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, we need to analyze the collisions between the spheres step by step. Let's denote the masses of the spheres and their velocities during the collisions.

### Step 1: Define the masses of the spheres
The masses of the spheres are given in the ratio \(1:3:9:\ldots:3^{n-1}\). We can define the masses as:
- \(m_1 = m\)
- \(m_2 = 3m\)
- \(m_3 = 9m = 3^2 m\)
- ...
- \(m_n = 3^{n-1} m\)

### Step 2: Initial conditions
The first sphere is projected towards the second sphere with an initial velocity \(v\). The initial velocities of the spheres are:
- \(u_1 = v\) (velocity of the first sphere)
- \(u_2 = 0\) (velocity of the second sphere)
- \(u_3 = 0\) (velocity of the third sphere)
- ...
- \(u_n = 0\) (velocity of the nth sphere)

### Step 3: Conservation of momentum for the first collision
Using the conservation of momentum for the first collision between the first and second spheres:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the values:
\[ m \cdot v + 3m \cdot 0 = m v_1 + 3m v_2 \]
This simplifies to:
\[ mv = mv_1 + 3mv_2 \]
Dividing through by \(m\):
\[ v = v_1 + 3v_2 \quad \text{(1)} \]

### Step 4: Coefficient of restitution
The coefficient of restitution \(e\) is given as \(\frac{1}{3}\). The formula for the coefficient of restitution is:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
The velocity of approach is \(u_1 - u_2 = v - 0 = v\) and the velocity of separation is \(v_2 - v_1\). Thus:
\[ \frac{1}{3} = \frac{v_2 - v_1}{v} \]
Rearranging gives:
\[ v_2 - v_1 = \frac{v}{3} \quad \text{(2)} \]

### Step 5: Solve equations (1) and (2)
From equation (1):
\[ v_1 = v - 3v_2 \]
Substituting this into equation (2):
\[ v_2 - (v - 3v_2) = \frac{v}{3} \]
This simplifies to:
\[ v_2 + 3v_2 - v = \frac{v}{3} \]
\[ 4v_2 - v = \frac{v}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 12v_2 - 3v = v \]
\[ 12v_2 = 4v \]
\[ v_2 = \frac{v}{3} \]

### Step 6: Generalizing for subsequent spheres
Following the same logic, we can find the velocity of the third sphere \(v_3\) after the second sphere collides with it:
\[ v_3 = \frac{v_2}{3} = \frac{v}{3^2} \]
Continuing this pattern, we find:
\[ v_n = \frac{v}{3^{n-1}} \]

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