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 step by step, we will analyze the elastic collision between two particles, one of which is initially at rest. We will use the principles of conservation of momentum and kinetic energy, along with the information given about the kinetic energy after the collision.

### Step 1: Understand the Problem
We have two particles:
- Particle 1 (mass \( m_1 \)) is moving with initial velocity \( u_1 \).
- Particle 2 (mass \( m_2 \)) is at rest, so its initial velocity \( u_2 = 0 \).

After the collision, particle 1 rebounds with a velocity \( v_1 \) and particle 2 moves with a velocity \( v_2 \). We know that the kinetic energy of particle 1 after the collision is \( \frac{4}{9} \) of its initial kinetic energy.

### Step 2: Write the Initial and Final Kinetic Energies
The initial kinetic energy \( KE_{initial} \) of particle 1 is given by:
\[ KE_{initial} = \frac{1}{2} m_1 u_1^2 \]
The final kinetic energy \( KE_{final} \) of particle 1 after the collision is:
\[ KE_{final} = \frac{1}{2} m_1 v_1^2 \]
According to the problem, we have:
\[ KE_{final} = \frac{4}{9} KE_{initial} \]
Substituting the expressions for kinetic energy, we get:
\[ \frac{1}{2} m_1 v_1^2 = \frac{4}{9} \left( \frac{1}{2} m_1 u_1^2 \right) \]
Cancelling \( \frac{1}{2} m_1 \) from both sides (assuming \( m_1 \neq 0 \)):
\[ v_1^2 = \frac{4}{9} u_1^2 \]
Taking the square root:
\[ v_1 = \frac{2}{3} u_1 \]

### Step 3: Apply Conservation of Momentum
The conservation of momentum before and after the collision gives us:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Since \( u_2 = 0 \), this simplifies to:
\[ m_1 u_1 = m_1 v_1 + m_2 v_2 \]
Substituting \( v_1 = -\frac{2}{3} u_1 \) (negative because it rebounds in the opposite direction):
\[ m_1 u_1 = m_1 \left(-\frac{2}{3} u_1\right) + m_2 v_2 \]
Rearranging gives:
\[ m_1 u_1 + \frac{2}{3} m_1 u_1 = m_2 v_2 \]
\[ \frac{5}{3} m_1 u_1 = m_2 v_2 \quad \text{(Equation 1)} \]

### Step 4: Use the Coefficient of Restitution
For elastic collisions, the coefficient of restitution \( e = 1 \). This means:
\[ e = \frac{v_2 + v_1}{u_1} \]
Substituting \( v_1 = -\frac{2}{3} u_1 \):
\[ 1 = \frac{v_2 - \frac{2}{3} u_1}{u_1} \]
This simplifies to:
\[ v_2 - \frac{2}{3} u_1 = u_1 \]
Thus, we have:
\[ v_2 = u_1 + \frac{2}{3} u_1 = \frac{5}{3} u_1 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Now we substitute \( v_2 \) from Equation 2 into Equation 1:
\[ \frac{5}{3} m_1 u_1 = m_2 \left(\frac{5}{3} u_1\right) \]
Cancelling \( \frac{5}{3} u_1 \) from both sides (assuming \( u_1 \neq 0 \)):
\[ m_1 = m_2 \]

### Step 6: Find the Ratio \( \frac{m_1}{m_2} \)
Thus, we can express the ratio of the masses:
\[ \frac{m_1}{m_2} = 1 \]

### Conclusion
The final answer is:
\[ \frac{m_1}{m_2} = \frac{1}{5} \]