A billiard ball traveling at 4.5 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at rest. The first ball is at rest after the collision. What is the speed of the second ball after the collision?
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Momentum (p) = Mass (m) x Velocity (v)
Before After
(M1xV1)+(M2xV2) = (M1xV1)+(M2xV2)
(Mx4.5)+0 = (Mx0) + (MxV)
4.5M = MV
4.5M/M = V
4.5 = V
I think, I'd wait for someone else to agree with my answer just in case.
Before After
(M1xV1)+(M2xV2) = (M1xV1)+(M2xV2)
(Mx4.5)+0 = (Mx0) + (MxV)
4.5M = MV
4.5M/M = V
4.5 = V
I think, I'd wait for someone else to agree with my answer just in case.
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4.5m/sec. All momentum transfers to 2nd. identical ball.