Question
A rubber ball of mass 18.0 g is dropped from a height of 1.75 m onto a floor. The velocity of the ball is reversed by the collision with the floor, and the ball rebounds to a height of 1.55 m. What impulse was applied to the ball during the collision?
Answers
You want the change of momentum.
First speed when hitting floor:
(1/2) m v^2 = m g h
v^2 = 2 g h
v^2 = 2*9.8 * 1.75
v = -5.86 (negative because down)
Now initial speed up
v^2 = 2 * 9.8 * 1.55
v = 5.28
change in velocity = 5.28 -(-5.86)
= 11.14
so change in momentum = .018*11.14 = 0.2 kg m/s
First speed when hitting floor:
(1/2) m v^2 = m g h
v^2 = 2 g h
v^2 = 2*9.8 * 1.75
v = -5.86 (negative because down)
Now initial speed up
v^2 = 2 * 9.8 * 1.55
v = 5.28
change in velocity = 5.28 -(-5.86)
= 11.14
so change in momentum = .018*11.14 = 0.2 kg m/s
The answer is 0,2kgm/s
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