Angle of Queue Ball after elastic collision

[luna02525]

Homework Statement

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.

Homework Equations

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

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!

 

[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).

 

[Vijay Bhatnagar]

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