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) Determine the magnitude and direction of the impulse delivered to the ball by the floor.

Answer 1

Answer:

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)