Question

Two particles start together from a point O and slide down along straight smooth wires inclined at ${30}^o$ and ${60}^o$ to the vertical plane and on the same side of vertical through O. The relative acceleration of second with respect to first will be of magnitude:

• A. $\frac{g}{2}$ in the vertical direction
• B. $\frac{\sqrt{3}g}{2}$ at ${45}^o$ with vertical
• C. $\frac{g}{\sqrt{3}}$ inclined at ${60}^o$ to the vertical
• D. $g$ in the vertical direction.

Answer 1

The correct option is A $\frac{g}{2}$ in the vertical direction

we can see only the gravitational force is acting in both cases but as the motion is along a diagonal plane only the component along plane will be having an effect on the particle and not the vertical component

so the component is $mgcos\theta$

so in case of 1st block it is $mgcos{60}^o$ and block 2 it is $mgcos{30}^o$

we know $a=\frac{F}{m}=gcos\theta$

so the a in case of block 2 is $gcos{60}^o=$ along the plane.

now we can calculate acceleration components along vertical and horizontal directions

so in case I verical component is $acos{60}^o=gcos{60}^o\times cos{60}^o=gco{s}^2{60}^o=\frac{g}{4}$ and horizontal component is $asin{60}^o=gsin{60}^{o}cos{60}^o=\frac{\sqrt{3}g}{4}$

similarly for block II vertical component is $gco{s}^2{30}^o=\frac{3g}{4}$ and horizontal is $gcos{30}^{o}sin{30}^o=\frac{\sqrt{3}g}{4}$

so particle in case II is going $\frac{g}{2}$ faster in vertical direction than in case I.