How Do You Calculate the Final Speeds of Pool Balls After Collision?

[Kennedy111]

Homework Statement

Mass of Cue stick : 595 g
Mass of White Ball: 170 g
Mass of Black Ball: 155 g

On the pool table, the white ball bounces off of the black ball and moves in the direction opposite to its original direction. If the speed of the white ball immediately after the collision is 3.13 m/s, then the speed of the black ball immediately after the collision is...?

Homework Equations

Σpi = Σpf

p =mΔv

The Attempt at a Solution

0 = (170g)(-3.13m/s) + (155g)(v)
532.1 = 155g(v)
v = 3.43 m/s

This is NOT the correct answer. Apparently the initial velocity of the white ball is not 0, but instead 3.13 m/s... I'm not sure how I find out what the final velocity of the white ball is, and without that information I don't know how to find the final velocity of the black ball... please help..

 

[Simon Bridge]

On the pool table, the white ball bounces off of the black ball and moves in the direction opposite to its original direction. If the speed of the white ball immediately after the collision is 3.13 m/s, then the speed of the black ball immediately after the collision is...?

... cannot be determined: not enough information.

Your calculation appears to assume that the balls are rolling towards each other at the same magnitude momentum before the collision but the problem statement says nothing about this.
[edit] read some more ...

You found out the speed of one ball before the collision, and you have the speed of one ball after the collision ... still not enough information.

To do conservation problems start out by writing the heading "before", under that heading, sketch the "before" situation ... draw the arrows for the vectors in the proper directions and label them.
Under that, use the diagram to write out the before numbers ... in this case:
$\vec{p}_{i} = m_W \vec{v}_{Wi} + m_B \vec{v}_{Bi}$

Repeat for the "after" situation ... then write down $\vec{p}_i=\vec{p}_f$

Then do the math.

Pool balls have extra physics that may be needed:
http://archive.ncsa.illinois.edu/Classes/MATH198/townsend/math.html

 

[rcgldr]

I think there's enough information if you assume that there are no energy losses, so that the total energy before and after the collision is the same.

 

[Simon Bridge]

The assumption of no energy losses would provide extra information... yes.

 

[Kennedy111]

That was litterally all of the information they gave to me. In the answer though they have:
m1v1 + m2v2 = m1v1` + m2v2`
(170g)(3.13m/s) + 0 = (170g)(-0.147 m/s) + (155g)(v2')
v2' = 3.59 m/s

I really just have no idea where they got the -0.147 m/s

 

[rcgldr]

On the pool table, the white ball bounces off of the black ball and moves in the direction opposite to its original direction.

This is only possible if the less massive black ball is initially moving towards the more massive white ball.

In the answer though they have:
m1v1 + m2v2 = m1v1` + m2v2`
(170g)(3.13m/s) + 0 = (170g)(-0.147 m/s) + (155g)(v2')
v2' = 3.59 m/s

This answer seems incorrect. If the black ball initially isn't moving, then the white ball continues to move in the same direction, but at a slower speed.

In the answer though they have:
m1v1 + m2v2 = m1v1` + m2v2`

If only momentum is conserved there should be multiple possible answers. If there are no energy losses, then you have another equation:

1/2 m1 (v1)² + 1/2 m2 (v2)² = 1/2 m1 (v1`)<sup>2</sup> + 1/2 m2 (v2`)²