posted by Bassel .
A billiard ball moving with a speed of +4 m/s runs head on into the back of another ball moving with a speed of +2 m/s. After the collision the first ball was measured to be moving with a speed of +2 m/s. Find the speed of the second ball.
formula = mivi+m2v2= mivi+m2v2
so far I have something like this: m(+4m/s)m2(+2m/s) = m2(+2m/s)+v2. I believe it is wrong.
The balls exchange speeds in an elastic head-on collision. Thus the second ball leaves at 4 m/s.
Note that total momentum and kinetic energy are conserved as a result. No other final velocity conserves both. The result than Ball #1 leaves at 2 m/s could have been predicted by solving both equations.