physics
posted by joy .
A 0.060 kg tennis ball, moving with a speed 2.50 m/s, collides head on with a 0.090kg ball initially moving away from it at a speed of 1.15 m/s. Assuming a perfectly elastic collision, what are the speed and direction of each ball after the collision?

Treat bthis as a onedimensional problem since it is a headon collision. There is no motion in perpendicular directions.
Problems of this type are most easily done in a coordinate system that moves with the center of mass. Then transform back to laboratory cordinates. You can also treat it only in lab coordinates as two equations in two unknowns, but it takes longer.
The speed of the center of mass of the two balls is Vcm
= (0.06*2.5 + 0.09*1.15)/0.15
= 1.69 m/s
In a coordinate system traveling with the CM, the smaller ball approaches with velocity 2.501.69 = 0.81 m/s and the larger ball approaches with a velocity 1.691.15 = 0.54 m/s
Both balls reverse direction and keep the same speed in the CM system. The preserves total momentum and kinetic energy.
In laboratory coordinates, the final velocity of the smaller ball is 1.69  0.81 = 0.88 m/s (forwards) and the final velocity of the larger ball is
0.54 + 0.81 = 1.35 m/s (also forwards) 
thanks :D but where did the 0.15 come from? :) what equation did you use?

0.15 (kg) is the total mass of both balls. You need it in the denominator to get the speed at which the center of mass moves.

My answer had some typos, as usual. "bthis" in the first paragraph and "The" in the nexttolast paragraph should be "this". Note that in the CM coordinate system, total momentum is zero, before and after the collision.
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