Question

A particle is moving in xy-plane with a constant speed $v_0$ such that its y displacement is given by $\text{y}=\alpha {\text{e}}^{\text{-}\frac{2{\text{v}}_{\text{x}}}{\sqrt{3}{\text{v}}_0}}\text{,}$ where v_x is component of velocity along the x-axis. If at some instant x component of it velocity is positive and the slope of the tangent on its path is $-\frac{1}{\sqrt{3}}\text{,}$ then the displacement of the particle in y-direction at the instant is

• A. $\alpha {\text{e}}^{-1}$
• B. $\alpha {\text{e}}^{-2}$
• C. Zero
• D. ${\alpha }^2\text{e}$

Answer 1

Answer: (A)

Here, particle is moving in two - dimensional (x - y) plane, with constant speed so slope of
graph $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{{\text{v}}_{\text{y}}}{{\text{v}}_{\text{x}}}$ and ${\text{v}}_0=\sqrt{{\text{v}}_{\text{x}}^2+{\text{v}}_{\text{y}}^2}\text{.}$
As slope, $\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{3}}=\frac{{\text{v}}_{\text{y}}}{{\text{v}}_{\text{x}}}$
So, ${\text{v}}_{\text{x}}=-{\text{v}}_{\text{y}}\sqrt{3}$
On squaring and adding
${\text{v}}_{\text{x}}^2+{\text{v}}_{\text{y}}^2={\text{v}}_0^2$
$3{\text{v}}_{\text{y}}^2+{\text{v}}_{\text{y}}^2={\text{v}}_0^2$
${\text{v}}_{\text{y}}=\pm \frac{{\text{v}}_0}{2}$
${\text{v}}_{\text{x}}=+-\frac{\sqrt{3}{\text{v}}_0}{2}$
So, $\text{y}=\alpha {\text{e}}^{\frac{-2}{\sqrt{3}{\text{v}}_0}\times \frac{\sqrt{3}{\text{v}}_0}{2}}$
$\text{y}=\alpha {\text{e}}^{-1}$