Question

In figure shown, the triangular wedge of mass $M$ is on smooth horizontal surface. No friction between block of mass $m$ and inclined surface of the wedge. If the wedge displaces towards right with an acceleration $a$. Then

• A. Acceleration of the block is $a$ relative to ground
• B. Acceleration of the block is $a\cos \alpha$ relative to wedge
• C. Acceleration of the block is $2a\sin \frac{\alpha }{2}$ relative to ground
• D. Acceleration of the block is $2a\cos \frac{\alpha }{2}$ relative to ground.

Answer 1

The correct option is C Acceleration of the block is $2a\sin \frac{\alpha }{2}$ relative to ground

$\begin{array}{c}\text{Correct option is (c)}\\ \text{Relative to wedge block has acceleration}\Rightarrow \\ {\overrightarrow{a}}_0=-(a\cos \alpha \hat{i}+a\sin \alpha \hat{j})\\ \text{since relative to ground acceleration of block}\\ \text{is}{\overrightarrow{a}}_G=a(1-\cos \alpha )\hat{i}-a\sin \alpha \hat{j}\\ \begin{array}{cc}\left|{\overrightarrow{a}}_G\right|& =a\sqrt{(1-\cos \alpha {)}^2+{\sin }^2\alpha }\\ & =a\sqrt{2-2\cos \alpha }\\ & =a\sqrt{2(1-\cos \alpha )}\\ & =2a\sin \alpha /2\end{array}\end{array}$