case is $a_2$
Now $\quad \frac{1}{2} a_{1} t^{2}=\frac{1}{2} a_{2}(2 t)^{2}$
$\Rightarrow \quad a_{2}=\frac{a_{1}}{4}$ $...(i)$
Clearly $a_{1}=\frac{\operatorname{mg} \sin \theta}{m}=g \sin \theta \quad$ $...(ii)$
and $a_{2}=\frac{m g \sin \theta-\mu m g \cos \theta}{m}$
$=g \sin \theta-\mu g \cos \theta$ $...(iii)$
From $(i),$ $(ii)$ and $(iii),$ we get
$\mu=0.75$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Reason $(R):$ The linear momentum of an isolated system remains constant.
