Question

The 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 is the value of $\theta$, so that the block $B$ moves vertically as seen from ground ?

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

Answer 1

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

$\overrightarrow{v_{{}_{B/A}}}=\overrightarrow{v_{{}_B}}-\overrightarrow{v_{{}_A}}=(-u\cos \theta )\hat{x}+\left(u\sin \theta \right)\hat{y}$

We know that $\overrightarrow{v_{{}_A}}=\left(v\right)\hat{x}$

$\Rightarrow \overrightarrow{v_{{}_B}}=(v-u\cos \theta )\hat{x}+\left(u\sin \theta \right)\widehat{(}y)$

But $B$ has only vertical velocity in the ground frame. Therefore, $x$ component is zero.

$\Rightarrow v=u\cos \theta \Rightarrow \cos \theta =\frac{v}{u}$

$\therefore \theta ={\cos }^{-1}\left(\frac{v}{u}\right)$

$\to QED.$