Question

If block $A$ is moving horizontally with velocity $v_A$, then find the velocity of block $B$ at the instant 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

The correct option is C $\frac{x{v}_A}{2\sqrt{x^2+h^2}}$


Let the vertical distance of block $B$ from the ground be $y$.
Hence, total length of string $L$ is:
$L=2h-2y+\sqrt{x^2+h^2}$
Differentiating the equation w.r.t time, we get:
$0=0-2\frac{dy}{dt}+\frac{1}{2\sqrt{x^2+h^2}}\times 2x\times \frac{dx}{dt}$
(since $h$ is constant)
$\Rightarrow \frac{dy}{dt}=\frac{x}{2\sqrt{x^2+h^2}}\frac{dx}{dt}....\left(i\right)$
But $\frac{dy}{dt}=v_B\&\frac{dx}{dt}=v_A....\left(ii\right)$
because $v_B=$ time rate of change of vertical distance $y$ between ground and block $B$
$v_A=$ time rate of change of horizontal distance $x$ between block $A$ and block $B$

From Eq. $\left(i\right)$ and Eq. $\left(ii\right)$:
$v_B=\frac{x{v}_A}{2\sqrt{x^2+h^2}}$
Hence option $\left(c\right)$ is correct.