A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Solution:
The concept is the same with Problem 2.56.
Part A
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\
v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right)} \\
v_{y_2} & = - \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\
v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part B
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1 \right) \\
\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right) \\
v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1 \right)} \\
v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2 \right) \left( 1.10 \ \text{m} - 0 \ \text{m} \right) }\\
v_{y_1} & = 4.65 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}Part C
\begin{align*}
a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{4.65 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{3.50 \times 10^{-3} \ \text{s}} \\
a & = 2877 \ \text{m/s}^2 \\
a & = 2.88 \times 10^{3}\ \text{m/s}^2 \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}Part D
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 2.88 \times 10^3 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.00510 \ \text{m} \\
\Delta y & = -5.10 \times 10^{-3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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