# Elastic Collision and speed of a ball

## Homework Statement

A 120 g ball moving to the right at 4.5 m/s catches up and collides with a 420 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 420 g ball after the collision?

## Homework Equations

$$m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f}$$

## The Attempt at a Solution

I manipulated the formula a little bit to solve for V2_f and came up with this:
$$(m_1v_{1i}+}m_2v_{2i} - m_1v_{1f}) / m_2$$

then I plugged in my numbers:
(.12*4.5 + .42*1.2 - .12*.63) / .42

simplified to this:
(.54 + .504 - .0756) / .42

and got 2.3057. Mastering physics said that is wrong however and I'm not sure what I'm doing wrong. Is the conservation of momentum the correct equation for this problem? I found the velocity of the 120 gram ball fine, which was .63 m/s, but am stuck on this one. Any help would be appreciated.

rl.bhat
Homework Helper
Hi Atlos, welcome to PF.
How did you get v1f = 0.63 m/s?

I came to find v1f by using the following equation:

$$[(m_1-m_2) / (m_1+m_2) ]v_{1i} + [2m_2 / (m_1+m_2]v_{2i}$$

A useful fact for straight line elastic collision is that

v_2b - v_1b = - (v_2a - v_1a)

v_1a = velocity of m1 before collision
v_1b = velocity of m1 after collision etc.

Relative velocity before the collision is the same as after, just opposite in sign.