A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

Answer 1

Impulse is change of momentum, integral of F dt
What you need to know is the speed of the ball down when it hits the floor, call that U (negative down), and the speed of the ball as it rebounds up, call that V (positive up)
then the impulse is m (V - U)
Those two speeds can be gotten from potential and kinetic energy relations.
the potential energy at 1.25 meters (mgh) is the kinetic energy at the floor (.5 m U^2)
so
9.8 * 1.25 = .5 U^2 (the mass on both sides cancels)
U^2 = 24.5
U = -4.95 (remember negative because down)
same deal for V but h in this case is .6 meters
9.8*.6 = .5 V^2
V = +3.43
V - U = 3.43 - (-4.95) = 8.38 m/s
That is the total change in velocity. Multiply by the mass to get the impulse or change in momentum
.120 * 8.38 = 1.01 kg m/s
U =

Answer 2

I thought I'd answered this already, but my answer seems to have disappeared. Please read Damon's response and ignore the nonsense posted by John.

Answer 3

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