# A particle moving in the x y plane experiences a velocity

Question:

A particle moving in the $x y$ plane experiences a velocity dependent force $\overrightarrow{\mathrm{F}}=k\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$, where $v_{x}$ and $v_{y}$ are $x$

and $y$ components of its velocity $\vec{v}$. if $\vec{a}$ is the acceleration of the particle, then which of the following statements is true for the particle?

1. (1) quantity $\vec{v} \times \vec{a}$ is constant in time

2. (2) $\vec{F}$ arises due to a magnetic field

3. (3) kinetic energy of particle is constant in time

4. (4) quantity $\vec{v} \cdot \vec{a}$ is constant in time

Correct Option: 1

Solution:

(1) Given, $\vec{F}=k\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$

$\therefore F_{x}=k v_{y} \hat{i}, F_{y}=k v_{x} \hat{j}$

$\frac{m d v_{x}}{d t}=k v_{y} \Rightarrow \frac{d v_{x}}{d t}=\frac{k}{m} v_{y}$

Similarly, $\frac{d v_{y}}{d t}=\frac{k}{m} v_{x}$

$\frac{d v_{y}}{d v_{x}}=\frac{v_{x}}{v_{y}} \Rightarrow \int v_{y} d v_{y}=\int v_{x} d v_{x}$

$v_{y}^{2}=v_{x}^{2}+C$

$v_{y}^{2}-v_{x}^{2}=\mathrm{constant}$

$\vec{v} \times \vec{a}=\left(v_{x} \hat{i}+v_{y} \hat{j}\right) \times \frac{k}{m}\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$

$=\left(v_{x}^{2} \hat{k}-v_{y}^{2} \hat{k}\right) \frac{k}{m}=\left(v_{x}^{2}-v_{y}^{2}\right) \frac{k}{m} \hat{k}=$ constant