In two separate collisions, the coefficient of restitutions `e_(1)` and `e_(2)` are in the ratio `3 : 1`. In the first collision the relative velocity of approach is twice the relative velocity of separation. Then, the ratio between relativevelocity of approach and relative velocity of separation in the second collision is

Options

A. `1 :6`
B. `2 : 3`
C. `3 : 2`
D. `6 : 1`

Explanation

To solve the problem, we need to analyze the information given about two separate collisions and use the definition of the coefficient of restitution.

### Step-by-Step Solution:

1. **Understanding the Coefficient of Restitution (e)**:
The coefficient of restitution \( e \) is defined as the ratio of the relative velocity of separation to the relative velocity of approach. Mathematically, this is expressed as:
\[ e = \frac{\text{Relative Velocity of Separation}}{\text{Relative Velocity of Approach}} \]

2. **Given Information**:
- The coefficients of restitution for the two collisions are in the ratio \( e_1 : e_2 = 3 : 1 \).
- In the first collision, the relative velocity of approach is twice the relative velocity of separation. This can be expressed as:
\[ \text{Relative Velocity of Approach} = 2 \times \text{Relative Velocity of Separation} \]
Let’s denote:
- Relative Velocity of Separation in the first collision as \( \Delta V_1 \)
- Relative Velocity of Approach in the first collision as \( \Delta U_1 \)

From the above relation, we can write:
\[ \Delta U_1 = 2 \Delta V_1 \]

3. **Applying the Coefficient of Restitution for the First Collision**:
Using the definition of \( e_1 \):
\[ e_1 = \frac{\Delta V_1}{\Delta U_1} \]
Substituting \( \Delta U_1 \):
\[ e_1 = \frac{\Delta V_1}{2 \Delta V_1} = \frac{1}{2} \]

4. **Finding \( e_2 \)**:
Given the ratio \( e_1 : e_2 = 3 : 1 \), we can express \( e_2 \) in terms of \( e_1 \):
\[ e_1 = 3k \quad \text{and} \quad e_2 = k \]
From \( e_1 = \frac{1}{2} \):
\[ 3k = \frac{1}{2} \implies k = \frac{1}{6} \]
Thus, we find:
\[ e_2 = k = \frac{1}{6} \]

5. **Applying the Coefficient of Restitution for the Second Collision**:
For the second collision, we have:
\[ e_2 = \frac{\Delta V_2}{\Delta U_2} \]
Substituting \( e_2 \):
\[ \frac{1}{6} = \frac{\Delta V_2}{\Delta U_2} \]

6. **Finding the Ratio of Relative Velocities**:
Rearranging gives:
\[ \Delta U_2 = 6 \Delta V_2 \]
Now, we want the ratio of the relative velocity of approach to the relative velocity of separation for the second collision:
\[ \frac{\Delta U_2}{\Delta V_2} = 6 \]

### Final Answer:
The ratio between the relative velocity of approach and the relative velocity of separation in the second collision is \( 6 : 1 \).

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