- #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.