Question

Potential energy (in joule) of a particle of mass $1kg$ moving in $x-y$ plane is $U=3x+4y$, here $x$ and $y$ are in meter. If at time $t=0$, particle is at rest at point $P(6m,4m).$ Then,

• A. acceleration of particle is $3\hat{i}+4\hat{j}\text{m/s}$
• B. time when it crosses $y-axis$ is $t=1\text{sec}$
• C. speed of particle when it crosses $y-axis$ is $10\text{m/s}$
• D. it crosses $y-axis$ at $y=-8\text{m}$

Answer 1

The correct option is C speed of particle when it crosses $y-axis$ is $10\text{m/s}$

We know, force is given by
$F_x=-\frac{\partial U}{\partial x}=-3N$,
$F_y=-\frac{\partial U}{\partial y}=-4N$,
$\overrightarrow{F}=(-3\hat{i}-4\hat{j})N$,
$\overrightarrow{a}=(-3\hat{i}-4\hat{j})m/s^2$

For motion along $x-$axis, when it crosses $y-$axis.
$u_x=0$, $a_x=-3m/s^2$
$S_x=-6m$, as particle was at $x=6m$ when $t=0$.
Using,
$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$-6=\frac{1}{2}(-3)t^2$
$\Rightarrow t=2\text{sec}$
Using, $v_x=u_x+a_{x}t$
$v_x=0-3\times 2$
$v_x=-6m/s$

For motion along $y-$axis, when it crosses $y-$axis
$u_y=0$, $a_y=-4m/s^2$ and $t=2\text{sec}$
Using,
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$S_y=\frac{1}{2}(-4)(2{)}^2=-8m$
$\Rightarrow y=-8+4=-4\text{m}$
$\Rightarrow v_y=u_y+a_{y}t=0-4\times 2v_y=-8\text{m/s}$

Speed of particle when it crosses $y-axis$ is
$v=\sqrt{v_x^2+v_y^2}=10\text{m/s}$