A basketball player drops a $0.60-k g$ basketball vertically so that it is traveling at $6.0 \mathrm{m} / \mathrm{s}$ when it reaches the floor. The ball rebounds upwardat a speed of $4.2 \mathrm{m} / \mathrm{s}$ . (a) Determine the magnitude and direction of the ball’s change in momentum. (b) Determine the average net force that the floor exerts on the ball if the collision lasts $0.12 \mathrm{s}$ .

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$\text{\color{#4257b2}Basketball dropped vertically (a)}$

The system, the basketball, is not isolated and the basketball has its momentum changed from; $-\alpha p$ to $\beta p$ with $0<\beta<\alpha$ .

The change in momentum is expressed by the generalized impulse-momentum equation;

\begin{gather*} \vec{p} _{\text{Bi}}+\vec{J} _{\text{onB}}=\vec{p} _{\text{Bf}}\\ p _{\text{Biy}}+J _{\text{onBy}}=p_{\text{Bfy}}\\ p _{\text{Bfy}}-p _{\text{Biy}}=J _{\text{onBy}}\\ m(v _{\text{Bfy}}-v _{\text{Biy}})=F _{\text{average}}(t_{\text{f}}-t_{\text{i}})=\Delta p \\ 0.6(4.2-(-6))=F_{\text{average}}(t_{\text{f}}-t_{\text{i}})=\Delta p \\ \Delta p=6.12\mathrm{\ kg\cdot m/s} \end{gather*}

$\text{\color{#4257b2}Basketball dropped vertically (b)}$

Applying (1) with the known momentum change and time interval;

\begin{gather*} 6.12\mathrm{\ kg\cdot m/s}=F _{\text{average}}(t_{\text{f}}-t_{\text{i}})\\ 6.12=-(F _{\text{EonB}}-N _{\text{GonB}})(t_{\text{f}}-t_{\text{i}})\\ 6.12=-(mg-N _{\text{GonB}})\cdot0.12\\ 51=-(0.6\cdot9.8-N _{\text{GonB}})\\ \text{Average force exerted by the ground;}\\ N _{\text{GonB}}=56.9 \mathrm{\ N} \end{gather*}

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