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March 6, 2015

March 6, 2015

Posted by **joy** on Thursday, August 19, 2010 at 9:03am.

- physics -
**drwls**, Thursday, August 19, 2010 at 9:51amTreat bthis as a one-dimensional problem since it is a head-on 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 2.50-1.69 = 0.81 m/s and the larger ball approaches with a velocity 1.69-1.15 = 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)

- physics -
**joy**, Thursday, August 19, 2010 at 11:27amthanks :D but where did the 0.15 come from? :) what equation did you use?

- physics -
**drwls**, Thursday, August 19, 2010 at 11:56am0.15 (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.

- physics -
**drwls**, Thursday, August 19, 2010 at 12:02pmMy answer had some typos, as usual. "bthis" in the first paragraph and "The" in the next-to-last paragraph should be "this". Note that in the CM coordinate system, total momentum is zero, before and after the collision.

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