Question

In the shown figure, the acceleration of $A$ is ${\overrightarrow{a}}_A=(15\hat{i}+15\hat{j})$, then the acceleration of $B$ is ($A$ remains in contact with $B$)

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

Answer 1

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

Plane is moving towards right so the acceleration of surface will be $-a\left(\hat{i}\right)$.
For, $a_A=(15\hat{i}+15\hat{j})$ and
$a_B=-a\left(\hat{i}\right)$
$a_{AB}=a_A-a_B=(15\hat{i}+15\hat{j})-(-a\hat{i})$
=$\left[\right(15+a)\hat{i}+15\hat{j}]$ .....(1)


For block $A$ acceleration,
$tan\left({37}^o\right)=\frac{3}{4}$ .....(2)
On comparing (1) & (2)
$\frac{3}{4}=\frac{15}{15+a}a=5m/s^2$
As the acceleration of surface is in $-\hat{i}$ direction so,
$a_B=5(-\hat{i})=-5\hat{i}$
Hence option (D) matches correctly.