A body of mass m1 collides head on perfectly elastically with a stationary body of mass `m_(2)`. If velocities of `m_(1)` before and after the collision are v and –v/3 respectively then the value of `m_(1)//m_(2)` is-

Options

A. 1
B. 2
C. 0.5
D. 4

Answer

To solve the problem, we will use the principles of conservation of momentum and conservation of kinetic energy, as the collision is perfectly elastic.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of the first body: \( m_1 \)
- Mass of the second body: \( m_2 \)
- Initial velocity of \( m_1 \): \( v \)
- Initial velocity of \( m_2 \): \( 0 \) (stationary)
- Final velocity of \( m_1 \): \( -\frac{v}{3} \)
- Final velocity of \( m_2 \): \( v_1 \) (unknown)

2. **Apply Conservation of Momentum:**
The total momentum before the collision equals the total momentum after the collision.
\[ m_1 v + m_2 \cdot 0 = m_1 \left(-\frac{v}{3}\right) + m_2 v_1 \]
Simplifying this, we have:
\[ m_1 v = -\frac{m_1 v}{3} + m_2 v_1 \]
Rearranging gives:
\[ m_1 v + \frac{m_1 v}{3} = m_2 v_1 \]
\[ \frac{4m_1 v}{3} = m_2 v_1 \quad \text{(Equation 1)} \]

3. **Apply Conservation of Kinetic Energy:**
The total kinetic energy before the collision equals the total kinetic energy after the collision.
\[ \frac{1}{2} m_1 v^2 = \frac{1}{2} m_1 \left(-\frac{v}{3}\right)^2 + \frac{1}{2} m_2 v_1^2 \]
Simplifying this, we have:
\[ \frac{1}{2} m_1 v^2 = \frac{1}{2} m_1 \frac{v^2}{9} + \frac{1}{2} m_2 v_1^2 \]
Multiplying through by 2 to eliminate the fractions:
\[ m_1 v^2 = \frac{m_1 v^2}{9} + m_2 v_1^2 \]
Rearranging gives:
\[ m_1 v^2 - \frac{m_1 v^2}{9} = m_2 v_1^2 \]
\[ \frac{8m_1 v^2}{9} = m_2 v_1^2 \quad \text{(Equation 2)} \]

4. **Substituting \( m_2 \) from Equation 1 into Equation 2:**
From Equation 1, we can express \( m_2 \):
\[ m_2 = \frac{4m_1 v}{3v_1} \]
Substituting this into Equation 2:
\[ \frac{8m_1 v^2}{9} = \left(\frac{4m_1 v}{3v_1}\right) v_1^2 \]
Simplifying gives:
\[ \frac{8m_1 v^2}{9} = \frac{4m_1 v v_1}{3} \]
Canceling \( m_1 \) (assuming \( m_1 \neq 0 \)):
\[ \frac{8v^2}{9} = \frac{4v v_1}{3} \]
Cross-multiplying gives:
\[ 8v^2 \cdot 3 = 4v v_1 \cdot 9 \]
\[ 24v^2 = 36v v_1 \]
Dividing both sides by \( 12v \) (assuming \( v \neq 0 \)):
\[ 2 = 3v_1 \]
Thus:
\[ v_1 = \frac{2}{3} v \]

5. **Finding the Ratio \( \frac{m_1}{m_2} \):**
Substitute \( v_1 \) back into the expression for \( m_2 \):
\[ m_2 = \frac{4m_1 v}{3 \cdot \frac{2}{3}v} = \frac{4m_1 v}{2v} = 2m_1 \]
Therefore:
\[ \frac{m_1}{m_2} = \frac{m_1}{2m_1} = \frac{1}{2} \]

### Final Answer:
\[ \frac{m_1}{m_2} = \frac{1}{2} \quad \text{or} \quad 0.5 \]