be appreciated!

Last edited: Oct 24, 2007
2. Oct 24, 2007

Dick

You can't determine the angle just by using the speeds. You have to consider that momentum is also conserved, which is a vector quantity. Choose x and y axes and split into components (each of which is conserved).

3. Oct 24, 2007

Vijay Bhatnagar

You have written equations for kinetic energy conservation. Write an equation for conservation of momentum also. Then solve.