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, $\text{y} = \alpha \text{e}^{\frac{- 2}{\sqrt{3} \text{v}_{0}} \times \frac{\sqrt{3} \text{v}_{0}}{2}}$
$\text{y} = \alpha \text{e}^{- 1}$