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 involving three spheres undergoing perfectly elastic collisions, we can break it down into a series of steps:

### Step 1: Define Initial Conditions
- Let the mass of sphere 1 be \( m_1 \) and its initial velocity be \( v_1 \).
- Let the mass of sphere 2 be \( m_2 \) and its initial velocity be \( 0 \) (at rest).
- Let the mass of sphere 3 be \( m_3 \) and its initial velocity also be \( 0 \) (at rest).

### Step 2: First Collision (Sphere 1 and Sphere 2)
- After the first collision, let the final velocities be \( v_2 \) for sphere 1 and \( v_3 \) for sphere 2.
- Apply the conservation of momentum:
\[ m_1 v_1 = m_1 v_2 + m_2 v_3 \quad \text{(Equation 1)} \]

- Since the collision is elastic, we also use the coefficient of restitution:
\[ v_1 - 0 = v_3 - v_2 \quad \text{(Equation 2)} \]
Rearranging gives:
\[ v_2 = v_3 - v_1 \]

### Step 3: Substitute Equation 2 into Equation 1
- Substitute \( v_2 \) from Equation 2 into Equation 1:
\[ m_1 v_1 = m_1 (v_3 - v_1) + m_2 v_3 \]
Simplifying this gives:
\[ m_1 v_1 = m_1 v_3 - m_1 v_1 + m_2 v_3 \]
Rearranging leads to:
\[ 2m_1 v_1 = (m_1 + m_2) v_3 \]
Thus, we find:
\[ v_3 = \frac{2m_1 v_1}{m_1 + m_2} \quad \text{(Equation 3)} \]

### Step 4: Second Collision (Sphere 2 and Sphere 3)
- After the second collision, let the final velocities be \( v_4 \) for sphere 2 and \( v_5 \) for sphere 3.
- Apply conservation of momentum again:
\[ m_2 v_3 = m_2 v_4 + m_3 v_5 \quad \text{(Equation 4)} \]

- The coefficient of restitution gives:
\[ v_3 - 0 = v_5 - v_4 \quad \text{(Equation 5)} \]
Rearranging gives:
\[ v_4 = v_5 - v_3 \]

### Step 5: Substitute Equation 5 into Equation 4
- Substitute \( v_4 \) from Equation 5 into Equation 4:
\[ m_2 v_3 = m_2 (v_5 - v_3) + m_3 v_5 \]
Simplifying leads to:
\[ m_2 v_3 = m_2 v_5 - m_2 v_3 + m_3 v_5 \]
Rearranging gives:
\[ m_2 v_3 + m_2 v_3 = (m_2 + m_3) v_5 \]
Thus, we find:
\[ v_5 = \frac{2m_2 v_3}{m_2 + m_3} \quad \text{(Equation 6)} \]

### Step 6: Substitute Equation 3 into Equation 6
- Substitute \( v_3 \) from Equation 3 into Equation 6:
\[ v_5 = \frac{2m_2 \left(\frac{2m_1 v_1}{m_1 + m_2}\right)}{m_2 + m_3} \]
Simplifying gives:
\[ v_5 = \frac{4m_1 m_2 v_1}{(m_1 + m_2)(m_2 + m_3)} \]

### Step 7: Maximize \( v_5 \)
- To find the maximum speed of sphere 3, we need to maximize \( v_5 \) with respect to \( m_2 \).
- Set the derivative of \( v_5 \) with respect to \( m_2 \) to zero and solve for \( m_2 \):
\[ m_2^2 = m_1 m_3 \implies m_2 = \sqrt{m_1 m_3} \]

### Step 8: Substitute Back to Find Maximum \( v_5 \)
- Substitute \( m_2 = \sqrt{m_1 m_3} \) back into the equation for \( v_5 \):
\[ v_5 = \frac{4m_1 \sqrt{m_1 m_3} v_1}{(\sqrt{m_1 m_3} + m_1)(\sqrt{m_1 m_3} + m_3)} \]
This simplifies to:
\[ v_5 = \frac{4m_1 v_1}{(\sqrt{m_1} + \sqrt{m_3})^2} \]

### Conclusion
The maximum speed of sphere 3 is given by:
\[ v_5 = \frac{4m_1 v_1}{(\sqrt{m_1} + \sqrt{m_3})^2} \]