Consider an elastic collision (ignoring friction and rotational motion).
A queue ball initially moving at 2.6 m/s strikes a stationary eight ball of the same size
and mass. After the collision, the queue ball’s final speed is 1.2 m/s.
Find the queue ball’s angle (theta) with respect to its original line of motion.
Answer in units of ◦.
I've attached a picture to make it a little clearer
kinetic energy = 1/2mv^2
momentum = mv
The Attempt at a Solution
I've tried a couple of things, but its really the math that's tripping me up I think.
First I found the velocity of the ball that was hit. I got 3.504...etc. using the kinetic energy equation.
The next thing I did was try to solve it in terms of the two angles, theta and phi. Since the two y-components must be equal to zero (because there was no y component to the original velocity),
Eq. 1: 0 = 1.2 sin(theta) + 3.504 sin(phi)
And because the x components must add up to the original x velocity,
Eq. 2: 2.6 = 1.2 cos(theta) + 3.504 cos(phi)
When I try to make these into a system, I get a very unruly equation that leaves me dumbfounded.
My alternative was to separate each of the two velocity vectors into their components, and this ended up getting me a four equation system which again, left me dumbfounded.
By similar logic to above, the four equations I came up with were:
vxcue + vxother = 2.6
vy cue + vy other = 0
(vx cue)2 + (vy cue)2 = 1.2
(vx other)2 + (vy other)2 = 3.504
Note: Everytime I typed 3.504 I actually used the full calculator value.
So my question, I guess, is whether or not there is a simpler way to solve this, and if not, if anyone could give me a hand with the math, its confusing me big time.