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October 31, 2014

October 31, 2014

Posted by **Jessica** on Wednesday, November 7, 2007 at 1:11pm.

- Physics -
**bobpursley**, Wednesday, November 7, 2007 at 2:38pmLinear momentum is conserved. Whatever the momentum A lost in the collision, B gained.

I will be happy to critique your thinking.

- Physics -
**Anonymous**, Wednesday, November 7, 2007 at 3:57pmBall A's original momentum is .00114 kgm/s. The change in Ball A's momentum would be .006 kg x (.19 - .08) m/s, correct? Ball B's change in momentum would be .009 kg x (.11 - v) m/s. The momentum of Ball B would be .009 x the change in velocity from the previous question. Ball B's speed would be the original velocity minus the change in velocity.

- Physics -
**Jessica**, Wednesday, November 7, 2007 at 4:50pmBall A's original momentum is .00114 kgm/s. The change in Ball A's momentum would be .006 kg x (.19 - .08) m/s, correct? Ball B's change in momentum would be .009 kg x (.11 - v) m/s. The momentum of Ball B would be .009 x the change in velocity from the previous question. Ball B's speed would be the original velocity minus the change in velocity.

- Physics -
**bobpursley**, Thursday, November 8, 2007 at 5:49amCorrect on the change of A momentum.

Ball B change in momentum is equal to the change in A momentum, not the change in A velocity.

New Ball B momentum:

.009*.11+ .006(.19-.08)

From that, you determine B velocity (by dividing the Momentum my mass B)

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