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 follow these steps:

### Step 1: Understand the Problem
We have two bodies:
- Body 1 (mass \( M \)) is moving with speed \( u \).
- Body 2 (mass \( m \)) is initially at rest.

We need to find the speed of body 2 after the collision when \( M \gg m \).

### Step 2: Apply Conservation of Momentum
In a perfectly elastic collision, the total momentum before the collision equals the total momentum after the collision.

**Initial Momentum:**
\[ p_{\text{initial}} = M \cdot u + m \cdot 0 = M \cdot u \]

**Final Momentum:**
Let \( V_1 \) be the final velocity of mass \( M \) and \( V_2 \) be the final velocity of mass \( m \).
\[ p_{\text{final}} = M \cdot V_1 + m \cdot V_2 \]

Setting initial momentum equal to final momentum:
\[ M \cdot u = M \cdot V_1 + m \cdot V_2 \quad \text{(Equation 1)} \]

### Step 3: Apply the Coefficient of Restitution
For a perfectly elastic collision, the coefficient of restitution \( e = 1 \). This gives us the relationship between the velocities after the collision:
\[ V_2 - V_1 = e \cdot (u - 0) \implies V_2 - V_1 = u \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
From Equation 2, we can express \( V_1 \) in terms of \( V_2 \):
\[ V_1 = V_2 - u \]

Substituting this expression for \( V_1 \) into Equation 1:
\[ M \cdot u = M \cdot (V_2 - u) + m \cdot V_2 \]

Expanding and rearranging:
\[ M \cdot u = M \cdot V_2 - M \cdot u + m \cdot V_2 \]
\[ M \cdot u + M \cdot u = V_2 \cdot (M + m) \]
\[ 2M \cdot u = V_2 \cdot (M + m) \]

### Step 5: Solve for \( V_2 \)
Now we can solve for \( V_2 \):
\[ V_2 = \frac{2M \cdot u}{M + m} \]

### Step 6: Simplify the Expression
Given that \( M \gg m \), we can approximate \( M + m \approx M \):
\[ V_2 \approx \frac{2M \cdot u}{M} = 2u \]

### Conclusion
The speed of the body of mass \( m \) after the collision is approximately \( 2u \).