Question

If block A is moving horizontally with velocity $v_A$, then find the velocity of block B at the instant as shown in $Fig.6.341$.

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

Answer 1

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

The total length of the starting is given by

$L=2h-2y+\sqrt{x^2+h^2}$ .........(1)

Differentiating equation (1) we get

$2\times \frac{dy}{dt}=\frac{2x}{2\sqrt{x^2+h^2}}\times \frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{2\sqrt{x^2+h^2}}\frac{dx}{dt}$

Since $V_B=\frac{dy}{dt}$ nas $V_A=\frac{dx}{dt}$

$\Rightarrow V_B=\frac{x{v}_A}{2\sqrt{x^2+h^2}}$