# Perfectly Elastic Collision Problem

1. Apr 26, 2008

### jonnejon

1. The problem statement, all variables and given/known data
A 110 g ball moving to the right at 4.0 m/s catches up and collides with a 400 g ball that is moving to the right at 1.2 m/s.
If the collision is perfectly elastic, what is the speed of the 110 g ball after the collision?

2. Relevant equations
KEi=KEf
1/2m1v1i^2 + 1/2m2v2i = 1/2m1v1f^2 + 1/2m2v2f^2

3. The attempt at a solution
Tried plugging it in but I don't really understand the concept so I am let with 2 final velocity. Please help.

110g < 400g so 110g should move left and 400g should move right, so does that make v2f=0?

2. Apr 26, 2008

### Hootenanny

Staff Emeritus
You have two unknowns and therefore need two simultaneous equations. What else is conserved besides energy?

3. Apr 26, 2008

### jonnejon

Momentum and kinetic energy?

4. Apr 26, 2008

### Hootenanny

Staff Emeritus
Correct! So you already have one equation (conservation of energy), can you now write a second equation using conservation of momentum? You should then have a system of two simultaneous equations.

5. Apr 27, 2008

### jonnejon

Nope, I don't understand how. There is an equation my text book derived for perfectly elastic collisions but it is only if the second mass is at rest.

6. Apr 27, 2008

### Hootenanny

Staff Emeritus
What does the principle of conservation of momentum state?

7. Apr 27, 2008

### jonnejon

Pi=Pf mivi=mfvf

8. Apr 27, 2008

### Hootenanny

Staff Emeritus
Correct, so can you now apply conservation of momentum to your problem?

9. Apr 27, 2008

### jonnejon

So, I solved for m2vf2 in the momentum equation and plug that into the energy equation and solve for vf1 but I got it wrong.

10. Apr 27, 2008

### Hootenanny

Staff Emeritus

11. Apr 27, 2008

### jonnejon

solve for v2f in momentum equation:
v2f = (m1v1i + m2v2i - m1v1f) / m2
v2f = (v1i/m2 + v2i - v1f/m2)
plug it in the energy equation:
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2(v1i/m2 + v2i - v1f/m2)^2
solve for v1f?

12. Apr 27, 2008

### Hootenanny

Staff Emeritus
Your step from the first line to the second is incorrect. Where did the m1 go?

13. Apr 27, 2008

### jonnejon

Opps, I thought it cancels out because of the subtraction sign.

1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1v1i/m2 + v2i - m1v1f/m2)^2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1^2v1i^2/m2^2 + v2i^2 - m1^2v1f^2/m2^2)
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2v1i^2 - m1^2v1f^2)/2m2 - m2v2i^2/2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2 (v1i^2 - v1f^2)/2m2) - m2v2i^2/2
(1/2m1v1i^2 + 1/2m2v2i^2)(2m2)/(m1^2) + m2v2i^2/2 = 1/2m1v1f^2 + (v1i^2 - v1f^2)

Man, this equation just getting to complicated. Am I even doing it right?

14. Apr 27, 2008

### Hootenanny

Staff Emeritus
Yes it is getting rather complicated. I really can't decipher what you've written, you could use latex to make things easier to read,

Conservation of Energy
$$m_1v_{1i}^2+m_2v_{2i}^2 = m_1v_{1f}^2+m_2v_{2f}^2$$

Conservation of Momentum
$$m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f}$$

15. Apr 29, 2008

### jonnejon

I tried doing that but it was just too messy for me. I actually found out how to do it. There are perfectly elastic collision equations in my text book for velocities at rest. But they said you can use the Galilean transformation of velocities to make the 2nd object's velocity into zero and then find the final velocity.

Used:
v' = v - V