Particle A makes a head on elastic collision with another stationary particle B. They fly apart in opposite directions with equal speeds. The mass ratio will be

Options

A. `1/3`
B. `1/2`
C. `1/4`
D. `2/3`

Explanation

To solve the problem of finding the mass ratio of two particles A and B after an elastic collision, we can follow these steps:

### Step 1: Understand the Problem
We have two particles:
- Particle A with mass \( M_A \) is moving with an initial velocity \( V \).
- Particle B with mass \( M_B \) is stationary (initial velocity \( 0 \)).

After the collision, both particles move in opposite directions with equal speeds \( V_A \) and \( V_B \).

### Step 2: Apply Conservation of Momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
\[ \text{Total Momentum} = M_A \cdot V + M_B \cdot 0 = M_A \cdot V \]

After the collision:
\[ \text{Total Momentum} = M_A \cdot (-V_A) + M_B \cdot V_B \]

Since \( V_A = V_B \) (both particles move apart with equal speeds), we can denote this common speed as \( V' \):
\[ \text{Total Momentum} = -M_A \cdot V' + M_B \cdot V' \]

Setting the total momentum before and after equal gives us:
\[ M_A \cdot V = -M_A \cdot V' + M_B \cdot V' \]

### Step 3: Rearranging the Momentum Equation
Rearranging the equation from Step 2:
\[ M_A \cdot V = V' (M_B - M_A) \]
This can be rewritten as:
\[ M_A \cdot V = V' \cdot (M_B - M_A) \]

### Step 4: Apply the Coefficient of Restitution
For elastic collisions, the coefficient of restitution \( e = 1 \). The formula for the coefficient of restitution is:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
In our case:
\[ 1 = \frac{V' + V'}{V - 0} = \frac{2V'}{V} \]
Thus, we can express \( V' \) in terms of \( V \):
\[ V' = \frac{V}{2} \]

### Step 5: Substitute \( V' \) Back into the Momentum Equation
Substituting \( V' = \frac{V}{2} \) into the momentum equation:
\[ M_A \cdot V = \frac{V}{2} \cdot (M_B - M_A) \]

### Step 6: Solve for the Mass Ratio
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ M_A = \frac{1}{2} (M_B - M_A) \]
Rearranging gives:
\[ M_A + \frac{1}{2} M_A = \frac{1}{2} M_B \]
\[ \frac{3}{2} M_A = \frac{1}{2} M_B \]
Multiplying through by 2:
\[ 3 M_A = M_B \]
Thus, the mass ratio \( \frac{M_A}{M_B} \) is:
\[ \frac{M_A}{M_B} = \frac{1}{3} \]

### Final Answer
The mass ratio \( \frac{M_A}{M_B} \) is \( \frac{1}{3} \).

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