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 the collision, is:

Let the mass of each particle be $$m$$. The incident particle has an initial speed $$v_0$$, while the target particle is at rest.

We first apply conservation of linear momentum, because no external force acts along the line of motion.

After the collision, let the speeds of the two particles be $$v_1$$ and $$v_2$$ (positive to the right). Then

$$\text{Initial momentum}=m\,v_0,$$

$$\text{Final momentum}=m\,v_1+m\,v_2.$$

Equating the two, we obtain

$$$m\,v_0=m\,v_1+m\,v_2 \;\;\Longrightarrow\;\; v_1+v_2=v_0.\tag1$$$

Next we use the information about kinetic energy. The total kinetic energy after the collision is said to be 50 % greater than the initial kinetic energy. Stating the formula first, the kinetic energy of a particle of mass $$m$$ moving with speed $$v$$ is $$\tfrac12 m v^2$$.

Initial kinetic energy:

$$K_i=\frac12 m v_0^{\,2}.$$

Final kinetic energy:

$$K_f=\frac12 m v_1^{\,2}+\frac12 m v_2^{\,2}.$$

According to the question,

$$K_f=1.5\,K_i.$$

Substituting the expressions,

$$\frac12 m\left(v_1^{\,2}+v_2^{\,2}\right)=1.5\left(\frac12 m v_0^{\,2}\right).$$

The common factor $$\tfrac12 m$$ cancels out, giving

$$$v_1^{\,2}+v_2^{\,2}=1.5\,v_0^{\,2}=\frac32\,v_0^{\,2}.\tag2$$$

Our goal is the magnitude of the relative speed after the collision, namely $$|v_1-v_2|$$. To find this, we compute $$(v_1-v_2)^2$$ and then take the square root.

We know the algebraic identity

$$(v_1-v_2)^2=(v_1+v_2)^2-4v_1 v_2.$$

First, from equation (1),

$$(v_1+v_2)^2=v_0^{\,2}.$$

Second, we find the product $$v_1 v_2$$. Starting with

$$(v_1+v_2)^2=v_1^{\,2}+v_2^{\,2}+2v_1 v_2,$$

we substitute the known values:

$$v_0^{\,2}= \frac32\,v_0^{\,2}+2v_1 v_2.$$

Rearranging,

$$2v_1 v_2=v_0^{\,2}-\frac32\,v_0^{\,2}=-\frac12\,v_0^{\,2},$$

so

$$$v_1 v_2=-\frac14\,v_0^{\,2}.\tag3$$$

Now we evaluate the relative speed:

$$(v_1-v_2)^2=(v_1+v_2)^2-4v_1 v_2=v_0^{\,2}-4\left(-\frac14 v_0^{\,2}\right)=v_0^{\,2}+v_0^{\,2}=2\,v_0^{\,2}.$$

Taking the positive square root (because speed is positive), we find

$$|v_1-v_2|=\sqrt{2}\;v_0.$$

Hence, the correct answer is Option C.