Thursday
March 30, 2017

Post a New Question

Posted by on .

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?

  • physics - ,

    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 =

  • physics - ,

    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 This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question