A ball of mass m approaches a wall of mass M (M>>m), with speed 2 `ms^(-1)` along the normal to the wall. The speed of wall is 1 `ms^(-1)` towards the ball. The speed of the ball after an elastic collision with the wall is-

Options

A. `5ms^(-1)` away from the wall
B. `9ms^(-1)` away from the wall
C. `3ms^(-1)` away from the wall
D. `4ms^(-1)` away from the wall

Answer

To solve the problem of finding the speed of the ball after an elastic collision with the wall, we can follow these steps:

### Step 1: Understand the initial conditions
- The mass of the ball is \( m \).
- The mass of the wall is \( M \) (where \( M \gg m \)).
- The initial speed of the ball is \( u_b = 2 \, \text{m/s} \) towards the wall.
- The initial speed of the wall is \( u_w = 1 \, \text{m/s} \) towards the ball.

### Step 2: Determine the velocity of approach
The velocity of approach is the sum of the speeds of both the ball and the wall since they are moving towards each other. Thus, we have:
\[ \text{Velocity of approach} = u_b + u_w = 2 + 1 = 3 \, \text{m/s} \]

### Step 3: Use the coefficient of restitution
For an elastic collision, the coefficient of restitution \( e \) is equal to 1. The formula for the coefficient of restitution is:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
Since \( e = 1 \), we can write:
\[ \text{Velocity of separation} = \text{Velocity of approach} = 3 \, \text{m/s} \]

### Step 4: Determine the final velocities
After the collision, the ball will rebound away from the wall. Let \( v \) be the final speed of the ball after the collision. The wall continues to move towards the ball at \( 1 \, \text{m/s} \).

The velocity of separation can be expressed as:
\[ v - (-u_w) = v + 1 \]
Setting this equal to the velocity of approach:
\[ v + 1 = 3 \]

### Step 5: Solve for \( v \)
Now, we can solve for \( v \):
\[ v + 1 = 3 \implies v = 3 - 1 = 2 \, \text{m/s} \]

### Step 6: Consider the direction
Since the ball is rebounding away from the wall, we need to consider the direction. The ball was initially moving towards the wall (positive direction) and after the collision, it will move in the opposite direction (negative direction). Therefore, the final speed of the ball will be:
\[ v = -2 \, \text{m/s} \]

### Conclusion
The speed of the ball after the elastic collision with the wall is \( 2 \, \text{m/s} \) away from the wall.

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