Particle A makes a perfectly elastic collision with anther particle B at rest. They fly apart in opposite direction with equal speeds. If the masses are `m_(A)&m_(B)` respectively, then

Options

A. `2m_(A)=m_(B)`
B. `3m_(A)=m_(B)`
C. `4m_(A)=m_(B)`
D. `sqrt3m_(A)=m_(B)`

Answer

To solve the problem of a perfectly elastic collision between particle A and particle B, we can follow these steps:

### Step 1: Understand the Initial Conditions
- Particle A has mass \( m_A \) and is moving with an initial velocity \( u \).
- Particle B has mass \( m_B \) and is initially at rest (velocity = 0).

### Step 2: Apply the Concept of Elastic Collision
In a perfectly elastic collision, both momentum and kinetic energy are conserved. The coefficient of restitution \( e \) for a perfectly elastic collision is equal to 1.

### Step 3: Write the Equation for Coefficient of Restitution
The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
For our case:
\[ 1 = \frac{v + v}{u} \]
Where \( v \) is the speed of both particles A and B after the collision (they fly apart with equal speeds in opposite directions).

### Step 4: Simplify the Coefficient of Restitution Equation
From the equation above, we can simplify:
\[ 1 = \frac{2v}{u} \]
This leads to:
\[ 2v = u \quad \Rightarrow \quad v = \frac{u}{2} \]

### Step 5: Apply Conservation of Momentum
The total momentum before and after the collision must be equal. Thus, we have:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
Before the collision:
\[ \text{Initial Momentum} = m_A \cdot u + m_B \cdot 0 = m_A \cdot u \]
After the collision:
\[ \text{Final Momentum} = -m_A \cdot v + m_B \cdot v \]
Setting the initial momentum equal to the final momentum:
\[ m_A \cdot u = -m_A \cdot v + m_B \cdot v \]

### Step 6: Substitute \( v \) in the Momentum Equation
Substituting \( v = \frac{u}{2} \) into the momentum equation:
\[ m_A \cdot u = -m_A \cdot \left(\frac{u}{2}\right) + m_B \cdot \left(\frac{u}{2}\right) \]
This simplifies to:
\[ m_A \cdot u = -\frac{m_A \cdot u}{2} + \frac{m_B \cdot u}{2} \]

### Step 7: Rearranging the Equation
Multiply through by 2 to eliminate the fraction:
\[ 2m_A \cdot u = -m_A \cdot u + m_B \cdot u \]
Rearranging gives:
\[ 2m_A \cdot u + m_A \cdot u = m_B \cdot u \]
\[ 3m_A \cdot u = m_B \cdot u \]

### Step 8: Solve for the Mass Relationship
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ 3m_A = m_B \]

### Conclusion
Thus, the relationship between the masses of particles A and B is:
\[ m_B = 3m_A \]