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

Options

A. `(1)/(2)` of its intial kinetic energy
B. `(1)/(9)` of its initial kinetic energy
C. `(8)/(9)` of its initial kinetic energy
D. `(1)/(4)` of its initial kinetic energy

Answer

To solve the problem of finding the loss of kinetic energy of a body of mass \( m \) moving with velocity \( v \) after a head-on elastic collision with another body of mass \( 2m \) that is initially at rest, we will follow these steps:

### Step 1: Understand the Initial Conditions
- We have two bodies:
- Body 1 (mass \( m \)) moving with velocity \( v \).
- Body 2 (mass \( 2m \)) initially at rest (velocity = 0).

### Step 2: Use Conservation of Momentum
- The total momentum before the collision must equal the total momentum after the collision.
- Initial momentum:
\[ p_{\text{initial}} = mv + 2m \cdot 0 = mv \]
- Let \( v_1 \) be the final velocity of mass \( m \) and \( v_2 \) be the final velocity of mass \( 2m \).
- Final momentum:
\[ p_{\text{final}} = mv_1 + 2mv_2 \]
- Setting initial momentum equal to final momentum:
\[ mv = mv_1 + 2mv_2 \quad \text{(1)} \]

### Step 3: Use the Coefficient of Restitution
- For an elastic collision, the coefficient of restitution \( e = 1 \).
- The equation for the coefficient of restitution is:
\[ e = \frac{\text{relative velocity after}}{\text{relative velocity before}} = \frac{v_2 - v_1}{v - 0} \]
- Therefore, we have:
\[ 1 = \frac{v_2 - v_1}{v} \quad \text{(2)} \]

### Step 4: Solve the Equations
- From equation (2), we can express \( v_2 \):
\[ v_2 = v + v_1 \quad \text{(3)} \]
- Substitute equation (3) into equation (1):
\[ mv = mv_1 + 2m(v + v_1) \]
\[ mv = mv_1 + 2mv + 2mv_1 \]
\[ mv = 3mv_1 + 2mv \]
- Rearranging gives:
\[ mv - 2mv = 3mv_1 \]
\[ -mv = 3mv_1 \]
\[ v_1 = -\frac{v}{3} \]

### Step 5: Find the Final Velocity of the Second Body
- Substitute \( v_1 \) back into equation (3):
\[ v_2 = v + \left(-\frac{v}{3}\right) = v - \frac{v}{3} = \frac{2v}{3} \]

### Step 6: Calculate Initial and Final Kinetic Energies
- Initial kinetic energy of body \( m \):
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 \]
- Final kinetic energy of body \( m \):
\[ KE_{\text{final}} = \frac{1}{2}m\left(-\frac{v}{3}\right)^2 = \frac{1}{2}m\frac{v^2}{9} = \frac{mv^2}{18} \]

### Step 7: Calculate the Loss of Kinetic Energy
- Loss of kinetic energy:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2}mv^2 - \frac{mv^2}{18} \]
- Finding a common denominator (which is 18):
\[ \Delta KE = \frac{9mv^2}{18} - \frac{mv^2}{18} = \frac{8mv^2}{18} = \frac{4mv^2}{9} \]

### Final Answer
The loss of kinetic energy of the colliding body (mass \( m \)) is:
\[ \Delta KE = \frac{4mv^2}{9} \]