In a collinear collision, a particle with an initial speed `v_0` 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_0)/(2)`
B. `(v_0)/(sqrt2)`
C. `(v_0)/(4)`
D. `sqrt(2)v_0`

Explanation

To solve the problem, we need to analyze the situation using the principles of conservation of momentum and kinetic energy. Here’s a step-by-step solution:

### Step 1: Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the moving particle can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} m v_0^2 \]
where \(m\) is the mass of the particle and \(v_0\) is its initial speed.

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

### Step 3: Expressing Final Kinetic Energy
After the collision, the final kinetic energy can also be expressed as:
\[ KE_{\text{final}} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
where \(v_1\) and \(v_2\) are the speeds of the two particles after the collision.

### Step 4: Setting Up the Equation
Equating the two expressions for the final kinetic energy, we have:
\[ \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{3}{4} m v_0^2 \]
Dividing through by \(\frac{1}{2} m\) (assuming \(m \neq 0\)):
\[ v_1^2 + v_2^2 = \frac{3}{2} v_0^2 \quad \text{(Equation 1)} \]

### Step 5: Conservation of Momentum
Using the conservation of momentum, we have:
\[ m v_0 = m v_1 + m v_2 \]
Dividing through by \(m\):
\[ v_0 = v_1 + v_2 \quad \text{(Equation 2)} \]

### Step 6: Squaring Equation 2
Now, we square Equation 2:
\[ v_0^2 = (v_1 + v_2)^2 = v_1^2 + v_2^2 + 2 v_1 v_2 \]

### Step 7: Substituting Equation 1 into Squared Equation
Substituting Equation 1 into this equation gives:
\[ v_0^2 = \frac{3}{2} v_0^2 + 2 v_1 v_2 \]
Rearranging this, we find:
\[ v_0^2 - \frac{3}{2} v_0^2 = 2 v_1 v_2 \]
\[ -\frac{1}{2} v_0^2 = 2 v_1 v_2 \]
Thus,
\[ v_1 v_2 = -\frac{1}{4} v_0^2 \quad \text{(Equation 3)} \]

### Step 8: Finding the Relative Velocity
Now we need to find the magnitude of the relative velocity \( |v_1 - v_2| \). We can express this as:
\[ |v_1 - v_2|^2 = v_1^2 + v_2^2 - 2v_1 v_2 \]
Substituting Equation 1 and Equation 3:
\[ |v_1 - v_2|^2 = \frac{3}{2} v_0^2 - 2\left(-\frac{1}{4} v_0^2\right) \]
\[ |v_1 - v_2|^2 = \frac{3}{2} v_0^2 + \frac{1}{2} v_0^2 = 2 v_0^2 \]

### Step 9: Taking the Square Root
Taking the square root gives:
\[ |v_1 - v_2| = \sqrt{2} v_0 \]

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