Three identical spheres lie at rest along a line on a smooth horizontal surface. The separation between any two adjacent spheres is L. The first sphere is moving with a velocity u towards the second sphere at time t = 0. The coefficient of restitution for a collision between any two blocks is `1/3` The correct statement is

Options

A. The third sphere will start moving at `t=(5L)/(2u)`
B. The third sphere will start moving at `t=(4L)/(u)`
C. The centre of mass of the system will have a final speed 2u
D. The centre of mass of the system will have of final speed u

Answer

To solve the problem step by step, we will analyze the motion of the spheres and apply the principles of conservation of momentum and the coefficient of restitution.

### Step 1: Understand the Initial Setup
We have three identical spheres:
- Sphere 1 is moving towards Sphere 2 with velocity \( u \).
- Spheres 2 and 3 are at rest.
- The distance between each pair of spheres is \( L \).

### Step 2: Collision Between Sphere 1 and Sphere 2
When Sphere 1 collides with Sphere 2, we can apply the conservation of momentum and the coefficient of restitution.

**Conservation of Momentum:**
Let \( v_1 \) be the final velocity of Sphere 1 and \( v_2 \) be the final velocity of Sphere 2 after the collision. The initial momentum is:
\[ m u + 0 + 0 = m v_1 + m v_2 \]
This simplifies to:
\[ u = v_1 + v_2 \quad \text{(Equation 1)} \]

**Coefficient of Restitution:**
The coefficient of restitution \( e \) is given as \( \frac{1}{3} \). The formula is:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
Here, the velocity of approach is \( u \) and the velocity of separation is \( v_2 - v_1 \):
\[ \frac{1}{3} = \frac{v_2 - v_1}{u} \]
This simplifies to:
\[ v_2 - v_1 = \frac{u}{3} \quad \text{(Equation 2)} \]

### Step 3: Solve Equations 1 and 2
Now we can solve Equations 1 and 2 simultaneously.

From Equation 1:
\[ v_1 = u - v_2 \]

Substituting \( v_1 \) into Equation 2:
\[ v_2 - (u - v_2) = \frac{u}{3} \]
This simplifies to:
\[ 2v_2 - u = \frac{u}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 6v_2 - 3u = u \]
\[ 6v_2 = 4u \]
\[ v_2 = \frac{2u}{3} \]

Substituting \( v_2 \) back into Equation 1 to find \( v_1 \):
\[ u = v_1 + \frac{2u}{3} \]
\[ v_1 = u - \frac{2u}{3} = \frac{u}{3} \]

### Step 4: Collision Between Sphere 2 and Sphere 3
Now Sphere 2 moves with velocity \( \frac{2u}{3} \) towards Sphere 3 (which is at rest). We apply the same principles again.

**Conservation of Momentum:**
Let \( v_2' \) be the final velocity of Sphere 2 after the second collision and \( v_3 \) be the final velocity of Sphere 3:
\[ m \frac{2u}{3} + 0 = m v_2' + m v_3 \]
This simplifies to:
\[ \frac{2u}{3} = v_2' + v_3 \quad \text{(Equation 3)} \]

**Coefficient of Restitution:**
Using the same coefficient of restitution:
\[ \frac{1}{3} = \frac{v_3 - v_2'}{\frac{2u}{3}} \]
This simplifies to:
\[ v_3 - v_2' = \frac{2u}{9} \quad \text{(Equation 4)} \]

### Step 5: Solve Equations 3 and 4
From Equation 3:
\[ v_3 = \frac{2u}{3} - v_2' \]

Substituting into Equation 4:
\[ \left(\frac{2u}{3} - v_2'\right) - v_2' = \frac{2u}{9} \]
This simplifies to:
\[ \frac{2u}{3} - 2v_2' = \frac{2u}{9} \]
Multiplying through by 9 to eliminate the fraction:
\[ 6u - 18v_2' = 2u \]
\[ 4u = 18v_2' \]
\[ v_2' = \frac{2u}{9} \]

Substituting \( v_2' \) back into Equation 3 to find \( v_3 \):
\[ v_3 = \frac{2u}{3} - \frac{2u}{9} \]
Finding a common denominator:
\[ v_3 = \frac{6u}{9} - \frac{2u}{9} = \frac{4u}{9} \]

### Step 6: Calculate the Time for the Third Sphere to Start Moving
The time taken for Sphere 1 to collide with Sphere 2:
\[ T_1 = \frac{L}{u} \]

The time taken for Sphere 2 to collide with Sphere 3 after Sphere 1 collides with Sphere 2:
\[ T_2 = \frac{L}{\frac{2u}{3}} = \frac{3L}{2u} \]

Total time until Sphere 3 starts moving:
\[ T = T_1 + T_2 = \frac{L}{u} + \frac{3L}{2u} = \frac{5L}{2u} \]

### Conclusion
The third sphere will start moving at time \( t = \frac{5L}{2u} \).