Chapter 6, Problem 52P

Part (a)

To determine

Fraction of its initial energy is lost during the bounce.

Answer to Problem 52P

Solution: 20%

Explanation of Solution

Given:

Initial height of ball $\text{=2}\text{.0}\text{m}$

Maximum height attained by ball after bounce $\text{=1}\text{.6}\text{m}$

Formula Used: P.E of a body at height from a point $\left(\text{reference point}\right)$ $=mgh$

Conservation of mechanical energy.

  $K{E}_2+P{E}_2=K{E}_1+P{E}_1$

When only conservative force is acting.

Calculation:

  

Initial height of ball $h_1=2.0\text{m}$

  $P.E_1=mg{h}_1$

Maximum height attained by ball is $h_2$

  $P{E}_2=mg{h}_2$

Loss in P.E $\left(P.E_2\right)=P.E_1-P.E_2$

  $\begin{array}{l}=mg{h}_1-mg{h}_2\\ =mg\left(h_1-h_2\right).\end{array}$

Fraction of its potential energy losses.

  $\begin{array}{l}=\frac{mg\left(h_1-h_2\right)}{mg{h}_1}\times 100\\ =\frac{h_1-h_2}{h_1}\times 100\\ =\frac{2-1.6}{2}\times 100\\ =\frac{0.4}{2}\times 100\\ =20\%\end{array}$

Conclusion:The fraction of energy lost during the bounce $\text{=20\%}$

Part (b)

To determine

The ball’s speed just before and just after the bounce.

Answer to Problem 52P

Solution: 6.3 m/s (just before bounce), 5.6 m/s (just after bounce)

Explanation of Solution

Given: Initial height of ball $\text{=2}\text{.0}\text{m}$

Maximum height attained by ball after bounce $\text{=1}\text{.6}\text{m}$

Formula Used: P.E of a body at height from a point $\left(\text{reference point}\right)$ $\text{=mgh}$

Conservation of mechanical energy.

  $K{E}_2+P{E}_2=K{E}_1+P{E}_1$

  $\left(\text{when only conservative force is acting}\right)$

Calculation:

  

Before and after bounce only gravitational force is acting on ball (conservative force) so apply conservation of energy before bounce and after bounces.

Initial mechanical energy of ball $\left({\text{E}}_{\text{i}}\right)$

  $\begin{array}{l}=P.E_1+K.E_1\\ =mg{h}_1+\frac{1}{2}m{\left(0\right)}^2\\ =mg{h}_1.\end{array}$

Mechanical energy of ball just before collision $\left({\text{E}}_{\text{f}}\right)$

  $\begin{array}{l}=P.E_2+K.E_2\\ =mg\left(0\right)+\frac{1}{2}m{\left(v_2\right)}^2\end{array}$

Reference position for potential energy is taken ground surface.

  $\begin{array}{l}{E}_i=E_f\\ mg{h}_1=\frac{1}{2}m{\left(v_2\right)}^2\\ v_2=\sqrt{2g{h}_1}\\ =\sqrt{2\times 9.80\times 2}\\ \text{ =6}\text{.26 m/s}\approx \text{6}\text{.3 m/s}\end{array}$

Speed of ball just before bounce $\text{=6}{\text{.26m/s}}^{\text{2}}$ after bounce

Initial mechanical energy of ball $\left({\text{E}}_{\text{i}}\right)$

  $\begin{array}{l}=P.E_1+K.E_1\\ =mg\left(0\right)+\frac{1}{2}m{v}_1{}^2\\ =\frac{1}{2}m{v}_1{}^2\end{array}$

Final mechanical energy of ball $\left({\text{E}}_{\text{f}}\right)$

  $\begin{array}{l}=P.E_2+K.E_2\\ =mg{h}_2+\frac{1}{2}m{\left(0\right)}^2\left(\text{at the heighest point velocity=0}\right)\\ =mg{h}_2\end{array}$

  $\begin{array}{l}\frac{1}{2}m{v}_2{}^2=mg{h}_2\left(\text{conservation of energy}\right)\\ v_2=\sqrt{2g{h}_2}\\ =\sqrt{2\times 9.80\times 1.60}\\ \text{ =5}\text{.6 m/s}\text{.}\end{array}$

Conclusion:

Speed of ball just after bounce $\text{=5}\text{.6 m/s}$

Part (c)

To determine

Where did the energy go

Answer to Problem 52P

Solution:Into heat and sound energy

Explanation of Solution

Loss in mechanical energy is converted into heat and sound energy during collision with Earth’s surface.

Conclusion:

Loss in mechanical energy is converted into sound and heat energy.