Question

A ball is dropped from some height and after first collision with the ground if it reaches $ \frac{3}{4} $th of its original height, then the % loss of its energy is: Options:

• A. \( 25 \)
• B. \( 75 \)
• C. \( 50 \)
• D. \( 55 \)

Hint:

The fraction of energy retained after a collision is related to the coefficient of restitution \( e \). The energy ratio is \( e^2 \), and the percentage loss is \( (1 - e^2) \times 100 \).

Answer 1

The Correct Option is A

We need to find the percentage loss of energy after the ball rebounds to \( \frac{3}{4} \) of its original height. Step 1: Determine the initial and final energies. The ball is dropped from height \( h \). Initial energy (just before collision) is: \[ E_{\text{initial}} = mgh \] After collision, it rebounds to \( \frac{3}{4}h \). Final energy (at the top of the rebound): \[ E_{\text{final}} = mg \left( \frac{3}{4}h \right) = \frac{3}{4} mgh \] Step 2: Calculate the fraction of energy retained. \[ \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{\frac{3}{4} mgh}{mgh} = \frac{3}{4} \] Step 3: Calculate the percentage loss of energy. Fraction of energy lost: \[ 1 - \frac{E_{\text{final}}}{E_{\text{initial}}} = 1 - \frac{3}{4} = \frac{1}{4} \] \[ \text{Percentage loss} = \frac{1}{4} \times 100 = 25% \] Final Answer: \[ \boxed{25} \]