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- Two blocks of equal mass have been placed on two faces of a fixed wedge as shown in figure. The blocks are released from position where centre of one block is at a height h above the centre of the other block. Find the time after which the centre of the two blocks will be at same horizontal level. There is no friction anywhere.
Q.
Two blocks of equal mass have been placed on two faces of a fixed wedge as shown in figure. The blocks are released from position where centre of one block is at a height $h$ above the centre of the other block. Find the time after which the centre of the two blocks will be at same horizontal level. There is no friction anywhere.
Laws of Motion
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Solution:
Acceleration of the two blocks,
$a=\frac{m g \sin 60^{\circ}-m g \sin 30^{\circ}}{3 m}=\left(\frac{\sqrt{3}-1}{4}\right) g$
If the two blocks move a distance $x$ along respective inclines in time $t, x=\frac{1}{2} a t^{2}$
The centre of two blocks will be at equal height if
$x \sin 60^{\circ}+x \sin 30^{\circ}=h$
$\therefore x\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)=h$
$\Rightarrow \frac{1}{2}\left(\frac{\sqrt{3}-1}{4}\right) g t^{2}\left(\frac{\sqrt{3}+1}{2}\right)=h$
$\therefore t^{2}=\frac{16 h}{(\sqrt{3}-1)(\sqrt{3}+1) g}=\frac{8 h}{g}$
$\Rightarrow t=2 \sqrt{2} \sqrt{\frac{h}{g}}$
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