Two balls with masses in the ratio of 1:2 moving in opposite direction have a head-on elastic collision. If their velocities before impact were in the ratio of 3:1, then velocities after impact will have the ratio :

Options

A. `5:3`
B. `7:5`
C. `4:5`
D. `2:3`

Answer

To solve the problem of two balls colliding elastically, we will follow these steps:

### Step 1: Understand the given data
We have two balls with masses in the ratio of \(1:2\) and their velocities before impact in the ratio of \(3:1\). We denote:
- Mass of ball 1, \(M_1 = m\)
- Mass of ball 2, \(M_2 = 2m\)
- Velocity of ball 1 before impact, \(U_1 = 3v\)
- Velocity of ball 2 before impact, \(U_2 = -v\) (negative because they are moving in opposite directions)

### Step 2: Write the formulas for final velocities after an elastic collision
For elastic collisions, the final velocities can be calculated using the following formulas:
1. For ball 1:
\[ V_1 = \frac{M_1 - M_2}{M_1 + M_2} U_1 + \frac{2M_2}{M_1 + M_2} U_2 \]
2. For ball 2:
\[ V_2 = \frac{M_2 - M_1}{M_1 + M_2} U_2 + \frac{2M_1}{M_1 + M_2} U_1 \]

### Step 3: Substitute the values into the equations
Substituting \(M_1 = m\), \(M_2 = 2m\), \(U_1 = 3v\), and \(U_2 = -v\) into the equations:

**For ball 1:**
\[ V_1 = \frac{m - 2m}{m + 2m} (3v) + \frac{2(2m)}{m + 2m} (-v) \]
\[ V_1 = \frac{-m}{3m} (3v) + \frac{4m}{3m} (-v) \]
\[ V_1 = -v - \frac{4}{3}v = -\frac{7}{3}v \]

**For ball 2:**
\[ V_2 = \frac{2m - m}{m + 2m} (-v) + \frac{2m}{m + 2m} (3v) \]
\[ V_2 = \frac{m}{3m} (-v) + \frac{2m}{3m} (3v) \]
\[ V_2 = -\frac{1}{3}v + 2v = \frac{5}{3}v \]

### Step 4: Find the ratio of the final velocities
Now, we need to find the ratio of the magnitudes of the final velocities \(V_1\) and \(V_2\):
\[ \text{Ratio} = \frac{|V_1|}{|V_2|} = \frac{\frac{7}{3}v}{\frac{5}{3}v} = \frac{7}{5} \]

### Final Answer
The ratio of the velocities after impact is \( \frac{7}{5} \).

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