A body of mass M moving with a speed u has a ‘head on’, perfectly elastic collision with a body of mass m initially at rest. If `M > > m` , the speed of the body of mass m after collision, will be nearly :

Options

A. `um//M`
B. `uM//m`
C. `u//2`
D. `2 u`

Explanation

To solve the problem of a perfectly elastic collision between two bodies, we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of body 1 (moving): \( M \)
- Initial speed of body 1: \( u \)
- Mass of body 2 (initially at rest): \( m \)
- Initial speed of body 2: \( 0 \)

2. **Apply the conservation of momentum**:
The total momentum before the collision is equal to the total momentum after the collision.
\[ Mu + 0 = MV_1 + mV_2 \]
where \( V_1 \) is the final velocity of mass \( M \) and \( V_2 \) is the final velocity of mass \( m \).

3. **Apply the conservation of kinetic energy**:
Since the collision is perfectly elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
\[ \frac{1}{2}Mu^2 = \frac{1}{2}MV_1^2 + \frac{1}{2}mV_2^2 \]

4. **Rearranging the momentum equation**:
From the momentum equation, we can express \( V_1 \):
\[ Mu = MV_1 + mV_2 \implies MV_1 = Mu - mV_2 \implies V_1 = \frac{Mu - mV_2}{M} \]

5. **Substituting \( V_1 \) in the kinetic energy equation**:
Substitute \( V_1 \) into the kinetic energy equation:
\[ Mu^2 = MV_1^2 + mV_2^2 \]
Substitute \( V_1 \):
\[ Mu^2 = M\left(\frac{Mu - mV_2}{M}\right)^2 + mV_2^2 \]

6. **Simplifying the equation**:
This will lead to a complex equation, but we can simplify by considering the case where \( M \gg m \). In this case, \( V_1 \) becomes very small compared to \( V_2 \).

7. **Using the elastic collision formula**:
For perfectly elastic collisions, we also have:
\[ V_2 - V_1 = u \]
Rearranging gives:
\[ V_1 = V_2 - u \]

8. **Substituting \( V_1 \) back**:
Substitute this back into the momentum equation:
\[ Mu = M(V_2 - u) + mV_2 \]
Rearranging gives:
\[ Mu = MV_2 - Mu + mV_2 \implies Mu + Mu = MV_2 + mV_2 \]
\[ 2Mu = V_2(M + m) \implies V_2 = \frac{2Mu}{M + m} \]

9. **Considering \( M \gg m \)**:
Since \( M \) is much larger than \( m \), we can approximate:
\[ V_2 \approx \frac{2Mu}{M} = 2u \]

### Final Result:
Thus, the speed of the body of mass \( m \) after the collision will be nearly \( 2u \).