Conservation of Energy — Question 2

Question 2

Energy accounting through a nonelastic bounce

A 0.50 kg ball is dropped from rest 1.8 m above a floor and rebounds to 0.80 m. Use g=10 m/s$^2$ and ignore air resistance. Include floor and Earth in the system; no energy leaves the boundary during the event. After vibrations settle, all nonmechanical energy is thermal.

1. Find mechanical energy before release and at the rebound apex.

2. Find the eventual thermal-energy increase.

3. Find the ratio of speeds just after and just before impact, and compare it with the mechanical-energy retention fraction.

Question 2 - Solution

Energy accounting through a nonelastic bounce

(a) Use the two zero-speed heights.

At the two height extrema, kinetic energy vanishes: $$\boxed{E_i=mgh_i=.50(10)(1.8)=9\ \mathrm J},\qquad\boxed{E_f=.50(10)(.80)=4\ \mathrm J}.$$

(b) Keep the missing mechanical energy inside the system.

No energy leaves the full system, so after vibrations settle, $$\Delta E_{\mathrm{mech}}+\Delta E_{\mathrm{th}}=0\Rightarrow\boxed{\Delta E_{\mathrm{th}}=9-4=5\ \mathrm J}.$$

(c) Remember the square in kinetic energy.

Speeds immediately before and after contact satisfy $v^2=2gh$: $$\boxed{\frac{v_+}{v_-}=\sqrt{\frac{h_f}{h_i}}=\sqrt{\frac{.80}{1.8}}=\frac23},$$ $$\boxed{\frac{E_f}{E_i}=\frac49=\left(\frac{v_+}{v_-}\right)^2}.$$ For the same mass, energy retention is the square of speed retention.