Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest the collision is head on and elastic in nature. After the collision the fraction of energy lost by colliding body A is :

Options

A. `(5)/(9)`
B. `(1)/(9)`
C. `(8)/(9)`
D. `(4)/(9)`

Explanation

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

### Step 1: Understand the Initial Conditions
- Mass of body A, \( m_A = 4m \)
- Mass of body B, \( m_B = 2m \)
- Initial speed of body A, \( u_A = u \)
- Initial speed of body B, \( u_B = 0 \) (at rest)

### Step 2: Write the Conservation of Momentum Equation
The total momentum before the collision must equal the total momentum after the collision:

\[ m_A \cdot u_A + m_B \cdot u_B = m_A \cdot v_A + m_B \cdot v_B \]

Substituting the values, we get:

\[ 4m \cdot u + 2m \cdot 0 = 4m \cdot v_1 + 2m \cdot v_2 \]

This simplifies to:

\[ 4u = 4v_1 + 2v_2 \quad \text{(1)} \]

### Step 3: Write the Conservation of Kinetic Energy Equation
The total kinetic energy before the collision must equal the total kinetic energy after the collision:

\[ \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 \]

Substituting the values, we get:

\[ \frac{1}{2} (4m) u^2 + 0 = \frac{1}{2} (4m) v_1^2 + \frac{1}{2} (2m) v_2^2 \]

This simplifies to:

\[ 2u^2 = 2v_1^2 + v_2^2 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1):

\[ 4u = 4v_1 + 2v_2 \implies 2u = 2v_1 + v_2 \implies v_2 = 2u - 2v_1 \quad \text{(3)} \]

Substituting equation (3) into equation (2):

\[ 2u^2 = 2v_1^2 + (2u - 2v_1)^2 \]

Expanding the square:

\[ 2u^2 = 2v_1^2 + (4u^2 - 8uv_1 + 4v_1^2) \]

Combining like terms:

\[ 2u^2 = 4u^2 - 6v_1^2 - 8uv_1 \]

Rearranging gives:

\[ 0 = 2u^2 - 6v_1^2 - 8uv_1 \]

Dividing through by 2:

\[ 0 = u^2 - 3v_1^2 - 4uv_1 \]

This is a quadratic equation in \( v_1 \). Solving using the quadratic formula:

\[ v_1 = \frac{-(-4u) \pm \sqrt{(-4u)^2 - 4(-3)(u^2)}}{2(-3)} \]

Calculating the discriminant:

\[ = \frac{4u \pm \sqrt{16u^2 + 12u^2}}{-6} = \frac{4u \pm \sqrt{28u^2}}{-6} = \frac{4u \pm 2\sqrt{7}u}{-6} = \frac{2u(2 \pm \sqrt{7})}{-6} = -\frac{u(2 \pm \sqrt{7})}{3} \]

Taking the positive root for \( v_1 \):

\[ v_1 = \frac{u(2 - \sqrt{7})}{3} \]

Substituting \( v_1 \) back into equation (3) to find \( v_2 \):

\[ v_2 = 2u - 2 \left(\frac{u(2 - \sqrt{7})}{3}\right) = 2u - \frac{2u(2 - \sqrt{7})}{3} \]

Calculating gives:

\[ v_2 = \frac{6u - 4u + 2u\sqrt{7}}{3} = \frac{2u + 2u\sqrt{7}}{3} = \frac{2u(1 + \sqrt{7})}{3} \]

### Step 5: Calculate the Initial and Final Kinetic Energies
- Initial kinetic energy of A:

\[ KE_{initial} = \frac{1}{2} (4m) u^2 = 2mu^2 \]

- Final kinetic energy of A:

\[ KE_{final} = \frac{1}{2} (4m) v_1^2 = 2m \left(\frac{u(2 - \sqrt{7})}{3}\right)^2 = 2m \cdot \frac{u^2(2 - \sqrt{7})^2}{9} \]

### Step 6: Calculate the Energy Lost
Energy lost by A:

\[ \Delta KE = KE_{initial} - KE_{final} \]

Substituting the values:

\[ \Delta KE = 2mu^2 - 2m \cdot \frac{u^2(2 - \sqrt{7})^2}{9} \]

### Step 7: Fraction of Energy Lost
The fraction of energy lost by A is:

\[ \text{Fraction} = \frac{\Delta KE}{KE_{initial}} = \frac{2mu^2 - 2m \cdot \frac{u^2(2 - \sqrt{7})^2}{9}}{2mu^2} \]

This simplifies to:

\[ = 1 - \frac{(2 - \sqrt{7})^2}{9} \]

Calculating gives:

\[ = 1 - \frac{4 - 4\sqrt{7} + 7}{9} = 1 - \frac{11 - 4\sqrt{7}}{9} = \frac{9 - (11 - 4\sqrt{7})}{9} = \frac{4\sqrt{7} - 2}{9} \]

### Final Answer
The fraction of energy lost by colliding body A is:

\[ \frac{8}{9} \]