Question

A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m. Then the body rebounds to a height of

• A. 1 m
• B. 0.36 m
• C. 0.4 m
• D. 0.6 m

Hint:

Logic Tip: A massive shortcut for bouncing ball problems is the formula $h_n = e^{2n} h_0$, where $h_n$ is the height after the $n$-th bounce. For the first bounce ($n=1$): $h_1 = e^2 h_0$. Substituting the values directly gives $h_1 = (0.6)^2 \times 1 = 0.36 \times 1 = 0.36 \text{ m}$ in less than 5 seconds!

Answer 1

The Correct Option is B

Concept:
The coefficient of restitution ($e$) for a collision between a falling body and a stationary floor is defined as the ratio of the speed of separation (rebound velocity, $v_f$) to the speed of approach (impact velocity, $v_b$): $$e = \frac{v_f}{v_b}$$ From kinematics, the velocity of an object falling from rest from a height $h$ is $v = \sqrt{2gh}$, and the maximum height it reaches when launched upwards with velocity $v$ is $h = \frac{v^2}{2g}$.
Step 1: Determine the impact velocity ($v_b$).
The body falls from an initial height $h = 1 \text{ m}$ with initial velocity $u = 0$. Using the third equation of motion ($v^2 = u^2 + 2gh$): $$v_b^2 = 0^2 + 2g(1)$$ $$v_b = \sqrt{2g} \text{ m/s}$$
Step 2: Determine the rebound velocity ($v_f$).
Using the coefficient of restitution formula $e = \frac{v_f}{v_b}$: $$v_f = e \times v_b$$ Substitute the given value $e = 0.6$: $$v_f = 0.6 \sqrt{2g}$$
Step 3: Calculate the rebound height ($h'$).
After rebounding, the body moves upwards until its final velocity becomes zero at the new maximum height $h'$. $$v^2 = u^2 - 2gh' \implies 0 = v_f^2 - 2gh'$$ $$h' = \frac{v_f^2}{2g}$$ Substitute the expression for $v_f$: $$h' = \frac{(0.6 \sqrt{2g})^2}{2g}$$ $$h' = \frac{0.36 \times 2g}{2g}$$ $$h' = 0.36 \text{ m}$$