Question

Two blocks $\text{A}$ and $\text{B}$ of equal masses are released from an inclined plane of inclination ${45}^{\circ }$ at $t=0$. Both the blocks are initially at rest. The coefficient of kinetic friction between the block $\text{A}$ and the inclined plane is $0.2$ while it is $0.3$ for block $\text{B}$. Initially the block $\text{A}$ is $\sqrt{2}\text{m}$ behind the block $\text{B}$. When will their front faces come in a line?

• A. $2\text{s}$
• B. $3\text{s}$
• C. $4\text{s}$
• D. $5\text{s}$

Answer 1

The correct option is A $2\text{s}$

Free body diagram of the blocks will be same.

Let, acceleration of $\text{A}$ is $a_A$ down the plane.

From $\text{FBD}$, we have

$a_A=g\sin \theta -{\mu }_{A}g\cos \theta$

Substituting, $\theta ={45}^{\circ }$ and ${\mu }_A=0.2$, we get

$a_A=g\sin {45}^{\circ }-0.2\times g\cos {45}^{\circ }$

$\Rightarrow a_A=10\times \frac{1}{\sqrt{2}}-0.2\times 10\times \frac{1}{\sqrt{2}}$

$\therefore a_A=4\sqrt{2}{\text{m/s}}^2$

Similarly for $\text{B}$,

$a_B=g\sin {45}^{\circ }-0.3\times g\cos {45}^{\circ }$

$\Rightarrow a_B=10\times \frac{1}{\sqrt{2}}-0.3\times 10\times \frac{1}{\sqrt{2}}$

$\therefore a_B=3.5\sqrt{2}{\text{m/s}}^2$

The front face of $\text{A}$ and $\text{B}$ will come in a line when,

$\text{Distance travelled by block A=Distance travelled by block B}+\sqrt{2}$

$S_A=S_B+\sqrt{2}$

$\Rightarrow \frac{1}{2}{a}_A{t}^2=\frac{1}{2}{a}_B{t}^2+\sqrt{2}$

$\Rightarrow \frac{1}{2}\times 4\sqrt{2}\times t^2=\frac{1}{2}\times 3.5\sqrt{2}\times t^2+\sqrt{2}$

$\Rightarrow t=2\text{s}$

Hence, option (a) is correct answer.