Question

A block B moves with a velocity $u$ relative to the wedge $A$. If the velocity of the wedge is $v$ as shown in figure, what should be the value of $\theta$ so that the block $B$ moves vertically as seen from the ground?

• A. ${\cos }^{-1}(\frac{u}{v})$
• B. ${\cos }^{-1}(\frac{v}{u})$
• C. ${\sin }^{-1}(\frac{u}{v})$
• D. ${\sin }^{-1}(\frac{v}{u})$

Answer 1

The correct option is B ${\cos }^{-1}(\frac{v}{u})$

Let ${\overrightarrow{v}}_A$ and ${\overrightarrow{v}}_B$ be the velocities of wedge and block respectively.
Then ${\overrightarrow{v}}_{BA}$ will be the velocity of block w.r.t wedge(i.e. $u$).
We know that,
${\overrightarrow{v}}_B={\overrightarrow{v}}_{BA}+{\overrightarrow{v}}_A$
Now
$\overrightarrow{v_{BA}}=-u\cos \theta \hat{i}+u\sin \theta \hat{j}$
$\overrightarrow{v_A}=v\hat{i}$
$\Rightarrow \overrightarrow{v_B}=-u\cos \theta \hat{i}+u\sin \theta \hat{j}+v\hat{i}$
$\Rightarrow \overrightarrow{v_B}=(-u\cos \theta +v)\hat{i}+u\sin \theta \hat{j}$
In the horizontal direction, the velocity of $B$ should be zero for it to move vertically.

$\Rightarrow 0=-u\cos \theta +v$
$\Rightarrow u\cos \theta =v$
$\Rightarrow \theta ={\cos }^{-1}(\frac{v}{u})$