- #1

jfnn

## Homework Statement

A billiard ball moves at a speed of 4.00 m/s and collides ELASTICALLY with an identical stationary ball. As a result, the stationary ball flies away at a speed of 1.69 m/s. Determine

a. the final speed and direction of the incoming ball after the collision

b. the direction of the stationary ball after the collision

## Homework Equations

[/B]

p = p'

px = px'

py = py'

## The Attempt at a Solution

I assume the net external force is zero, therefore, momentum is conserved in both the x and y direction.

I can say:

m1*v1x + m2*v2x = m1*v1x' + m2*v2x'

Since the two balls are identical, masses cancel. Furthermore, the v2x is 0 because it is stationary.

Therefore,

v1x = v1x' + v2x' --> I want v1x' therefore, v1x' = v1x - v2x'

I know v1x is 4.00 m/s (the initial velocity is only in x direction) so I can write:

__v1x' = 4 - v2x' (this is equation one)__I did the same for the y-direction --> py = py'

The initial velocity for both particles is 0 (particle one is only horizontal and then particle two is stationary), I can now say:

0 = m1*v1y' + m2*v2y' --> This becomes m1*v1y' = -m2*v2y'

The masses cancel because they are identical, and I am left with

__v1y' = -v2y' (this is equation two)__Then, since the collision is elastic, I know that momentum is conserved as well as kinetic energy. So I can write,

1/2m1*v1^2 + 1/2m2*v2^2 = 1/2m1*v1'^2 + 1/2m2*v2'^2

Again, it condenses to:

1/2m1*v1^2 = 1/2m1*v1'^2 + 1/2m2*v2'^2 (masses cancel and 1/2 cancels to give):

v1^2 = v1'^2 + v2'^2 (I know v1 and also v2', so can plug it into the equation to give):

16 = v1'^2 + 2.8561

__Therefore, v1' = 3.63 m/s__

After getting the speed of particle one after the collision, i am lost and can not seem to get the x and y components to do the tangent relationship to find the direction?b) I am lost here too...

Any help will be GREATLY appreciated. Thank you in advance.