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 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:
he2
h(1+2e2)
h(1-2e2)
h(1+e2)
The Correct Option is B
Approach Solution - 1
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 $h(1 + 2e^2)$.
Approach Solution -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)
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Concepts Used:
Laws of Motion
The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.
Newton’s First Law of Motion
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion
Newton's 2nd law of motion deals with the relation between force and acceleration. According to the second law of motion, the acceleration of an object as built by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Newton’s Third Law of Motion
Newton's 3rd law of motion states when a body applies a force on another body that there is an equal and opposite reaction for every action.