In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

Options

A. `(v_(@))/(sqrt(2))`
B. `(v_(@))/(4)`
C. `sqrt(2)v_(@)`
D. `(v_(@))/(2)`

Answer

To solve the problem, we need to analyze the collinear collision between two particles of equal mass. Let's denote the mass of each particle as \( m \), the initial speed of the moving particle as \( v_0 \), and the final speeds of the two particles after the collision as \( v_1 \) and \( v_2 \).

### Step-by-Step Solution:

1. **Initial Kinetic Energy Calculation**:
The initial kinetic energy (KE_initial) of the moving particle is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m v_0^2 \]

2. **Final Kinetic Energy Calculation**:
According to the problem, the final total kinetic energy is 50% greater than the initial kinetic energy:
\[ KE_{\text{final}} = KE_{\text{initial}} + 0.5 \times KE_{\text{initial}} = 1.5 \times KE_{\text{initial}} = 1.5 \times \frac{1}{2} m v_0^2 = \frac{3}{4} m v_0^2 \]

3. **Using Conservation of Momentum**:
The conservation of linear momentum states that the total momentum before the collision equals the total momentum after the collision:
\[ m v_0 = m v_1 + m v_2 \]
Simplifying, we get:
\[ v_0 = v_1 + v_2 \quad (1) \]

4. **Using the Kinetic Energy Relation**:
The final kinetic energy can also be expressed in terms of \( v_1 \) and \( v_2 \):
\[ KE_{\text{final}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (v_1^2 + v_2^2) \]
Setting this equal to the expression for \( KE_{\text{final}} \):
\[ \frac{1}{2} m (v_1^2 + v_2^2) = \frac{3}{4} m v_0^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ v_1^2 + v_2^2 = \frac{3}{2} v_0^2 \quad (2) \]

5. **Substituting Equation (1) into Equation (2)**:
From equation (1), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_0 - v_1 \]
Substituting this into equation (2):
\[ v_1^2 + (v_0 - v_1)^2 = \frac{3}{2} v_0^2 \]
Expanding the second term:
\[ v_1^2 + (v_0^2 - 2v_0 v_1 + v_1^2) = \frac{3}{2} v_0^2 \]
Combining like terms:
\[ 2v_1^2 - 2v_0 v_1 + v_0^2 = \frac{3}{2} v_0^2 \]
Rearranging gives:
\[ 2v_1^2 - 2v_0 v_1 + v_0^2 - \frac{3}{2} v_0^2 = 0 \]
Simplifying:
\[ 2v_1^2 - 2v_0 v_1 - \frac{1}{2} v_0^2 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 4v_1^2 - 4v_0 v_1 - v_0^2 = 0 \]

6. **Solving the Quadratic Equation**:
Using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -4v_0, c = -v_0^2 \):
\[ v_1 = \frac{4v_0 \pm \sqrt{(-4v_0)^2 - 4 \cdot 4 \cdot (-v_0^2)}}{2 \cdot 4} \]
Simplifying:
\[ v_1 = \frac{4v_0 \pm \sqrt{16v_0^2 + 16v_0^2}}{8} = \frac{4v_0 \pm \sqrt{32v_0^2}}{8} = \frac{4v_0 \pm 4\sqrt{2}v_0}{8} \]
\[ v_1 = \frac{v_0(1 \pm \sqrt{2})}{2} \]

7. **Finding \( v_2 \)**:
Using \( v_2 = v_0 - v_1 \):
\[ v_2 = v_0 - \frac{v_0(1 \pm \sqrt{2})}{2} = \frac{v_0(2 - (1 \pm \sqrt{2}))}{2} \]

8. **Calculating the Relative Velocity**:
The relative velocity between the two particles after the collision is given by:
\[ |v_1 - v_2| = |(v_0(1 \pm \sqrt{2})/2) - (v_0(2 - (1 \pm \sqrt{2}))/2)| \]
This simplifies to:
\[ |v_1 - v_2| = \sqrt{2} v_0 \]

### Final Answer:
The magnitude of the relative velocity between the two particles after the collision is:
\[ \sqrt{2} v_0 \]