A particle of mass `m` is moving horizontally with a constant velocity `v` towards a rigid wall that is moving in opposite direction with a constant speed `u`. Assuming elastic impact between the particle and wall, the work done by the wall in reflecting the particle is equal to

Options

A. `(1/2)m(u+v)^(2)`
B. `(1/2)m(u+v)`
C. `(1/2)"mu"v`
D. none of these

Explanation

To solve the problem, we need to determine the work done by the wall in reflecting the particle after an elastic collision. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the initial conditions
The particle of mass `m` is moving towards a wall with a velocity `v`, while the wall is moving in the opposite direction with a velocity `u`.

### Step 2: Determine the velocity of approach
Before the collision, the total velocity of approach is the sum of the velocities of the particle and the wall:
\[ \text{Velocity of approach} = v + u \]

### Step 3: Use the principle of elastic collision
In an elastic collision, the coefficient of restitution (e) is equal to 1. This means that the velocity of separation after the collision will equal the velocity of approach before the collision.

### Step 4: Write the equation for velocity of separation
Let \( v' \) be the velocity of the particle after the collision. The velocity of separation can be expressed as:
\[ \text{Velocity of separation} = v' - (-u) = v' + u \]
According to the principle of restitution:
\[ v' + u = v + u \]
This simplifies to:
\[ v' = 2u + v \]

### Step 5: Calculate the change in kinetic energy
The work done by the wall on the particle is equal to the change in kinetic energy of the particle. The initial kinetic energy (KE_initial) and final kinetic energy (KE_final) can be expressed as:
\[ \text{KE_initial} = \frac{1}{2} m v^2 \]
\[ \text{KE_final} = \frac{1}{2} m (v')^2 = \frac{1}{2} m (2u + v)^2 \]

### Step 6: Calculate the work done
The work done (W) by the wall is the change in kinetic energy:
\[ W = \text{KE_final} - \text{KE_initial} \]
Substituting the expressions for kinetic energy:
\[ W = \frac{1}{2} m (2u + v)^2 - \frac{1}{2} m v^2 \]

### Step 7: Simplify the expression
Using the formula \( a^2 - b^2 = (a - b)(a + b) \):
\[ W = \frac{1}{2} m \left( (2u + v)^2 - v^2 \right) \]
\[ = \frac{1}{2} m \left( (4u^2 + 4uv + v^2) - v^2 \right) \]
\[ = \frac{1}{2} m (4u^2 + 4uv) \]
\[ = 2m(2u^2 + 2uv) \]

Thus, the work done by the wall in reflecting the particle is:
\[ W = 2m(u^2 + uv) \]

### Final Answer
The work done by the wall in reflecting the particle is:
\[ W = 2m(u^2 + uv) \]