1. Since the track is frictionless, mechanical energy is conserved.

2. The particle is gently pushed, so its initial speed at $A$ can be taken as zero.

3. From the figure, point $B$ is lower than point $A$ by a vertical height of $h=0.5\ \text{m}$

4. Apply conservation of mechanical energy between $A$ and $B$: $mgh=\frac{1}{2}mv^2$

5. Substitute $g=10\ \text{m/s}^2$ and $h=0.5\ \text{m}$: $m(10)(0.5)=\frac{1}{2}mv^2$ $5m=\frac{1}{2}mv^2$

6. Cancel $m$: $5=\frac{v^2}{2}$ $v^2=10$ $v=\sqrt{10}\ \text{m/s}$

7. Now compare with the options:

• A: $2\sqrt{10}$
• B: $10$
• C: $\sqrt{10}$
• D: $20$

So the correct option is: $\boxed{\text{C }\; \sqrt{10}\ \text{m/s}}$