A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of

Question

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that $40 \%$ of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature(in mK) of the ball in the collision. Heat capacity of the ball is $800 \mathrm{~J} \mathrm{~K}^{-1}$.

Professor PC Thomas Classes · Answer

To calculate the rise in temperature, we equate the mechanical energy lost during the collision to the heat energy absorbed by the ball.
Mechanical Energy Lost ( $\Delta E$ )
The loss in potential energy as the ball rebounds from 2.0 m to 1.5 m is:
$
\Delta E=m g\left(h_1-h_2ight)
$
Using $g=9.8 \mathrm{~m} / \mathrm{s}^2$ :
$
\Delta E=m 	imes 9.8 	imes(2.0-1.5)=4.9 m \mathrm{~J}
$
Thermal Energy Absorbed ( $Q$ )
Since $40 \%$ of the lost energy is converted into heat within the ball:
$
Q=0.40 	imes \Delta E=0.40 	imes 4.9 \mathrm{~m}=1.96 \mathrm{~m} \mathrm{~J}
$
Temperature Rise ( $\Delta T$ )
Assuming the provided "heat capacity" ( $800 \mathrm{~J} \mathrm{~K}^{-1}$ ) refers to the specific heat capacity ( $s$ ) of the ball's material:
$
\begin{gathered}
Q=m s \Delta T \
1.96 m=m 	imes 800 	imes \Delta T \
\Delta T=\frac{1.96}{800}=0.00245 \mathrm{~K}
\end{gathered}
$