Question

A ball falls freely from a height h on a rigid horizontal plane. If the coefficient of resolution is e, then the total distance travelled by the ball before hitting the plane second time is: 

• A. he²
• B. h(1+2e²)
• C. h(1-2e²)
• D. h(1+e²)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the total distance travelled by a ball before it hits the plane for the second time after falling freely from a height $h$ onto a rigid horizontal surface, given the coefficient of restitution $e$.

1. Understanding the Motion:
- When the ball is dropped from a height $h$, it travels a distance $h$ downward.
- It then rebounds. The height it reaches after the first bounce is $h_1 = e^2 h$ (since $e = \frac{v_{\text{rebound}}}{v_{\text{initial}}}$ and height is proportional to square of velocity).
- After reaching this new height, it falls back down covering the same distance $e^2 h$.

2. Total Distance Travelled:
Total distance before hitting the ground second time =
Descent (initial fall) + Ascent (after first bounce) + Descent (after reaching max height)
$D = h + e^2 h + e^2 h = h + 2e^2 h = h(1 + 2e^2)$

Final Answer:
The total distance travelled by the ball before hitting the plane the second time is $h(1 + 2e^2)$.

Answer 2

The correct option is: (B) : h(1+2e2).

Prior to the initial collision, the following relationship exists:

mgh = (1/2)mv₁²

This can be further expressed as:

v₁ = √(2gh)

Following the collision, the velocity becomes v₂. Hence, the relationship can be described as:

v₁/v₂ = e Or, v₂ = ev₁

Considering the subsequent calculation of the ball's reached height:

mgh₂ = (1/2)mv₂²

This leads to the equation:

h₂ = (1/2)e²h

The total distance covered before the second impact is determined by:

h + (1/2)h₂ = h + (1/2)e²h = h(1 + 2e²)