Body of mass M is much heavier than the other body of mass m,.The heavier body with speed v collides with thelighter body which was at rest initially elastically The speed of lighter body after collisioin is

Options

A. 2V
B. 3V
C. V
D. `V/2`

Answer

To solve the problem of an elastic collision between two bodies of different masses, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's denote the masses and velocities as follows:

- Mass of the heavier body: \( M \)
- Mass of the lighter body: \( m \)
- Initial velocity of the heavier body: \( v \)
- Initial velocity of the lighter body: \( 0 \) (since it is at rest)
- Final velocity of the heavier body after collision: \( V_1 \)
- Final velocity of the lighter body after collision: \( V_2 \)

### Step 1: Write down the conservation of momentum equation.

The total momentum before the collision must equal the total momentum after the collision. Therefore, we can write:

\[ M \cdot v + m \cdot 0 = M \cdot V_1 + m \cdot V_2 \]

This simplifies to:

\[ M \cdot v = M \cdot V_1 + m \cdot V_2 \quad \text{(1)} \]

### Step 2: Write down the conservation of kinetic energy equation.

Since the collision is elastic, the total kinetic energy before the collision must equal the total kinetic energy after the collision. Therefore, we can write:

\[ \frac{1}{2} M v^2 + \frac{1}{2} m \cdot 0^2 = \frac{1}{2} M V_1^2 + \frac{1}{2} m V_2^2 \]

This simplifies to:

\[ \frac{1}{2} M v^2 = \frac{1}{2} M V_1^2 + \frac{1}{2} m V_2^2 \quad \text{(2)} \]

### Step 3: Analyze the situation for a much heavier mass.

Since \( M \) is much larger than \( m \), we can assume that the velocity of the heavier body \( V_1 \) does not change significantly after the collision. Thus, we can approximate \( V_1 \approx v \).

### Step 4: Substitute \( V_1 \) back into the momentum equation.

Substituting \( V_1 \approx v \) into equation (1):

\[ M \cdot v = M \cdot v + m \cdot V_2 \]

This simplifies to:

\[ 0 = m \cdot V_2 \]

### Step 5: Solve for \( V_2 \).

Since \( m \) is not zero, we can conclude that:

\[ V_2 = 0 \]

This means that the lighter body does not move after the collision, which is not realistic for an elastic collision.

### Step 6: Use the elastic collision formula.

For elastic collisions involving two bodies where one is much heavier, the speed of the lighter body after the collision can be approximated as:

\[ V_2 = \frac{2M}{M+m} \cdot v \]

Since \( M \) is much larger than \( m \), we can simplify this to:

\[ V_2 \approx v \]

### Final Answer:

Thus, the speed of the lighter body after the collision is approximately equal to the speed of the heavier body before the collision:

\[ V_2 \approx v \]