The centres of three spheres 1,2 and 3 lies on a single straight line . Sphere 1 is moving with an initial speed ` v_(1)` directed along this line towards sphere 2. Spheres 2 and 3 are initially at rest. Acquiring some speed after collision, sphere 2 hits sphere 3. Sphere 1 and 3 have masses `m_(1)` and `m_(3)` , respectively , and all the collisions are perfectly elastic and head on . Then.

Options

A. `m_(3)` will acquire maximum speed when `m_(2) = sqrt(m_(1) m_(3))`
B. `m_(3)` will acquire maximum speed equal to `( 4m_(1)v_(1))/( (sqrt(m_(1))+sqrt(m_(3)))^(2))`
C. `m_(3)` will acquire maximum speed when `m_(2) = ( m_(1) + m_(3))/( 2)`
D. `m_(3)` will acquire maximum speed equal to `( m_(1) m_(3))/( 2)`

Answer

To solve the problem, we will analyze the collisions between the spheres step by step, applying the principles of conservation of momentum and kinetic energy, since the collisions are perfectly elastic.

### Step 1: Analyze the first collision between Sphere 1 and Sphere 2

1. **Initial momentum** of the system before the collision:
\[ p_{initial} = m_1 v_1 + m_2 \cdot 0 = m_1 v_1 \]

2. **Final momentum** of the system after the collision:
\[ p_{final} = m_1 v_1' + m_2 v_2' \]

3. **Conservation of momentum** gives us:
\[ m_1 v_1 = m_1 v_1' + m_2 v_2' \quad \text{(1)} \]

4. For perfectly elastic collisions, we also have:
\[ v_2' - v_1' = v_1 - 0 \quad \text{(2)} \]

### Step 2: Solve the equations

From equation (2):
\[ v_2' = v_1 + v_1' \quad \text{(3)} \]

Substituting equation (3) into equation (1):
\[ m_1 v_1 = m_1 v_1' + m_2 (v_1 + v_1') \]
Rearranging gives:
\[ m_1 v_1 = m_1 v_1' + m_2 v_1 + m_2 v_1' \]
\[ m_1 v_1 - m_2 v_1 = (m_1 + m_2) v_1' \]
\[ v_1' = \frac{(m_1 - m_2)v_1}{m_1 + m_2} \quad \text{(4)} \]

### Step 3: Find the velocity of Sphere 2 after the first collision

Substituting equation (4) back into equation (3):
\[ v_2' = v_1 + \frac{(m_1 - m_2)v_1}{m_1 + m_2} \]
\[ v_2' = \frac{(m_1 + m_2)v_1}{m_1 + m_2} + \frac{(m_1 - m_2)v_1}{m_1 + m_2} \]
\[ v_2' = \frac{(2m_1)v_1}{m_1 + m_2} \quad \text{(5)} \]

### Step 4: Analyze the second collision between Sphere 2 and Sphere 3

1. **Initial momentum** of the second collision:
\[ p_{initial,2} = m_2 v_2' + m_3 \cdot 0 = m_2 v_2' \]

2. **Final momentum** after the collision:
\[ p_{final,2} = m_2 v_2'' + m_3 v_3'' \]

3. **Conservation of momentum** gives:
\[ m_2 v_2' = m_2 v_2'' + m_3 v_3'' \quad \text{(6)} \]

4. For the second elastic collision:
\[ v_3'' - v_2'' = v_2' - 0 \quad \text{(7)} \]

### Step 5: Solve the equations for the second collision

From equation (7):
\[ v_3'' = v_2' + v_2'' \quad \text{(8)} \]

Substituting equation (8) into equation (6):
\[ m_2 v_2' = m_2 v_2'' + m_3 (v_2' + v_2'') \]
\[ m_2 v_2' = m_2 v_2'' + m_3 v_2' + m_3 v_2'' \]
Rearranging gives:
\[ m_2 v_2' - m_3 v_2' = (m_2 + m_3) v_2'' \]
\[ v_2'' = \frac{(m_2 - m_3)v_2'}{m_2 + m_3} \quad \text{(9)} \]

### Step 6: Find the velocity of Sphere 3 after the second collision

Substituting equation (9) back into equation (8):
\[ v_3'' = v_2' + \frac{(m_2 - m_3)v_2'}{m_2 + m_3} \]
\[ v_3'' = \frac{(m_2 + m_3)v_2'}{m_2 + m_3} + \frac{(m_2 - m_3)v_2'}{m_2 + m_3} \]
\[ v_3'' = \frac{(2m_2)v_2'}{m_2 + m_3} \quad \text{(10)} \]

### Step 7: Substitute \(v_2'\) from equation (5) into equation (10)

Substituting \(v_2'\) from equation (5):
\[ v_3'' = \frac{(2m_2) \cdot \frac{(2m_1)v_1}{m_1 + m_2}}{m_2 + m_3} \]
\[ v_3'' = \frac{4m_1 m_2 v_1}{(m_1 + m_2)(m_2 + m_3)} \quad \text{(11)} \]

### Step 8: Find the condition for maximum speed of Sphere 3

To maximize \(v_3''\), we need to minimize the denominator \( (m_1 + m_2)(m_2 + m_3) \).

Taking the derivative and setting it to zero will give us the optimal mass \(m_2\) for maximum speed of Sphere 3.

### Final Result

After performing the necessary calculations, we find that the maximum speed of Sphere 3 occurs when:
\[ m_2 = \sqrt{m_1 m_3} \]

Thus, the correct option is:
**B: Sphere 3 will acquire maximum speed equal to \( \frac{4m_1 v_1}{\sqrt{m_1} + \sqrt{m_3}}^2 \)**.