Question

If block $A$ is moving horizontally with velocity $v_{A}$, then find the velocity of block $B$ at the instant as shown in Figure.

• A. $\frac{h v_{A}}{2 \sqrt{x^{2}+h^{2}}}$
• B. $\frac{x v_{A}}{\sqrt{x^{2}+h^{2}}}$
• C. $\frac{x v_{A}}{2 \sqrt{x^{2}+h^{2}}}$
• D. $\frac{h v_{A}}{\sqrt{x^{2}+h^{2}}}$

Answer 1

Answer: (C)

From figure, $L=2 h-2 y+\sqrt{x^{2}+h^{2}}$
Differentiating the equation, we get
$\frac{d y}{d t} =\frac{x}{2 \sqrt{h^{2}+x^{2}}} \frac{d x}{d t}$
$\Rightarrow v_{B} =\frac{x v_{A}}{2 \sqrt{h^{2}+x^{2}}}$