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 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 \( \sqrt{2gh} \).

Answer 2

Step 1: Given data.
The ball is dropped from a height \( h \).
After rebounding, it reaches a height \( \frac{h}{2} \).
We are asked to find:
1. The percentage loss of total mechanical energy of the system.
2. The velocity of the ball just before it strikes the ground.

Step 2: Velocity before striking the ground.
Using the equation of motion for free fall under gravity:
\[ v^2 = 2gh \quad \Rightarrow \quad v = \sqrt{2gh}. \] Thus, the velocity of the ball before impact is \( \sqrt{2gh} \).

Step 3: Energy before and after collision.
The potential energy of the ball before it falls = \( E_1 = mgh. \)
After rebounding, it reaches a height \( \frac{h}{2} \), so its new potential energy = \( E_2 = mg \cdot \frac{h}{2} = \frac{1}{2}mgh. \)

Step 4: Percentage loss in total energy.
\[ \text{Percentage loss} = \frac{E_1 - E_2}{E_1} \times 100 = \frac{mgh - \frac{1}{2}mgh}{mgh} \times 100 = \frac{1}{2} \times 100 = 50\%. \]

Final Answers:
1. Percentage loss of total energy = \( 50\% \)
2. Velocity before striking the ground = \( \sqrt{2gh} \)

Final Answer:
\[ \boxed{50\%, \; \sqrt{2gh}} \]