1.
After a 0.320-kg rubber ball is dropped from a height of 18.5 m, it bounces off a concrete floor
and rebounds to a height of 15.0 m. (a) Determine the magnitude and direction of the impulse
delivered to the ball by the floor. magnitude kg m/s directionSelect. (b) Estimate the time the ball
is in contact with the floor to be 0.04 seconds. Calculate the average force the floor exerts on the
ball magnitude directionSelect
Solution
(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(9.8 m/s²)(18.5 m)]
v = 19.04 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(9.8 m/s²)(15 m)]
v = 17.15 m/s
impulse = change in momentum

2.
impulse = (0.320 kg)(17.15 m/s - (-19.04 m/s))
impulse = 11.58 kg-m/s (upward)
(b)
impulse = (force exerted)(time)
11.58 kg-m/s = F(0.04 s)
F = 289.5 N (upward)