A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision what could be minimum and the maximum value of u.

Explanation

To solve the problem, we need to find the minimum and maximum values of the initial speed \( u \) of a particle of mass 100 g (0.1 kg) that collides with another particle of the same mass at rest, given that the total kinetic energy after the collision is 0.2 J.

### Step-by-Step Solution

1. **Convert Mass to Kilograms**:
The mass of each particle is given as 100 g. We convert this to kilograms:
\[ m = 100 \, \text{g} = 0.1 \, \text{kg} \]

2. **Write the Expression for Final Kinetic Energy**:
The total kinetic energy after the collision is given as 0.2 J. The final kinetic energy can be expressed as:
\[ KE_f = \frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 = 0.2 \, \text{J} \]
Substituting \( m = 0.1 \, \text{kg} \):
\[ 0.1 \left( \frac{1}{2} V_1^2 + \frac{1}{2} V_2^2 \right) = 0.2 \]
Simplifying gives:
\[ V_1^2 + V_2^2 = 4 \quad \text{(Equation 1)} \]

3. **Apply Conservation of Momentum**:
The initial momentum of the system is:
\[ P_{initial} = m u = 0.1 u \]
The final momentum is:
\[ P_{final} = m V_1 + m V_2 = 0.1 V_1 + 0.1 V_2 \]
Setting the initial momentum equal to the final momentum:
\[ 0.1 u = 0.1 (V_1 + V_2) \]
This simplifies to:
\[ u = V_1 + V_2 \quad \text{(Equation 2)} \]

4. **Relate Velocities Using Coefficient of Restitution**:
The coefficient of restitution \( e \) relates the velocities before and after the collision:
\[ e = \frac{V_2 - V_1}{u} \]
Rearranging gives:
\[ V_2 - V_1 = e u \quad \text{(Equation 3)} \]

5. **Solve the System of Equations**:
We have three equations:
- \( V_1 + V_2 = u \) (Equation 2)
- \( V_1^2 + V_2^2 = 4 \) (Equation 1)
- \( V_2 - V_1 = e u \) (Equation 3)

From Equation 2, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = u - V_1 \]
Substituting this into Equation 1:
\[ V_1^2 + (u - V_1)^2 = 4 \]
Expanding gives:
\[ V_1^2 + (u^2 - 2uV_1 + V_1^2) = 4 \]
Simplifying:
\[ 2V_1^2 - 2uV_1 + u^2 - 4 = 0 \]
This is a quadratic equation in \( V_1 \).

6. **Use the Discriminant**:
For \( V_1 \) to have real solutions, the discriminant must be non-negative:
\[ D = (-2u)^2 - 4 \cdot 2 \cdot (u^2 - 4) \geq 0 \]
Simplifying:
\[ 4u^2 - 8(u^2 - 4) \geq 0 \]
\[ 4u^2 - 8u^2 + 32 \geq 0 \]
\[ -4u^2 + 32 \geq 0 \]
\[ u^2 \leq 8 \]
Thus:
\[ u \leq \sqrt{8} = 2\sqrt{2} \]

7. **Find Minimum Value of \( u \)**:
The minimum value of \( u \) occurs when \( e = 1 \):
\[ u = 2 \quad \text{(minimum)} \]

8. **Find Maximum Value of \( u \)**:
The maximum value of \( u \) occurs when \( e = 0 \):
\[ u = 2\sqrt{2} \quad \text{(maximum)} \]

### Final Answer
- Minimum value of \( u = 2 \, \text{m/s} \)
- Maximum value of \( u = 2\sqrt{2} \, \text{m/s} \)