# Why conservation of energy here vs momentum?

1. Feb 22, 2014

### oddjobmj

1. The problem statement, all variables and given/known data

A billiard ball moving at 2.58 m/s collides elastically with an identical ball initially at rest. After the collision the speed of one ball is 1.36 m/s. What is the speed of the other?

2. Relevant equations

K1+K2=K1'+K2'

or

P1+P2=P1'+P2'

3. The attempt at a solution

I get the correct answer using conservation of energy and the wrong answer using conservation of momentum. My understanding is that conservation of momentum problems work for both elastic and inelastic problems whereas conservation of energy work for elastic problems only (unless energy lost is known).

In this case the masses cancel out so both methods (momentum/energy) actually look the same when solved symbolically other than the energy equation yields squared velocities and the momentum equation does not.

How do I know to use conservation of energy here instead of conservation of momentum? Shouldn't momentum also be conserved?

Thank you!

2. Feb 22, 2014

### Latsabb

Out of curiosity, what are the two answers that you are getting? Being that they are identical, and I am assuming you mean that the collision is completely elastic, conservation of momentum would give you a speed of 1.22, correct?

3. Feb 22, 2014

### oddjobmj

Yes, conservation of momentum yields 1.22 m/s. Conservation of energy yields 2.19 m/s. The correct answer is 2.19 m/s.

I guess it does not say 'perfectly elastically' but just elastically.

4. Feb 22, 2014

### Latsabb

2.19 m/s would assume an increase in overall kinetic energy, which seems very unlikely without the mass of the first ball being larger than that of the second.

Edit: Sorry, I fat fingered this on my calculator. 2.19 by kinetic energy balance is correct, without an increase in overall energy.

Last edited: Feb 22, 2014
5. Feb 22, 2014

### oddjobmj

Nah, that is the result when using conservation of energy. Because (1/2)'s and masses cancel out:

v1i2=v1f2+v2f2

v1i2-v1f2=v2f2

v2f=$\sqrt{v_{1i}^2-v_{1f}^2}$

6. Feb 22, 2014

### Staff: Mentor

Momentum is conserved! But remember that it is a vector. How can you calculate the final momentum of each ball if you don't know the angle between the trajectories of the two balls after collision?

7. Feb 22, 2014

### oddjobmj

Ah, find the magnitude of the resulting vector using the Pythagorean theorem which ends up looking the same as the energy equation.

Strange though that there is no indication or reason to believe they wouldn't both be travelling in the initial direction.

edit: maybe there is? I guess if they were identical balls and the resulting velocity was in the direction of the initial velocity the first ball would stop and the second ball would travel at the initial velocity, right?

8. Feb 22, 2014

### Latsabb

Think of it as one ball hitting the other ball off center. If you have the first ball moving only in the x direction, and the ball nicks the bottom of the second ball, then both balls will end up with y components.

9. Feb 22, 2014

### Staff: Mentor

If a problem involves billiard balls, you should never assume that you can consider the system to be one-dimensional unless specifically told otherwise.

10. Feb 22, 2014

### D H

Staff Emeritus
No, it doesn't. You don't have enough information to solve for the velocity of the other ball using conservation of momentum.

That is correct.

"Perfectly elastic" and "elastic" are synonyms. The term means the collision conserves kinetic energy.

11. Feb 22, 2014

### PeroK

You obviously never seen a game of billiards!

12. Feb 22, 2014

### oddjobmj

I am happy to proceed with the assumption that they are not moving in the same direction if it is not explicitly stated so. However, I guess we do actually have enough information to say that they definitely are not moving in the same direction simply because they are both moving. If the collision were perfectly elastic and it was a one dimensional problem the first ball would stop and the second ball would proceed at the initial ball's velocity.

edit: Hah, PeroK, thanks.

13. Feb 22, 2014

### D H

Staff Emeritus
Exactly.