Question

Two blocks $A$ and $B$ 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 $A$ and the inclined plane is $0.2$ while it is $0.3$ for block $B$. Initially the block $A$ is $\sqrt{2}\text{m}$ behind the block $B$. When and where from the position of A, front faces of both the blocks will come in a line? (Take $g=10m/s^2)$

• A. $2\text{s},8\sqrt{2}\text{m}$
• B. $1\text{s},8\sqrt{2}\text{m}$
• C. $2\text{s},10\sqrt{2}\text{m}$
• D. $1\text{s},10\sqrt{2}\text{m}$

Answer 1

The correct option is A $2\text{s},8\sqrt{2}\text{m}$

Acceleration of $A$ down the plane.

$a_A=g\sin {45}^{\circ }-{\mu }_{Ag}\cos {45}^{\circ }$
$=\left(10\right)\left(\frac{1}{\sqrt{2}}\right)-\left(0.2\right)\left(10\right)\left(\frac{1}{\sqrt{2}}\right)=4\sqrt{2}\text{m}/s^2$
Similarly acceleration of $B$ down the plane.
$a_B=g\sin {45}^{\circ }-{\mu }_{Bg}\cos {45}^{\circ }$
$=\left(10\right)\left(\frac{1}{\sqrt{2}}\right)-\left(0.3\right)\left(10\right)\left(\frac{1}{\sqrt{2}}\right)=3.5\sqrt{2}\text{m}/s^2$
The front face of $A$ and $B$ will come in a line when,
$s_A=s_B+\sqrt{2}$
[where $s_A$ and $s_B$ are the distance travelled by the block $A$ and $B$ in $m$ respectively]
or $\frac{1}{2}{a}_A{t}^2=\frac{1}{2}{a}_B{t}^2+\sqrt{2}$
$\frac{1}{2}\times 4\sqrt{2}\times t^2=\frac{1}{2}\times 3.5\sqrt{2}\times t^2+\sqrt{2}$
Solving this equation, we get $t=2\text{s}$
Further, $s_A=\frac{1}{2}{a}_A{t}^2=\frac{1}{2}\times 4\sqrt{2}\times (2{)}^2=8\sqrt{2}m$
Hence, both the blocks will come in a line after $A$ has travelled a distance $8\sqrt{2}\text{m}$ down the plane.