Question:
A ball of mass 0.5 kg is dropped freely from a point A which is at a height of 10 m from the ground. Between second and third collisions with the ground, the linear momentum of the ball becomes zero at a point B. If the coefficient of restitution between the ball and the ground is 0.5, then the percentage loss of the potential energy of the ball when it reaches point B is
A ball of mass 0.5 kg is dropped freely from a point A which is at a height of 10 m from the ground. Between second and third collisions with the ground, the linear momentum of the ball becomes zero at a point B. If the coefficient of restitution between the ball and the ground is 0.5, then the percentage loss of the potential energy of the ball when it reaches point B is
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After $n$ rebounds, energy scales as $e^{2n}$.
Updated On: Jun 22, 2026
- 23.25
- 83.75
- 6.25
- 93.75 \bigskip
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The Correct Option is D
Solution and Explanation
Concept:
Each collision reduces velocity by coefficient of restitution $e$. Energy is proportional to square of velocity, so energy decreases by factor $e^2$ after each bounce.
Step 1: Velocity before first impact.
From height $h=10$: \[ v = \sqrt{2gh} = \sqrt{200} \]
Step 2: After first rebound.
\[ v_1 = ev = 0.5\sqrt{200} \]
Step 3: After second rebound.
\[ v_2 = e^2 v = 0.25\sqrt{200} \]
Step 4: Energy comparison.
\[ E \propto v^2 \Rightarrow E_2 = e^4 E_0 \] \[ E_2 = (0.5)^4 = \frac{1}{16} \] So remaining energy = $6.25%$
Step 5: Percentage loss.
\[ = 100 - 6.25 = 93.75% \]
Step 1: Velocity before first impact.
From height $h=10$: \[ v = \sqrt{2gh} = \sqrt{200} \]
Step 2: After first rebound.
\[ v_1 = ev = 0.5\sqrt{200} \]
Step 3: After second rebound.
\[ v_2 = e^2 v = 0.25\sqrt{200} \]
Step 4: Energy comparison.
\[ E \propto v^2 \Rightarrow E_2 = e^4 E_0 \] \[ E_2 = (0.5)^4 = \frac{1}{16} \] So remaining energy = $6.25%$
Step 5: Percentage loss.
\[ = 100 - 6.25 = 93.75% \]
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