Question

1. 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

Explanation:

a.

Step1: Define the formula for impulse

Impulse \(J=\Delta p = m(v_f - v_i)\)

Step2: Assign values

Let the initial velocity \(v_i = 18.0\space m/s\) (forward), final velocity \(v_f=- 13.5\space m/s\) (backward), and mass \(m = 0.530\space kg\)

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The magnitude of the impulse is \(16.7\space kg\cdot m/s\)

b.

Step1: Use the formula \(J = F\Delta t\)

We can solve for force \(F=\frac{J}{\Delta t}\)

Step2: Substitute values

We know \(J=-16.7\space kg\cdot m/s\) and \(\Delta t = 0.100\space s\)

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The magnitude of the force is \(167\space N\)

Answer:

a. The impulse is \(16.7\space kg\cdot m/s\)
b. The force is \(167\space N\)