A particle moving with kinetic energy `= 3J` makes an elastic head-on collision with a stationary particle which has twice its mass. During the impact :-

Options

A. the minimum kinetic energy of the system is `1 J`
B. the change in elastic potential energy of the system is `2 J`
C. momentum and total energy are conserved at every instant.
D. the ratio of kinetic energy to potential energy of the system first decreases and then increases.

Explanation

To solve the problem step by step, we will analyze the elastic collision between two particles, one of which is moving and the other is stationary.

### Step 1: Understand the Given Information
- A particle with kinetic energy \( KE_1 = 3 \, J \) is moving.
- The mass of this particle is \( m \).
- The second particle, which is stationary, has a mass of \( 2m \).
- We need to analyze the situation during the elastic collision.

### Step 2: Calculate the Initial Velocity of the Moving Particle
The kinetic energy of the moving particle is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
Setting this equal to the given kinetic energy:
\[ \frac{1}{2} m u^2 = 3 \]
From this, we can solve for \( u^2 \):
\[ u^2 = \frac{6}{m} \]
Thus, the initial velocity \( u \) of the first particle is:
\[ u = \sqrt{\frac{6}{m}} \]

### Step 3: Apply Conservation of Momentum
In an elastic collision, momentum is conserved. The initial momentum of the system is:
\[ p_{initial} = m u + 2m \cdot 0 = m u \]
Let \( v_1 \) be the final velocity of the first particle and \( v_2 \) be the final velocity of the second particle. The momentum conservation equation becomes:
\[ m u = m v_1 + 2m v_2 \]
Dividing through by \( m \):
\[ u = v_1 + 2v_2 \quad \text{(1)} \]

### Step 4: Apply Conservation of Kinetic Energy
In an elastic collision, kinetic energy is also conserved. Thus:
\[ KE_{initial} = KE_{final} \]
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (2m) v_2^2 \]
Simplifying gives:
\[ u^2 = v_1^2 + 2v_2^2 \quad \text{(2)} \]

### Step 5: Solve the System of Equations
We have two equations (1) and (2):
1. \( u = v_1 + 2v_2 \)
2. \( u^2 = v_1^2 + 2v_2^2 \)

Substituting \( v_1 = u - 2v_2 \) from equation (1) into equation (2):
\[ u^2 = (u - 2v_2)^2 + 2v_2^2 \]
Expanding the equation:
\[ u^2 = u^2 - 4uv_2 + 4v_2^2 + 2v_2^2 \]
This simplifies to:
\[ 0 = -4uv_2 + 6v_2^2 \]
Factoring out \( v_2 \):
\[ v_2(6v_2 - 4u) = 0 \]
Thus, either \( v_2 = 0 \) (which is not possible since the second particle moves after the collision) or:
\[ 6v_2 = 4u \implies v_2 = \frac{2u}{3} \]

### Step 6: Find \( v_1 \)
Substituting \( v_2 \) back into equation (1):
\[ u = v_1 + 2\left(\frac{2u}{3}\right) \]
This gives:
\[ u = v_1 + \frac{4u}{3} \]
Rearranging:
\[ v_1 = u - \frac{4u}{3} = -\frac{1u}{3} \]

### Step 7: Calculate the Kinetic Energies After Collision
The kinetic energy of the first particle after the collision:
\[ KE_1' = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left(-\frac{u}{3}\right)^2 = \frac{1}{2} m \frac{u^2}{9} = \frac{1}{2} m \frac{6}{m \cdot 9} = \frac{1}{3} \, J \]
The kinetic energy of the second particle after the collision:
\[ KE_2' = \frac{1}{2} (2m) v_2^2 = m \left(\frac{2u}{3}\right)^2 = m \frac{4u^2}{9} = m \frac{4 \cdot 6}{9m} = \frac{8}{3} \, J \]

### Step 8: Total Kinetic Energy After Collision
The total kinetic energy after the collision is:
\[ KE_{total}' = KE_1' + KE_2' = \frac{1}{3} + \frac{8}{3} = 3 \, J \]

### Conclusion
The minimum kinetic energy of the system is \( 1 \, J \) and the change in the elastic potential energy is \( 2 \, J \).