Question

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 1

Step-by-step explanation:

mass of ball, m = 0.530 kg

initial sped, u = 18 m/s

final speed, v = - 13.5 m/s

time, t = 0.1 s

(a) Impulse is defined as the change in momentum.

I = change in momentum

I = m ( v - u)

I = 0.530 x ( - 13.5 - 18)

I = 16.7 N s

(b) Force is defined as the rate of change of momentum.

F = m ( v - u) / t

F = 16.7 / 0.1 = 167 N

Answer 2

Step-by-step explanation:

Change in momentum = m ( v -u )

= .53 ( 18+13.5 )

= 16.7 kg m/s

Impulse = change in momentum

= 16.7 kg m/s

b ) force x time = 16.7

force = 16.7 / .1

= 167 N

force of the wall on the ball

= 167 N.