Ask Question
1 April, 13:26

After a 0.260-kg rubber ball is dropped from a height of 19.5 m, it bounces off a concrete floor and rebounds to a height of 15.5 m. (a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

+4
Answers (1)
  1. 1 April, 13:28
    0
    a) impulse = 10.7296 kg-m/s (upward)

    b) F = 268.24 N (upward)

    Explanation:

    (a)

    velocity of ball before it strikes the floor:

    initial gravitational potential energy = final kinetic energy

    mgh = (1/2) mv²

    v = sqrt (2gh)

    v = sqrt[2 (9.81 m/s²) (19.5 m) ]

    v = 19.5599 m/s

    velocity of ball after striking the floor:

    initial kinetic energy = final gravitational potential energy

    (1/2) mv² = mgh

    v = sqrt (2gh)

    v = sqrt[2 (9.81 m/s²) (15.5m) ]

    v = 17.4387 m/s

    impulse = change in momentum

    impulse = (0.290 kg) (17.4387 m/s - (-19.5599 m/s))

    impulse = 10.7296 kg-m/s (upward)

    (b)

    impulse = (force exerted) (time)

    10.7296 kg-m/s = F (0.04 s)

    F = 268.24 N (upward)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “After a 0.260-kg rubber ball is dropped from a height of 19.5 m, it bounces off a concrete floor and rebounds to a height of 15.5 m. (a) ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers