Question

In figure the block moves downwards with velocity $v_1$, the wedge moves rightwards with velocity $v_2$.
The correct relation between $v_1$ and $v_2$ is

• A. $v_2=v_1$
• B. $v_2=v_1\sin \theta$
• C. $2v_2\sin \theta =v_1$
• D. $v_2(1+\sin \theta )=v_1$

Answer 1

The correct option is D $v_2(1+\sin \theta )=v_1$

Consider the problem

From constraint relation

$x+\sqrt{x^2=h^2}+y=l$

$\frac{dx}{dt}+\frac{1.\left(2x\right)}{2\sqrt{x^2+h^2}}.\frac{dx}{dt}+\frac{dy}{dt}=0$

$-v_2=\frac{x}{\sqrt{x^2=h^2}}(-v_2)+v_1=0$

$v_1=v_2(1+\sin \theta )$

$v_2(1+\sin \theta )=v_1$

Hence option $D$ is correct answer.