Question

If a rubber ball falls from a height h and rebounds upto the height of h/2. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

• A. $50\%, \frac{\sqrt{gh}}{2}$
• B. $50\%, \sqrt{gh}$
• C. $40\%, \sqrt{2gh}$
• D. $50\%, \sqrt{2gh}$

Answer 1

Answer: (D)

$50\%, \sqrt{2gh}$

Answer 2

Initial Potential Energy of the Ball:
When the ball is at height $h$, its initial potential energy $PE_{\text{initial}}$ is:
$PE_{\text{initial}} = mgh$
where $m$ is the mass of the ball and $g$ is the acceleration due to gravity.

Potential Energy After Rebound:
After rebounding to a height of $\frac{h}{2}$, the potential energy $PE_{\text{final}}$ of the ball is:
$PE_{\text{final}} = mg \left( \frac{h}{2} \right) = \frac{mgh}{2}.$

Calculate the Percentage Loss of Energy:
The loss in energy $\Delta E$ is given by the difference between the initial and final potential energies:
$\Delta E = PE_{\text{initial}} - PE_{\text{final}} = mgh - \frac{mgh}{2} = \frac{mgh}{2}.$

The percentage loss in energy is:
$\text{Percentage loss} = \frac{\Delta E}{PE_{\text{initial}}} \times 100 = \frac{\frac{mgh}{2}}{mgh} \times 100 = 50\%.$

Calculate the Velocity Before Striking the Ground:
Using energy conservation, the velocity $v$ of the ball just before it strikes the ground can be found from the initial potential energy:
$PE_{\text{initial}} = KE_{\text{impact}}$

$mgh = \frac{1}{2} mv^2$ Solving for $v$:

$v = \sqrt{2gh}.$

Conclusion:
The percentage loss of total energy is 50% , and the velocity of the ball before it strikes the ground is $\sqrt{2gh}$.