A particle of mass `m_1` makes a head - on elastic collision with another particle of mass `m_2` at rest. `m_1` rebounds straight back with `4/9` of its initial kinetic energy . Then `m_1/m_2` is :

Options

A. `2/3`
B. `1/5`
C. `3/2`
D. `2/5`

Answer

To solve the problem, we will follow these steps systematically:

### Step 1: Understand the Problem
We have two particles:
- Particle 1 (mass \( m_1 \)) is moving with an initial velocity \( u \).
- Particle 2 (mass \( m_2 \)) is at rest (velocity = 0).
After the collision, particle 1 rebounds with a velocity \( v_1 \) and retains \( \frac{4}{9} \) of its initial kinetic energy.

### Step 2: Write Down the Conservation of Momentum
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Initial momentum:
\[ P_{\text{initial}} = m_1 u + m_2 \cdot 0 = m_1 u \]

Final momentum:
\[ P_{\text{final}} = m_1 (-v_1) + m_2 v_2 \]

Setting these equal gives us:
\[ m_1 u = -m_1 v_1 + m_2 v_2 \quad \text{(1)} \]

### Step 3: Write Down the Kinetic Energy Relation
The initial kinetic energy of particle 1 is:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u^2 \]

The final kinetic energy of particle 1 is:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 \]

According to the problem, the final kinetic energy is \( \frac{4}{9} \) of the initial kinetic energy:
\[ \frac{1}{2} m_1 v_1^2 = \frac{4}{9} \left( \frac{1}{2} m_1 u^2 \right) \]

Cancelling \( \frac{1}{2} m_1 \) from both sides (assuming \( m_1 \neq 0 \)):
\[ v_1^2 = \frac{4}{9} u^2 \]

Taking the square root gives:
\[ v_1 = \frac{2}{3} u \quad \text{(2)} \]

### Step 4: Substitute \( v_1 \) into the Momentum Equation
Now we substitute \( v_1 \) from equation (2) into equation (1):
\[ m_1 u = -m_1 \left( \frac{2}{3} u \right) + m_2 v_2 \]

This simplifies to:
\[ m_1 u + \frac{2}{3} m_1 u = m_2 v_2 \]
\[ \frac{5}{3} m_1 u = m_2 v_2 \quad \text{(3)} \]

### Step 5: Use the Coefficient of Restitution
For an elastic collision, the coefficient of restitution \( e = 1 \) gives us the relation:
\[ v_2 - (-v_1) = e (u - 0) \]
This simplifies to:
\[ v_2 + v_1 = u \]

Substituting \( v_1 = \frac{2}{3} u \):
\[ v_2 + \frac{2}{3} u = u \]
\[ v_2 = u - \frac{2}{3} u = \frac{1}{3} u \quad \text{(4)} \]

### Step 6: Substitute \( v_2 \) into the Momentum Equation
Now substitute \( v_2 \) from equation (4) into equation (3):
\[ \frac{5}{3} m_1 u = m_2 \left( \frac{1}{3} u \right) \]

Cancelling \( u \) (assuming \( u \neq 0 \)):
\[ \frac{5}{3} m_1 = \frac{1}{3} m_2 \]
Multiplying through by 3 gives:
\[ 5 m_1 = m_2 \]

### Step 7: Find the Ratio \( \frac{m_1}{m_2} \)
Rearranging gives:
\[ \frac{m_1}{m_2} = \frac{1}{5} \]

### Final Answer
Thus, the ratio \( \frac{m_1}{m_2} \) is:
\[ \frac{m_1}{m_2} = \frac{1}{5} \]