Asked by Sarah
A 0.400-kg ball is dropped from rest at a point 1.80 m above the floor. The ball rebounds straight upward to a height of 0.730 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?
Answers
Answered by
drwls
Get the velocity before floor impact from the height H from which it is dropped. Call it V1 (positive down)
V1 = sqrt(2H/g)
Get the velocity just after floor impact from the height H' to which it rises. Call that V2 (positive up)
V2 = sqrt(2H'/g)
The momentum change, which is the impulse, is
M*(V1 + V2)
The impulse on the ball is up, since the momentum increases in the upward direction.
V1 = sqrt(2H/g)
Get the velocity just after floor impact from the height H' to which it rises. Call that V2 (positive up)
V2 = sqrt(2H'/g)
The momentum change, which is the impulse, is
M*(V1 + V2)
The impulse on the ball is up, since the momentum increases in the upward direction.
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