Question

Two fixed frictionless inclined plane making an angle ${30}^{\circ }$ and ${60}^{\circ }$ with the vertical as shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$?

• A. $4.9{\text{ms}}^{-2}$ in horizontal direction
• B. $9.8{\text{ms}}^{-2}$ in vertical direction
• C. Zero
• D. $4.9{\text{ms}}^{-2}$ in vertical direction

Answer 1

The correct option is D $4.9{\text{ms}}^{-2}$ in vertical direction


From FBD of mass $A$ and $B$, we get
${\overrightarrow{a}}_A=g\sin {60}^{\circ }\cos {60}^{\circ }\hat{i}-g{\sin }^2{60}^{\circ }\hat{j}$
${\overrightarrow{a}}_A=\frac{\sqrt{3}}{4}g\hat{i}-\frac{3}{4}g\hat{j}$
${\overrightarrow{a}}_B=g\sin {30}^{\circ }\cos {30}^{\circ }\hat{i}-g{\sin }^2{30}^{\circ }\hat{j}$
${\overrightarrow{a}}_B=\frac{\sqrt{3}}{4}g\hat{i}-\frac{1}{4}g\hat{j}$
The relative acceleration of block $A$ with respect to $B$.
${\overrightarrow{a}}_{A/B}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B=\frac{-g}{2}\hat{j}$
$\Rightarrow {\overrightarrow{a}}_{A/B}=4.9{\text{ms}}^{-2}$ in vertical direction