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

The Correct Option is D

Initial Potential Energy:
At height \( h \), the ball's initial potential energy \( PE_{\text{initial}} \) is calculated as:
\[ PE_{\text{initial}} = mgh \]
where \( m \) is the ball's mass and \( g \) is the acceleration due to gravity.

Potential Energy Post-Rebound:
After rebounding to a height of \( \frac{h}{2} \), the ball's potential energy \( PE_{\text{final}} \) becomes:
\[ PE_{\text{final}} = mg \left( \frac{h}{2} \right) = \frac{mgh}{2}. \]

Energy Loss Calculation:
The energy loss \( \Delta E \) is the difference between 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 determined by:
\[ \text{Percentage loss} = \frac{\Delta E}{PE_{\text{initial}}} \times 100 = \frac{\frac{mgh}{2}}{mgh} \times 100 = 50\%. \] 

Velocity Pre-Impact Calculation:
Applying energy conservation, the ball's velocity \( v \) just before striking the ground is derived from the initial potential energy:
\[ PE_{\text{initial}} = KE_{\text{impact}} \]

\[ mgh = \frac{1}{2} mv^2 \] Solving for \( v \):

\[ v = \sqrt{2gh}. \] 

Summary:
The total energy loss is calculated to be 50%. The ball's velocity immediately before impact with the ground is \( \sqrt{2gh} \).