Question

In figure, the block moves downwards with velocity $u_1$, the wedge moves rightwards with velocity $u_2$. The correct relation between $u_1$ and $u_2$ is

• A. $u_2=u_1$
• B. $u_2=u_1\sin \alpha$
• C. $2u_2\sin \alpha =u_1$
• D. $u_2(1+\sin \alpha )=u_1$

Answer 1

The correct option is D $u_2(1+\sin \alpha )=u_1$



$x_1+\sqrt{x_1^2+h^2}+y_1=$ constant
differentiating both sides
$\Rightarrow \frac{d{x}_1}{dt}+\frac{1}{2\sqrt{x_1^2+h^2}}2x_1\times \frac{d{x}_1}{dt}+\frac{d{y}_1}{dt}=0$
$-u_2-\sin \alpha u_2+u_1=0$
$\therefore u_2(1+\sin \alpha )=u_1$