1. Given

• Two particles collide collinearly.
• Masses are equal: let each mass be $m$.
• Initial velocities:
  • First particle: $u_1 = v_0$
  • Second particle: $u_2 = 0$

So the initial total kinetic energy is

$K_i = \frac12 m v_0^2$

The problem says the final total kinetic energy is 50% greater than the initial kinetic energy. Therefore,

$K_f = 1.5 K_i = \frac32 \cdot \frac12 m v_0^2 = \frac34 m v_0^2$

2. Use momentum conservation

Let final velocities be $v_1$ and $v_2$.

Since no external force acts during collision,

$m v_0 + m(0) = m v_1 + m v_2$

So,

$v_1 + v_2 = v_0$

3. Use kinetic energy relation

Final kinetic energy is

$K_f = \frac12 m v_1^2 + \frac12 m v_2^2 = \frac12 m (v_1^2 + v_2^2)$

Given that

$\frac12 m (v_1^2 + v_2^2) = \frac34 m v_0^2$

Cancelling $m$,

$v_1^2 + v_2^2 = \frac32 v_0^2$

4. Find relative speed after collision

We need

$|v_2 - v_1|$

Use the identity:

$(v_1 + v_2)^2 = v_1^2 + v_2^2 + 2v_1v_2$

Substitute known values:

$v_0^2 = \frac32 v_0^2 + 2v_1v_2$

Hence,

$2v_1v_2 = -\frac12 v_0^2$

$v_1v_2 = -\frac14 v_0^2$

Now,

$(v_2 - v_1)^2 = v_1^2 + v_2^2 - 2v_1v_2$

Substitute:

$(v_2 - v_1)^2 = \frac32 v_0^2 - 2\left(-\frac14 v_0^2\right)$

$(v_2 - v_1)^2 = \frac32 v_0^2 + \frac12 v_0^2 = 2v_0^2$

Therefore,

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

5. Check options

• A: $\dfrac{v_0}{\sqrt2}$
• B: $\dfrac{v_0}{4}$
• C: $\sqrt2\,v_0$
• D: $\dfrac{v_0}{2}$

So the correct option is:

$\boxed{\text{C }\; \sqrt2\,v_0}$