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Motion along a rough inclined plane - class-XI
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A block released from rest from the top of a smooth inclined plane of angle $\theta _1$ reaches the bottom in time $t _1$. The same block released from rest from the top of another smooth inclined plane of angle $\theta _2$, reaches the bottom in time $t _2$. If the two inclined planes have the same height, the relation between $t _1$ and $t _2$ is
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A
$\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)$
💡 Explanation:
From the figure
we see that
down the incline acceleration us $a=g\sin \theta$
distance $s=\dfrac {h}{\sin \theta _1}$
using $s=ut+\dfrac {1}{2} at^2 \ ;\ u=0$ (initially rest)
gives $t=\sqrt {\dfrac {2s}{a}}$
For $\theta _1 \quad t _1=\sqrt {\dfrac {2\ h}{\sin \theta _1 \times g\sin \theta _1}}=\dfrac {1}{\sin \theta _1} \sqrt {\dfrac {2\ h}{g}}$
so for $O _2, \ t _2=\dfrac {1}{\sin \theta _2}\sqrt {\dfrac {2\ h}{g}}$
$\Rightarrow \ \dfrac {t _2}{t _1}=\dfrac {\sin \theta _1}{\sin \theta _2}$