57. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.
Answer: Answer: (a) -16.7 N s; (b) -167 N
Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s
Find: (a) Impulse, (b) Force
(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)
Impulse = -16.7 kg•m/s = -16.7 N•s
where the "-" indicates that the impulse was opposite the original direction of motion.
(Note that a kg•m/s is equivalent to a N•s)
(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.
Impulse = F•t
F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N
where the "-" indicates that the impulse was opposite the original direction of motion
Answers & Comments
57. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.
Answer: Answer: (a) -16.7 N s; (b) -167 N
Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s
Find: (a) Impulse, (b) Force
(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)
Impulse = -16.7 kg•m/s = -16.7 N•s
where the "-" indicates that the impulse was opposite the original direction of motion.
(Note that a kg•m/s is equivalent to a N•s)
(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.
Impulse = F•t
F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N
where the "-" indicates that the impulse was opposite the original direction of motion