A body of mass 2 m moving with velocity v makes a head - on elastic collision with another body of mass m which is initially at rest. Loss of kinetic energy of the colliding body (mass 2 m) is

Options

A. `(1)/(9)` of its initial kinetic energy
B. `(1)/(6)` of its initial kinetic energy
C. `(1)/(2)` of its initial kinetic energy
D. `(8)/(9)` of its initial kinetic energy

Answer

To solve the problem of finding the loss of kinetic energy of the colliding body (mass 2m) after a head-on elastic collision with another body of mass m, we can follow these steps:

### Step 1: Calculate Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the body with mass 2m moving with velocity v is given by the formula:
\[ KE_{\text{initial}} = \frac{1}{2} \times (2m) \times v^2 = mv^2 \]

### Step 2: Apply Conservation of Momentum
In an elastic collision, momentum is conserved. The equation for conservation of momentum before and after the collision can be written as:
\[ 2m \cdot v + m \cdot 0 = 2m \cdot v_1 + m \cdot v_2 \]
This simplifies to:
\[ 2mv = 2mv_1 + mv_2 \]
Dividing through by m, we get:
\[ 2v = 2v_1 + v_2 \quad \text{(1)} \]

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, kinetic energy is also conserved. The equation for conservation of kinetic energy is:
\[ \frac{1}{2} \cdot (2m) \cdot v^2 = \frac{1}{2} \cdot (2m) \cdot v_1^2 + \frac{1}{2} \cdot m \cdot v_2^2 \]
This simplifies to:
\[ mv^2 = mv_1^2 + \frac{1}{2}mv_2^2 \]
Dividing through by m, we have:
\[ v^2 = v_1^2 + \frac{1}{2}v_2^2 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1), we can express \(v_2\) in terms of \(v_1\):
\[ v_2 = 2v - 2v_1 \]
Substituting this expression for \(v_2\) into equation (2):
\[ v^2 = v_1^2 + \frac{1}{2}(2v - 2v_1)^2 \]
Expanding and simplifying:
\[ v^2 = v_1^2 + \frac{1}{2}(4v^2 - 8vv_1 + 4v_1^2) \]
\[ v^2 = v_1^2 + 2v^2 - 4vv_1 + 2v_1^2 \]
Combining like terms:
\[ 0 = 3v_1^2 - 4vv_1 + v^2 \]

### Step 5: Solve the Quadratic Equation
This is a quadratic equation in \(v_1\):
\[ 3v_1^2 - 4vv_1 + v^2 = 0 \]
Using the quadratic formula \(v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ v_1 = \frac{4v \pm \sqrt{(-4v)^2 - 4 \cdot 3 \cdot v^2}}{2 \cdot 3} \]
\[ = \frac{4v \pm \sqrt{16v^2 - 12v^2}}{6} \]
\[ = \frac{4v \pm 2v}{6} \]
This gives two possible solutions:
\[ v_1 = v \quad \text{or} \quad v_1 = \frac{v}{3} \]
Since the first case is not possible (the two bodies cannot have the same velocity after collision), we take:
\[ v_1 = \frac{v}{3} \]

### Step 6: Calculate Final Kinetic Energy
Now we can calculate the final kinetic energy (KE_final) of the body with mass 2m:
\[ KE_{\text{final}} = \frac{1}{2} \cdot (2m) \cdot \left(\frac{v}{3}\right)^2 = \frac{1}{2} \cdot (2m) \cdot \frac{v^2}{9} = \frac{mv^2}{9} \]

### Step 7: Calculate Loss of Kinetic Energy
The loss of kinetic energy (ΔKE) is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = mv^2 - \frac{mv^2}{9} = mv^2 \left(1 - \frac{1}{9}\right) = mv^2 \cdot \frac{8}{9} \]
Thus, the loss of kinetic energy of the colliding body (mass 2m) is:
\[ \Delta KE = \frac{8}{9} mv^2 \]

### Final Answer
The loss of kinetic energy of the colliding body (mass 2m) is \(\frac{8}{9} mv^2\).