Question

In the figure shown , the acceleration of $\text{A}$ is ${\overrightarrow{a}}_A=(15\hat{i}+15\hat{j}){\text{m/s}}^2$ and block $\text{A}$ remains in contact with wedge $\text{B}$. The acceleration of $\text{B}$ $\left(\text{in}{\text{m/s}}^2\right)$ will be.

• A. $6\hat{i}$
• B. $-15\hat{i}$
• C. $-10\hat{i}$
• D. $-5\hat{i}$

Answer 1

The correct option is D $-5\hat{i}$

Let us suppose, the wedge move towards $+ve$ $\text{x-}$axis


From wedge constraints the components of acceleration of $A$ and $B$ along the normal direction must be equal.

$\Rightarrow (a_A{)}_{\perp }=(a_B{)}_{\perp }$

$\Rightarrow (a_A{)}_x\cos {53}^{\circ }-(a_A{)}_y\cos {37}^{\circ }=a_B\cos {53}^{\circ }...\left(1\right)$

Given,
${\overrightarrow{a}}_A=(15\hat{i}+15\hat{j}){\text{m/s}}^2$

So, $(a_A{)}_x=15{\text{m/s}}^2,(a_A{)}_y=15{\text{m/s}}^2$

Substituting the values in equation $\left(1\right)$,

$\Rightarrow (15\times \frac{3}{5})-(15\times \frac{4}{5})=a_B\left(\frac{3}{5}\right)$

$\Rightarrow 9-12=\frac{3a_B}{5}$

$\therefore a_B=-5$

Therefore wedge will move along $\text{-ve x-direction}$.

$\overrightarrow{a_B}=-5\hat{i}{\text{m/s}}^2$

Hence, option (d) is the correct answer.

Why this question? Tip:-When two rigid bodies are in contact, the component of acceleration perpendicular to contact surface must be same, because there is no relative motion between bodies along that direction.