n smooth identical spheres are placed in a row and their masses are in the ratio `1:3:9:"_________":3^(n-1)` . Coefficient of restitution between any two successive collisions in `1//3`. If first sphere is projected with a velocity v towards the second sphere, then calculate the velocity of the last sphere after `( n - 1)^(th)` sphere strikes it.

Options

A. `(v)/( 3^(n))`
B. `3^(n)v`
C. `(v)/( 3^(n-1))`
D. `3^(n-1) v`

Answer

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

### Step 1: Understand the Mass Ratios
The masses of the spheres are given in the ratio:
- \( m_1 : m_2 : m_3 : \ldots : m_n = 1 : 3 : 9 : \ldots : 3^{n-1} \)

We can express the masses as:
- \( m_1 = 1 \)
- \( m_2 = 3 \)
- \( m_3 = 9 \)
- \( m_n = 3^{n-1} \)

### Step 2: Initial Velocities
The first sphere (mass \( m_1 \)) is projected towards the second sphere with an initial velocity \( v \):
- \( v_1 = v \)
- \( v_2 = 0 \) (initially at rest)

### Step 3: Apply Conservation of Momentum for the First Collision
Using the conservation of momentum for the collision between \( m_1 \) and \( m_2 \):
\[ m_1 v_1 + m_2 v_2 = m_1 v_{f1} + m_2 v_{f2} \]
Substituting the known values:
\[ 1 \cdot v + 3 \cdot 0 = 1 \cdot v_{f1} + 3 \cdot v_{f2} \]
This simplifies to:
\[ v = v_{f1} + 3v_{f2} \quad \text{(Equation 1)} \]

### Step 4: Apply Coefficient of Restitution
The coefficient of restitution \( e \) is given as \( \frac{1}{3} \):
\[ e = \frac{v_{f2} - v_{f1}}{v_{1} - v_{2}} \]
Substituting the values:
\[ \frac{1}{3} = \frac{v_{f2} - v_{f1}}{v - 0} \]
This leads to:
\[ v_{f2} - v_{f1} = \frac{v}{3} \quad \text{(Equation 2)} \]

### Step 5: Solve Equations 1 and 2
From Equation 2:
\[ v_{f2} = v_{f1} + \frac{v}{3} \]
Substituting this into Equation 1:
\[ v = v_{f1} + 3(v_{f1} + \frac{v}{3}) \]
Simplifying gives:
\[ v = v_{f1} + 3v_{f1} + v = 4v_{f1} + v \]
Thus:
\[ 0 = 4v_{f1} \implies v_{f1} = 0 \]
Then substituting back into Equation 2:
\[ v_{f2} = 0 + \frac{v}{3} = \frac{v}{3} \]

### Step 6: Second Collision (m2 and m3)
Now, we consider the collision between \( m_2 \) and \( m_3 \):
Using conservation of momentum:
\[ m_2 v_{f2} + m_3 v_3 = m_2 v_{f2}' + m_3 v_{f3} \]
Substituting:
\[ 3 \cdot \frac{v}{3} + 9 \cdot 0 = 3v_{f2}' + 9v_{f3} \]
This simplifies to:
\[ v = 3v_{f2}' + 9v_{f3} \quad \text{(Equation 3)} \]

### Step 7: Coefficient of Restitution for the Second Collision
Using the coefficient of restitution again:
\[ \frac{1}{3} = \frac{v_{f3} - v_{f2}'}{v_{f2} - 0} \]
This leads to:
\[ v_{f3} - v_{f2}' = \frac{1}{3} \cdot \frac{v}{3} = \frac{v}{9} \quad \text{(Equation 4)} \]

### Step 8: Solve Equations 3 and 4
From Equation 4:
\[ v_{f3} = v_{f2}' + \frac{v}{9} \]
Substituting into Equation 3:
\[ v = 3v_{f2}' + 9(v_{f2}' + \frac{v}{9}) \]
This simplifies to:
\[ v = 3v_{f2}' + 9v_{f2}' + v = 12v_{f2}' + v \]
Thus:
\[ 0 = 12v_{f2}' \implies v_{f2}' = 0 \]
Then substituting back into Equation 4:
\[ v_{f3} = 0 + \frac{v}{9} = \frac{v}{9} \]

### Step 9: Generalize for n Collisions
Continuing this process for \( n-1 \) collisions, we find that the velocity of the last sphere after \( n-1 \) collisions is:
\[ v_n = \frac{v}{3^{n-1}} \]

### Final Answer
The velocity of the last sphere after \( (n-1)^{th} \) collision is:
\[ \boxed{\frac{v}{3^{n-1}}} \]