# Elastic Collision with a Softball

1. Mar 6, 2012

### PeachBanana

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

A softball of mass 0.220 kg that is moving with a speed of 8.4 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.4 m/s.

2. Relevant equations

KE final = KE initial

3. The attempt at a solution

0.5(-3.4 m/s)^2 + 0.5(V)^2 = 0.5(8.4 m/s)^2 + 0.5(0 m/s)^2
5.78 m^2/s^2 + 0.5(V)^2 = 35.28 m^2/s^2
0.5 V^2 = 29.5 m^2/s^2
v^2 = 59 m^2/s^2
v = 7.68 m/s

Which is incorrect and I'm not sure why.

2. Mar 6, 2012

### Staff: Mentor

You seem to have assumed that the second ball has the same mass as the first baseball. The problem doesn't explicitly say that, so I think you have to assume that they are not the same.

If I may make a small suggestion; In problems where you need to apply both conservation laws (momentum, energy) to figure out the unknowns, consider this equivalent rule to the conservation of energy: For elastic collisions the relative speed of the colliding bodies after collision is equal to the negative of their relative speeds before collision. So if v1 and v2 represent the initial speeds of bodies 1 and 2, and if u1 and u2 represent their speeds after collision, then

(v1 - v2) = -(u1 - u2)

This may make your mathematical life easier

3. Mar 6, 2012

### PeachBanana

Thank you for the advice. I tried something else and got the right answer.