Question

A ball falls freely from rest on to a hard horizontal floor and repeatedly bounces. If the velocity of the ball just before the first bounce is 7 m/s and the coefficient of restitution is 0.75, the total distance travelled by the ball before it comes to rest (acceleration due to gravity = 10 ms\(^{-2}\)) is

• A. 10.75 m
• B. 9.75 m
• C. 8.75 m
• D. 11.75 m

Hint:

For problems involving bouncing objects, use the coefficient of restitution to find the height after each bounce. The total distance is the sum of these heights and the fall distances.

Answer 1

The Correct Option is C

Step 1: We are given: - Initial velocity before first bounce = 7 m/s - Coefficient of restitution \( e = 0.75 \) - Acceleration due to gravity \( g = 10 \ \text{m/s}^2 \) We need to calculate the total distance travelled by the ball before it comes to rest. 

Step 2: Height Reached After First Bounce 
From the kinematic equation: \[ v = \sqrt{2gh} \] Since the velocity before the first impact is 7 m/s, \[ h = \frac{v^2}{2g} = \frac{7^2}{2 \times 10} = \frac{49}{20} = 2.45 \ \text{m} \]

 Step 3: Height After Subsequent Bounces 
By the law of restitution, - After the first bounce, the ball's velocity is \( e \times v = 0.75 \times 7 = 5.25 \ \text{m/s} \) Height reached after the first bounce: \[ h_1 = \frac{(5.25)^2}{2 \times 10} = \frac{27.5625}{20} = 1.378 \ \text{m} \] - After the second bounce, the ball's velocity is \( e \times 5.25 = 0.75 \times 5.25 = 3.9375 \) Height after the second bounce: \[ h_2 = \frac{(3.9375)^2}{2 \times 10} = \frac{15.5}{20} = 0.775 \ \text{m} \] - Each subsequent bounce follows a geometric progression (GP) with first term \( 2h_1 = 2 \times 1.378 = 2.756 \) and common ratio \( e^2 = (0.75)^2 = 0.5625 \).

 Step 4: Total Distance Travelled 
Total distance travelled is: \[ \text{Total Distance} = 2h + 2h_1 + 2h_1 e^2 + 2h_1 e^4 + \ldots \] Using the sum of an infinite GP, \[ S = 2h + 2h_1 \left( \frac{1}{1 - e^2} \right) \] \[ S = 2 \times 2.45 + 2 \times 1.378 \left( \frac{1}{1 - 0.5625} \right) \] \[ S = 4.9 + 2.756 \times \frac{1}{0.4375} \] \[ = 4.9 + 2.756 \times 2.2857 \] \[ = 4.9 + 6.3 = 8.75 \ \text{m} \]

 Final Answer: (3) 8.75 m