Question

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.

• A. $2 \sqrt{2} \sqrt{\frac{h}{g}}$
• B. $2 \sqrt{2} \sqrt{\frac{h}{g}}$
• C. $2 \sqrt{2} \sqrt{\frac{h}{g}}$
• D. $2 \sqrt{2} \sqrt{\frac{h}{g}}$

Answer 1

Answer: (A)

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
$x \sin 60^{\circ}+x \sin 30^{\circ}=h$
$\therefore x\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)=h$
$\Rightarrow \frac{1}{2}\left(\frac{\sqrt{3}-1}{4}\right) g t^{2}\left(\frac{\sqrt{3}+1}{2}\right)=h$
$\therefore t^{2}=\frac{16 h}{(\sqrt{3}-1)(\sqrt{3}+1) g}=\frac{8 h}{g}$
$\Rightarrow t=2 \sqrt{2} \sqrt{\frac{h}{g}}$