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 of the ball in the collision. Heat capacity of the ball is 800 J K⁻¹.

Answer 1

Height of the floor from which ball is dropped, h₁ = 2.0 m
Height to which the ball rises after collision, h₂ = 1.5 m
Let the mass of ball be m kg.
Let the speed of the ball when it falls from h₁ and h₂ be v₁ and v₂, respectively.

$$\begin{gathered} v_1=\sqrt{2g{h}_1}=\sqrt{2\times 10\times 2}=\sqrt{40}\mathrm{m}/\mathrm{s} \\ {v}_2=\sqrt{2g{h}_2}=\sqrt{2\times 10\times 1.5}=\sqrt{30}\mathrm{m}/\mathrm{s} \end{gathered}$$

Change in kinetic energy is given by
$$\begin{gathered} ∆K=\frac{1}{2}\times \mathrm{m}\times 40-\left(\frac{1}{2}\mathrm{m}\right)\times 30=\left(\frac{10}{2}\right)\mathrm{m} \\ \Rightarrow ∆K=5\mathrm{m} \end{gathered}$$

If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,
Loss in PE = 0
The change in kinetic energy is utilised in increasing the temperature of the ball.

Let the change in temperature be ΔT. Then,
$$\begin{gathered} \left(\frac{40}{100}\right)\times ∆K=m\times 800\times ∆T \\ \left(\frac{40}{100}\right)\times \frac{10}{2}m=m\times 800\times ∆T \\ \Rightarrow ∆T=\frac{1}{400}=0.0025 \\ =2.5\times {10}^{-3}°\mathrm{C} \end{gathered}$$