The masses of five balls at rest and lying at equal distance in a straight line are in geometrical progression with ratio 2 and their coefficients of restitution are each 2/3 . If the first ball is started towards the second with velocity u, then the velocity communicated to `5^(th)` ball is

Options

A. `(5)/(9)u`
B. `((5)/(9))^(2)u`
C. `((5)/(9))^(3)u`
D. `((5)/(9))^(4)u`

Answer

To solve the problem step by step, we will first analyze the situation involving the five balls and their masses in geometrical progression, followed by calculating the velocities after each collision.

### Step 1: Determine the masses of the balls
Given that the masses of the five balls are in geometrical progression with a ratio of 2, we can denote the masses as follows:
- \( m_1 = m \) (first ball)
- \( m_2 = 2m \) (second ball)
- \( m_3 = 4m \) (third ball)
- \( m_4 = 8m \) (fourth ball)
- \( m_5 = 16m \) (fifth ball)

### Step 2: Coefficient of restitution
The coefficient of restitution (e) for the balls is given as \( \frac{2}{3} \).

### Step 3: Calculate the velocity of the second ball after the first collision
Using the formula for the velocity of the second ball after the collision:
\[ u_2 = u_1 \left( \frac{m_1 + e m_2}{m_1 + m_2} \right) + u_2 \left( \frac{m_2 - e m_1}{m_1 + m_2} \right) \]
Substituting \( u_1 = u \), \( m_1 = m \), and \( m_2 = 2m \):
\[ u_2 = u \left( \frac{m + \frac{2}{3} \cdot 2m}{m + 2m} \right) + 0 \left( \frac{2m - \frac{2}{3} m}{m + 2m} \right) \]
This simplifies to:
\[ u_2 = u \left( \frac{m + \frac{4}{3} m}{3m} \right) = u \left( \frac{\frac{7}{3} m}{3m} \right) = u \left( \frac{7}{9} \right) \]

### Step 4: Calculate the velocity of the third ball after the second collision
Now, we will calculate the velocity of the third ball after the second collision using the same formula:
\[ u_3 = u_2 \left( \frac{m_2 + e m_3}{m_2 + m_3} \right) \]
Substituting \( u_2 = \frac{7}{9}u \), \( m_2 = 2m \), and \( m_3 = 4m \):
\[ u_3 = \frac{7}{9}u \left( \frac{2m + \frac{2}{3} \cdot 4m}{2m + 4m} \right) = \frac{7}{9}u \left( \frac{2m + \frac{8}{3} m}{6m} \right) \]
This simplifies to:
\[ u_3 = \frac{7}{9}u \left( \frac{\frac{14}{3} m}{6m} \right) = \frac{7}{9}u \left( \frac{14}{18} \right) = \frac{7}{9}u \cdot \frac{7}{9} = \frac{49}{81}u \]

### Step 5: Calculate the velocity of the fourth ball after the third collision
Using the same approach:
\[ u_4 = u_3 \left( \frac{m_3 + e m_4}{m_3 + m_4} \right) \]
Substituting \( u_3 = \frac{49}{81}u \), \( m_3 = 4m \), and \( m_4 = 8m \):
\[ u_4 = \frac{49}{81}u \left( \frac{4m + \frac{2}{3} \cdot 8m}{4m + 8m} \right) = \frac{49}{81}u \left( \frac{4m + \frac{16}{3} m}{12m} \right) \]
This simplifies to:
\[ u_4 = \frac{49}{81}u \left( \frac{\frac{28}{3} m}{12m} \right) = \frac{49}{81}u \cdot \frac{28}{36} = \frac{49 \cdot 28}{81 \cdot 36}u = \frac{1372}{2916}u \]

### Step 6: Calculate the velocity of the fifth ball after the fourth collision
Finally, we calculate the velocity of the fifth ball:
\[ u_5 = u_4 \left( \frac{m_4 + e m_5}{m_4 + m_5} \right) \]
Substituting \( u_4 = \frac{1372}{2916}u \), \( m_4 = 8m \), and \( m_5 = 16m \):
\[ u_5 = \frac{1372}{2916}u \left( \frac{8m + \frac{2}{3} \cdot 16m}{8m + 16m} \right) = \frac{1372}{2916}u \left( \frac{8m + \frac{32}{3} m}{24m} \right) \]
This simplifies to:
\[ u_5 = \frac{1372}{2916}u \cdot \frac{\frac{56}{3} m}{24m} = \frac{1372}{2916}u \cdot \frac{56}{72} = \frac{1372 \cdot 56}{2916 \cdot 72}u \]

### Final Answer
Thus, the velocity communicated to the fifth ball is:
\[ u_5 = \left( \frac{5}{9} \right)^4 u = \frac{625}{6561} u \]