Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of $45^o$ . Their coefficients of kinetic friction are $\mu _A = 0.2$ and $\mu _B = 0.3$ respectively. At $t = 0$ , both the blocks are at rest and block $A$ is $\sqrt 2$ metre behind block $B$ . The time and distance from the initial position where the front faces of the blocks come in line on the inclined plane as shown in figure. (Use $g = 10\, ms^{-2}$ )
Difficult
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$a_{A}=g\left[\sin 45-\mu_{A} \cos 45\right]=\frac{8}{\sqrt{2}}$
$a_{B}=g\left[\sin 45-\mu_{B} \cos 45\right]=\frac{7}{\sqrt{2}}$
$a_{A B}=a_{A}-a_{B}=g\left(\mu_{B}-\mu_{A}\right) \cos 45=\frac{1}{\sqrt{2}}$
$s_{A B}=\sqrt{2}$
Now $s_{\mathrm{AB}}=\frac{1}{2} a_{\mathrm{AB}} \mathrm{t}^{2} \Rightarrow \sqrt{2}=\frac{1}{2} \times \frac{1}{\sqrt{2}} \mathrm{t}^{2}$
$\Rightarrow \quad t=2 \mathrm{sec}$
Again $s_{A}=\frac{1}{2} a_{A} t^{2}=\frac{1}{2}\left(\frac{8}{\sqrt{2}}\right) 4$
$\Rightarrow \quad s_{A}=8 \sqrt{2} \mathrm{m}$
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