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 given information step by step.

### Step 1: Understand the Initial Conditions
We have two particles of equal mass \( m \). One particle is moving with an initial speed \( v_0 \) while the other is stationary. Therefore, the initial kinetic energy \( KE_i \) of the system is given by:
\[ KE_i = \frac{1}{2} m v_0^2 \]

### Step 2: Determine the Final Kinetic Energy
According to the problem, the final total kinetic energy is 50% greater than the initial kinetic energy. Thus, we can express the final kinetic energy \( KE_f \) as:
\[ KE_f = 1.5 \times KE_i = 1.5 \times \frac{1}{2} m v_0^2 = \frac{3}{4} m v_0^2 \]

### Step 3: Set Up the Equation for Final Kinetic Energy
After the collision, let the velocities of the two particles be \( v_1 \) and \( v_2 \). The final kinetic energy can also be expressed as:
\[ KE_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]
Since the masses are the same, we can cancel \( m \) and \( \frac{1}{2} \):
\[ v_1^2 + v_2^2 = \frac{3}{2} v_0^2 \tag{1} \]

### Step 4: Apply Conservation of Momentum
Since there are no external forces acting on the system in the horizontal direction, we can apply the conservation of momentum:
\[ m v_0 = m v_1 + m v_2 \]
Again, cancel \( m \):
\[ v_0 = v_1 + v_2 \tag{2} \]

### Step 5: Solve the System of Equations
We now have two equations (1) and (2):
1. \( v_1^2 + v_2^2 = \frac{3}{2} v_0^2 \)
2. \( v_0 = v_1 + v_2 \)

From equation (2), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_0 - v_1 \]

Substituting \( v_2 \) into equation (1):
\[ v_1^2 + (v_0 - v_1)^2 = \frac{3}{2} v_0^2 \]
Expanding the second term:
\[ v_1^2 + (v_0^2 - 2v_0v_1 + v_1^2) = \frac{3}{2} v_0^2 \]
Combining like terms:
\[ 2v_1^2 - 2v_0v_1 + v_0^2 = \frac{3}{2} v_0^2 \]
Rearranging gives:
\[ 2v_1^2 - 2v_0v_1 + v_0^2 - \frac{3}{2} v_0^2 = 0 \]
This simplifies to:
\[ 2v_1^2 - 2v_0v_1 - \frac{1}{2} v_0^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Multiplying through by 2 to eliminate the fraction:
\[ 4v_1^2 - 4v_0v_1 - v_0^2 = 0 \]
Using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -4v_0, c = -v_0^2 \):
\[ v_1 = \frac{4v_0 \pm \sqrt{(-4v_0)^2 - 4 \cdot 4 \cdot (-v_0^2)}}{2 \cdot 4} \]
\[ = \frac{4v_0 \pm \sqrt{16v_0^2 + 16v_0^2}}{8} \]
\[ = \frac{4v_0 \pm \sqrt{32v_0^2}}{8} \]
\[ = \frac{4v_0 \pm 4v_0\sqrt{2}}{8} \]
\[ = \frac{v_0(1 \pm \sqrt{2})}{2} \]

### Step 7: Find \( v_2 \)
Using \( v_2 = v_0 - v_1 \):
\[ v_2 = v_0 - \frac{v_0(1 \pm \sqrt{2})}{2} \]
This gives us two possible values for \( v_2 \).

### Step 8: Calculate the Relative Velocity
The relative velocity \( v_{rel} \) is given by:
\[ v_{rel} = |v_2 - v_1| \]
Substituting the values of \( v_1 \) and \( v_2 \) will yield the final answer.

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