Two block $A$ and $B$ of equal masses are placed on rough inclined plane as shown in figure. When and where will the two blocks come on the same line on the inclined plane if they are released simultaneously? Initially the block $A$ is $\sqrt{2}\,\text{m}$ behind the block $B$. Co-efficient of kinetic friction for the blocks $A$ and $B$ are $0.2$ and $0.3$ respectively ($g=10\,\text{m/s}^{2}$).

Step 1: Identify the formula for acceleration down a rough inclined plane.

For a block of mass $m$ on an inclined plane with angle $\theta$ and coefficient of kinetic friction $\mu$, the equation of motion along the incline is:

$$m a = m g \sin\theta - f$$

The kinetic friction force is given by:

$$f = \mu N = \mu m g \cos\theta$$

Substituting $f$ into the equation of motion yields the acceleration:

$$a = g(\sin\theta - \mu\cos\theta)$$

Step 2: Calculate the acceleration of block $A$.

Substitute the given values $g = 10\,\text{m/s}^{2}$, $\theta = 45^\circ$, and $\mu_A = 0.2$ into the acceleration formula.

$$a_A = 10(\sin 45^\circ - 0.2\cos 45^\circ)$$ $$a_A = 10\left(\frac{1}{\sqrt{2}} - 0.2 \cdot \frac{1}{\sqrt{2}}\right)$$ $$a_A = 4\sqrt{2}\,\text{m/s}^{2}$$

Step 3: Calculate the acceleration of block $B$.

Substitute the given values $g = 10\,\text{m/s}^{2}$, $\theta = 45^\circ$, and $\mu_B = 0.3$ into the acceleration formula.

$$a_B = 10(\sin 45^\circ - 0.3\cos 45^\circ)$$ $$a_B = 10\left(\frac{1}{\sqrt{2}} - 0.3 \cdot \frac{1}{\sqrt{2}}\right)$$ $$a_B = 3.5\sqrt{2}\,\text{m/s}^{2}$$

Step 4: Determine the relative acceleration between the two blocks.

Since block $A$ is behind block $B$ and has a higher acceleration, it will catch up to block $B$.

$$a_{\text{rel}} = a_A - a_B$$ $$a_{\text{rel}} = 4\sqrt{2} - 3.5\sqrt{2} = 0.5\sqrt{2}\,\text{m/s}^{2}$$

Step 5: Calculate the time taken for block $A$ to catch block $B$.

Using the relative kinematic equation with the initial separation $s_{\text{rel}} = \sqrt{2}\,\text{m}$:

$$s_{\text{rel}} = \frac{1}{2} a_{\text{rel}} t^{2}$$ $$\sqrt{2} = \frac{1}{2} \left(0.5\sqrt{2}\right) t^{2}$$ $$t^{2} = 4$$ $$t = 2\,\text{s}$$

Step 6: Calculate the distance traveled by each block.

Using the equations of motion from rest for each block over $t = 2\,\text{s}$:

$$s_B = \frac{1}{2} a_B t^{2} = \frac{1}{2} \left(3.5\sqrt{2}\right) (2)^{2} = 7\sqrt{2}\,\text{m}$$ $$s_A = \frac{1}{2} a_A t^{2} = \frac{1}{2} \left(4\sqrt{2}\right) (2)^{2} = 8\sqrt{2}\,\text{m}$$

Answer: The two blocks will come on the same line after $2\,\text{s}$, at a distance of $7\sqrt{2}\,\text{m}$ from the initial position of block $B$ (or $8\sqrt{2}\,\text{m}$ from the initial position of block $A$).