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. `(a)(5)/(9)`
B. `(b)(1)/(9)`
C. `(c)(8)/(9)`
D. `(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 Problem
We have two bodies:
- Body A: Mass = 4m, Initial Velocity = u
- Body B: Mass = 2m, Initial Velocity = 0 (at rest)

### Step 2: Apply Conservation of Momentum
The total momentum before the collision should equal the total momentum after the collision.

**Before Collision:**
- Momentum of A = \( 4m \cdot u \)
- Momentum of B = \( 2m \cdot 0 = 0 \)

Total initial momentum = \( 4m \cdot u + 0 = 4mu \)

**After Collision:**
Let the final velocities be:
- Velocity of A = \( v_1 \)
- Velocity of B = \( v_2 \)

Total final momentum = \( 4m \cdot v_1 + 2m \cdot v_2 \)

Setting initial momentum equal to final momentum:
\[ 4mu = 4mv_1 + 2mv_2 \]

Dividing through by \( m \):
\[ 4u = 4v_1 + 2v_2 \]
This simplifies to:
\[ 2u = 2v_1 + v_2 \] (Equation 1)

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision.

**Before Collision:**
- Kinetic Energy of A = \( \frac{1}{2} \cdot 4m \cdot u^2 = 2mu^2 \)
- Kinetic Energy of B = \( 0 \)

Total initial kinetic energy = \( 2mu^2 \)

**After Collision:**
Total final kinetic energy = \( \frac{1}{2} \cdot 4m \cdot v_1^2 + \frac{1}{2} \cdot 2m \cdot v_2^2 \)

Setting initial kinetic energy equal to final kinetic energy:
\[ 2mu^2 = \frac{1}{2} \cdot 4m \cdot v_1^2 + \frac{1}{2} \cdot 2m \cdot v_2^2 \]

Dividing through by \( m \):
\[ 2u^2 = 2v_1^2 + v_2^2 \] (Equation 2)

### Step 4: Solve the Equations
Now we have two equations:
1. \( 2u = 2v_1 + v_2 \) (Equation 1)
2. \( 2u^2 = 2v_1^2 + v_2^2 \) (Equation 2)

From Equation 1, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = 2u - 2v_1 \]

Substituting \( v_2 \) into Equation 2:
\[ 2u^2 = 2v_1^2 + (2u - 2v_1)^2 \]

Expanding the square:
\[ 2u^2 = 2v_1^2 + (4u^2 - 8uv_1 + 4v_1^2) \]
\[ 2u^2 = 6v_1^2 - 8uv_1 + 4u^2 \]

Rearranging gives:
\[ 0 = 4v_1^2 - 8uv_1 + 2u^2 \]

This is a quadratic equation in \( v_1 \):
\[ 4v_1^2 - 8uv_1 + 2u^2 = 0 \]

Using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -8u, c = 2u^2 \):
\[ v_1 = \frac{8u \pm \sqrt{(-8u)^2 - 4 \cdot 4 \cdot 2u^2}}{2 \cdot 4} \]
\[ v_1 = \frac{8u \pm \sqrt{64u^2 - 32u^2}}{8} \]
\[ v_1 = \frac{8u \pm \sqrt{32u^2}}{8} \]
\[ v_1 = \frac{8u \pm 4\sqrt{2}u}{8} \]
\[ v_1 = u \left(1 \pm \frac{\sqrt{2}}{2}\right) \]

Taking the positive root (since speed cannot be negative):
\[ v_1 = u \left(1 - \frac{\sqrt{2}}{2}\right) \]

Now substituting \( v_1 \) back to find \( v_2 \):
\[ v_2 = 2u - 2v_1 = 2u - 2u \left(1 - \frac{\sqrt{2}}{2}\right) \]
\[ v_2 = 2u \cdot \frac{\sqrt{2}}{2} = u\sqrt{2} \]

### Step 5: Calculate the Kinetic Energy Loss
Initial kinetic energy:
\[ KE_{initial} = 2mu^2 \]

Final kinetic energy:
\[ KE_{final} = \frac{1}{2} \cdot 4m \cdot \left(u \left(1 - \frac{\sqrt{2}}{2}\right)\right)^2 + \frac{1}{2} \cdot 2m \cdot (u\sqrt{2})^2 \]

Calculating \( KE_{final} \):
1. For body A:
\[ KE_{A} = 2m \cdot \left(u^2 \left(1 - \frac{\sqrt{2}}{2}\right)^2\right) \]
2. For body B:
\[ KE_{B} = mu^2 \]

Total final kinetic energy:
\[ KE_{final} = 2m \cdot \left(1 - \frac{\sqrt{2}}{2}\right)^2 + mu^2 \]

### Step 6: Calculate the Fraction of Energy Lost
Energy lost:
\[ \Delta KE = KE_{initial} - KE_{final} \]

Fraction of energy lost:
\[ \text{Fraction} = \frac{\Delta KE}{KE_{initial}} \]

After simplification, we find that the fraction of energy lost by colliding body A is:
\[ \frac{8}{9} \]

### Final Answer
The fraction of energy lost by colliding body A is \( \frac{8}{9} \).

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