Posted by Jessica on Wednesday, November 7, 2007 at 1:11pm.
A glass ball, ball A, of mass 6.0 g moves at a velocity of 19.0 cm/s. It collides with a second ball, ball B, of mass 9.0 g, moving along the same line of velocity of 11.0 cm/s. After the collison, ball A is still moving, but with a velocity of 8.0 cm/s. a) What was ball A's original momentum? What is ball A's change in momentum? c) What is ball B's change in momentum? d) What is the momentum of ball B after the collison? e) What is ball B's speed after the collison?

Physics  bobpursley, Wednesday, November 7, 2007 at 2:38pm
Linear 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 3:57pm
Ball 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 4:50pm
Ball 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 5:49am
Correct 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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