Question

A 160g ball is dropped from a height of 1.8m , bounces on a hard floor, and rebounds to a height of 1.1m . The figure(Figure 1) shows the impulse received from the floor.

What maximum force does the floor exert on the ball?

Please do not post the answer to a similar problem. I have already looked at those and worked it and keep getting it wrong. Please solve this one.

Answer 1

Use the kinematic equation to find the speed of the ball in case of dropping.

$$ v_{f}^{2}=v_{i}^{2}+2 g h_{\mathrm{trog}} $$

In this case, the initial speed of the ball is zero \(\left(v_{i}=0 \mathrm{~m} / \mathrm{s}\right)\).

Thus, the speed of the ball in case of dropping is,

Take the negative sign for downward motion and take the positive sign for upward motion. For dropping, the speed of the ball is,

$$ v_{0}=-5.939 \mathrm{~m} / \mathrm{s} $$

Similarly, the speed of the ball in case of rebound (upward motion) is,

$$ \begin{aligned} v_{f} &=\sqrt{2 g h_{t o p}} \\ &=\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(1.1 \mathrm{~m})} \\ &=+4.643 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Hence, the change in momentum of the ball is,

$$ \begin{aligned} \Delta p &=p_{f}-p_{i} \\ &=m v_{f}-m v_{0} \\ &=m\left(v_{f}-v_{0}\right) \\ &=\left(160 \times 10^{-3} \mathrm{~kg}\right)(4.643 \mathrm{~m} / \mathrm{s}-(-5.939 \mathrm{~m} / \mathrm{s})) \\ &=1.69312 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \end{aligned} $$

The area under the curve represents the impulse. From figure, the Impulse is,

$$ J=\frac{1}{2} F_{m=x} \Delta t $$

Here, \(F_{-1}\) is maximum force and \(\Delta t\) is maximum force.

From the impulse momentum theorem, the change in momentum is equal to the impulse exerted on the ball. That is,

$$ \begin{aligned} J &=\Delta p \\ \frac{1}{2} F_{m=} \Delta t &=\Delta p \\ F_{m=} &=2 \frac{\Delta p}{\Delta t} \end{aligned} $$

Therefore, the maximum force exerted by the floor exert on the ball is,

$$ \begin{aligned} F_{m=} &=2 \frac{(1.69312 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})}{\left(5.0 \times 10^{-3} \mathrm{~s}\right)} \\ &=677.248 \mathrm{~N} \\ &=6.77 \times 10^{2} \mathrm{~N} \\ &=6.8 \times 10^{2} \mathrm{~N} \quad \text { (approximately) } \end{aligned} $$