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 the loss of kinetic energy in a head-on elastic collision between two bodies, we will follow these steps:

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

### Step 2: Write the conservation of momentum equation
The total momentum before the collision must equal the total momentum after the collision.

Initial momentum:
\[ p_{initial} = (2m)v + (m)(0) = 2mv \]

Let \( v_1 \) be the final velocity of the body with mass \( 2m \) and \( v_2 \) be the final velocity of the body with mass \( m \).

Final momentum:
\[ p_{final} = (2m)v_1 + (m)v_2 \]

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

Dividing through by \( m \):
\[ 2v = 2v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 3: Write the conservation of kinetic energy equation
In an elastic collision, kinetic energy is also conserved.

Initial kinetic energy:
\[ KE_{initial} = \frac{1}{2}(2m)v^2 = mv^2 \]

Final kinetic energy:
\[ KE_{final} = \frac{1}{2}(2m)v_1^2 + \frac{1}{2}(m)v_2^2 = mv_1^2 + \frac{1}{2}mv_2^2 \]

Setting initial kinetic energy equal to final kinetic energy:
\[ mv^2 = mv_1^2 + \frac{1}{2}mv_2^2 \]

Dividing through by \( m \):
\[ v^2 = v_1^2 + \frac{1}{2}v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Use the coefficient of restitution
For elastic collisions, the coefficient of restitution \( e = 1 \). This gives us:
\[ v_2 - v_1 = e(v - 0) = v \quad \text{(Equation 3)} \]

### Step 5: Solve the equations
We now have three equations:
1. \( 2v = 2v_1 + v_2 \) (Equation 1)
2. \( v^2 = v_1^2 + \frac{1}{2}v_2^2 \) (Equation 2)
3. \( v_2 - v_1 = v \) (Equation 3)

From Equation 3, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v + v_1 \]

Substituting \( v_2 \) into Equation 1:
\[ 2v = 2v_1 + (v + v_1) \implies 2v = 3v_1 + v \implies 2v - v = 3v_1 \implies v = 3v_1 \implies v_1 = \frac{v}{3} \]

Now substituting \( v_1 \) back into Equation 3 to find \( v_2 \):
\[ v_2 = v + v_1 = v + \frac{v}{3} = \frac{4v}{3} \]

### Step 6: Calculate the loss of kinetic energy
Now we can calculate the initial and final kinetic energies:

Initial kinetic energy:
\[ KE_{initial} = mv^2 \]

Final kinetic energy:
\[ KE_{final} = mv_1^2 + \frac{1}{2}mv_2^2 = m\left(\frac{v}{3}\right)^2 + \frac{1}{2}m\left(\frac{4v}{3}\right)^2 \]
\[ = m\left(\frac{v^2}{9}\right) + \frac{1}{2}m\left(\frac{16v^2}{9}\right) = \frac{mv^2}{9} + \frac{8mv^2}{9} = \frac{9mv^2}{9} = mv^2 \]

Loss of kinetic energy:
\[ \text{Loss} = KE_{initial} - KE_{final} = mv^2 - \left(\frac{mv^2}{9} + \frac{8mv^2}{9}\right) = mv^2 - mv^2 = \frac{8mv^2}{9} \]

### Final Result
The loss of kinetic energy of the colliding body (mass \( 2m \)) is:
\[ \text{Loss} = \frac{8}{9} KE_{initial} \]