Question

A ball falls freely from rest from a height of 6.25 m onto a hard horizontal surface. If the ball reaches a height of 81 cm after the second bounce from the surface, the coefficient of restitution is:

• A. \( 0.3 \)
• B. \( 0.45 \)
• C. \( 0.75 \)
• D. \( 0.6 \)

Hint:

The coefficient of restitution \( e \) determines how much energy is conserved in a collision. It is calculated using \( h_n = e^{2n} h_0 \) for multiple bounces.

Answer 1

The Correct Option is D

Step 1: Understanding coefficient of restitution
The coefficient of restitution \( e \) is given by the relation: \[ h_n = e^{2n} h_0. \] where: - \( h_0 \) is the initial height,
- \( h_n \) is the height after \( n \) bounces,
- \( e \) is the coefficient of restitution.
Given:
- Initial height \( h_0 = 6.25 \) m,
- Height after second bounce \( h_2 = 81 \) cm = \( 0.81 \) m.
Step 2: Applying the formula
For the second bounce: \[ h_2 = e^4 h_0. \] Substituting values: \[ 0.81 = e^4 \times 6.25. \] Step 3: Solving for \( e \)
\[ e^4 = \frac{0.81}{6.25}. \] \[ e^4 = 0.1296. \] Taking the fourth root: \[ e = \sqrt[4]{0.1296}. \] \[ e = 0.6. \] Step 4: Conclusion
Thus, the coefficient of restitution is: \[ 0.6. \]