Q59GP In a typical squash game (Fig. 1... [FREE SOLUTION]

Chapter 14: Q59GP (page 390)

In a typical squash game (Fig. 14–21), two people hit a soft rubber ball at a wall. Assume that the ball hits the wall at a velocity of 22 m/s and bounces back at a velocity of 12 m/s, and that the kinetic energy lost in the process heats the ball. What will be the temperature increase of the ball after one bounce? (The specific heat of rubber is about\(1200\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}\).)

Short Answer

Expert verified

The increase in temperature of the ball is \(0.14^\circ {\rm{C}}\).

Step by step solution

Given data

The initial velocity of the ball is\({v_{\rm{i}}} = 22\;{\rm{m/s}}\).

The final velocity of the ball is\({v_{\rm{f}}} = 12\;{\rm{m/s}}\).

The specific heat of the rubber is \(c = 1200\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}\).

Understanding loss of kinetic energy

In this problem, to find the increase in temperature, consider that the loss of kinetic energy is converted into heat, which changes the temperature of the ball.

Calculation of the change in temperature

The relation to calculate temperature is given by:

\(\begin{aligned}{c}KE = Q\\\frac{1}{2}m{\left( {{v_{\rm{i}}} - {v_{\rm{f}}}} \right)^2} = mc\Delta T\\\frac{1}{2}{\left( {{v_{\rm{i}}} - {v_{\rm{f}}}} \right)^2} = c\Delta T\end{aligned}\)

Here,\(\Delta T\)is the change in temperature.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}\frac{1}{2}\left( {{{\left( {22\;{\rm{m/s}}} \right)}^2} - {{\left( {12\;{\rm{m/s}}} \right)}^2}} \right) = \left( {1200\;{\rm{J/kg}} \cdot ^\circ {\rm{C}}} \right)\Delta T\\\Delta T = 0.14^\circ {\rm{C}}\end{aligned}\)

Thus, \(\Delta T = 0.14^\circ {\rm{C}}\) is the required change in temperature.