A particle of mass m moving with velocity v makes head-on elastic collision with a sationary particle of mass 2m. Kinetic energy lost by the lighter particle during period of deformation is

Options

A. `(1)/(2)mv^(2)`
B. `(1)/(18)mv^(2)`
C. `(4)/(9)mv^(2)`
D. `(8)/(9)mv^(2)`

Answer

To solve the problem of the kinetic energy lost by the lighter particle during the period of deformation in a head-on elastic collision, we can follow these steps:

### Step 1: Understand the initial conditions
- We have a particle of mass \( m \) moving with velocity \( v \).
- There is another particle of mass \( 2m \) that is stationary (velocity = 0).

### Step 2: Apply the principle of conservation of momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

The initial momentum \( p_i \) is given by:
\[ p_i = mv + 2m \cdot 0 = mv \]

Let \( V_0 \) be the common velocity of both masses after the collision. The final momentum \( p_f \) is:
\[ p_f = (m + 2m)V_0 = 3mV_0 \]

Setting initial momentum equal to final momentum:
\[ mv = 3mV_0 \]

### Step 3: Solve for the common velocity \( V_0 \)
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ v = 3V_0 \implies V_0 = \frac{v}{3} \]

### Step 4: Calculate the initial and final kinetic energies of the lighter particle
- The initial kinetic energy \( KE_i \) of the lighter particle (mass \( m \)) is:
\[ KE_i = \frac{1}{2}mv^2 \]

- The final kinetic energy \( KE_f \) of the lighter particle after the collision (moving with velocity \( V_0 = \frac{v}{3} \)) is:
\[ KE_f = \frac{1}{2}m\left(\frac{v}{3}\right)^2 = \frac{1}{2}m\frac{v^2}{9} = \frac{mv^2}{18} \]

### Step 5: Calculate the kinetic energy lost
The kinetic energy lost \( \Delta KE \) by the lighter particle is:
\[ \Delta KE = KE_i - KE_f = \frac{1}{2}mv^2 - \frac{mv^2}{18} \]

To simplify this, we need a common denominator:
\[ \Delta KE = \frac{9}{18}mv^2 - \frac{1}{18}mv^2 = \frac{8}{18}mv^2 = \frac{4}{9}mv^2 \]

### Final Answer
The kinetic energy lost by the lighter particle during the period of deformation is:
\[ \Delta KE = \frac{4}{9}mv^2 \]

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