Question

A particle is moving in xy-plane with a constant speed $v_{0}$ such that its y displacement is given by $y=\alpha e^{- \frac{2 v_{x}}{\sqrt{3} v_{0}}},$ 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 $-\frac{1}{\sqrt{3}},$ then the displacement of the particle in $y-$ direction at the instant is

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

Answer 1

Answer: (A)

Here, particle is moving in two - dimensional $(x-y)$ plane, with constant speed so slope of
graph $\frac{d y}{d x}=\frac{v_{y}}{v_{x}}$ and $v_{0}=\sqrt{ v _{ x }^{2}+ v _{ y }^{2}}$.
As slope, $\frac{d y}{d x}=-\frac{1}{\sqrt{3}}=\frac{v_{y}}{v_{x}}$
So, $v _{ x }=- v _{ y } \sqrt{3}$
On squaring and adding
$ v _{ x }^{2}+ v _{ y }^{2}= v _{0}^{2}$
$3 v _{ y }^{2}+ v _{ y }^{2}= v _{0}^{2} $
$ v _{ y }=\pm \frac{ v _{0}}{2} $
$v _{ x }=\mp \frac{\sqrt{3} v _{0}}{2}$
So, $ y =\alpha e ^{\frac{-2}{\sqrt{3} v _{0}}} \times \frac{\sqrt{3} v _{0}}{2} $
$y =\alpha e ^{-1}$