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 will follow these steps:

### Step 1: Understand the scenario
- Particle A is moving with an initial speed \( u \) and has a mass \( m_A \).
- Particle B is at rest and has a mass \( m_B \).
- After the collision, both particles move in opposite directions with equal speeds \( v \).

### Step 2: Apply conservation of momentum
The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision.

**Before collision:**
- Momentum of A = \( m_A \cdot u \)
- Momentum of B = \( m_B \cdot 0 = 0 \)

Total initial momentum = \( m_A \cdot u \)

**After collision:**
- Momentum of A = \( -m_A \cdot v \) (negative because it moves in the opposite direction)
- Momentum of B = \( m_B \cdot v \)

Total final momentum = \( -m_A \cdot v + m_B \cdot v \)

Setting the initial momentum equal to the final momentum gives us:
\[ m_A \cdot u = m_B \cdot v - m_A \cdot v \]
This simplifies to:
\[ m_A \cdot u = (m_B - m_A) \cdot v \quad \text{(Equation 1)} \]

### Step 3: Apply the coefficient of restitution
For a perfectly elastic collision, the coefficient of restitution \( E \) is defined as:
\[ E = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
Since both particles move apart with equal speeds after the collision, the velocity of separation is \( v + v = 2v \) and the velocity of approach is \( u \).

Setting the coefficient of restitution equal to 1 gives us:
\[ 1 = \frac{2v}{u} \]
From this, we can express \( u \) in terms of \( v \):
\[ u = 2v \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Now we substitute \( u = 2v \) into Equation 1:
\[ m_A \cdot (2v) = (m_B - m_A) \cdot v \]
This simplifies to:
\[ 2m_A \cdot v = m_B \cdot v - m_A \cdot v \]
Dividing through by \( v \) (assuming \( v \neq 0 \)):
\[ 2m_A = m_B - m_A \]
Rearranging gives:
\[ 2m_A + m_A = m_B \]
Thus:
\[ 3m_A = m_B \]

### Final Result
The relationship between the masses is:
\[ m_B = 3m_A \]