A ball of mass m is dropped from height h on a horizontal floor which collides with it with speed u. If coefficient of restitution in floor is e, then impulse imparted to ball by the floor on its second rebounce is

Options

A. `meu`
B. `meu (e+1)`
C. `me^(2)u(e+1)`
D. `me^(2)u(e -1)`

Answer

To solve the problem of finding the impulse imparted to the ball by the floor on its second rebound, we can follow these steps:

### Step 1: Understand the scenario
A ball of mass \( m \) is dropped from a height \( h \) and collides with the floor. The coefficient of restitution between the ball and the floor is \( e \). We need to determine the impulse imparted to the ball by the floor on its second rebound.

### Step 2: Determine the velocity before the first collision
When the ball is dropped from height \( h \), it gains speed due to gravitational acceleration. The speed \( u \) just before it hits the floor can be calculated using the equation of motion:
\[ u = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Apply the coefficient of restitution
The coefficient of restitution \( e \) relates the velocities before and after the collision. The velocity of the ball after the first collision (upward) can be expressed as:
\[ v = e \cdot u \]
where \( v \) is the velocity of the ball after the first rebound.

### Step 4: Determine the velocity before the second collision
After the first rebound, the ball will rise to a certain height and then fall back down. The velocity just before the second collision (downward) can be calculated using the same principle:
\[ v' = e \cdot v = e \cdot (e \cdot u) = e^2 \cdot u \]

### Step 5: Calculate the impulse during the second collision
Impulse is defined as the change in momentum. The momentum change during the second collision can be calculated as:
\[ \text{Impulse} = m(v' - (-v)) = m(v' + v) \]
Substituting \( v' \) and \( v \):
\[ \text{Impulse} = m(e^2 \cdot u + e \cdot u) \]
Factoring out \( u \):
\[ \text{Impulse} = m \cdot u \cdot (e^2 + e) \]

### Step 6: Final expression for impulse
Thus, the impulse imparted to the ball by the floor on its second rebound is:
\[ \text{Impulse} = m \cdot u \cdot e (e + 1) \]

### Summary
The impulse imparted to the ball by the floor on its second rebound is given by:
\[ \text{Impulse} = m \cdot u \cdot e (e + 1) \]