Question

The potential energy of a particle of mass 5 kg moving in $x-y$ plane is given as $U=7x+24y$ J, and $x$ and $y$ being in metre. Initially at $t=0$, the particle is at origin $(0,0)$ moving with a velocity of $(8.6\hat{i}+23.2\hat{j})m{s}^{-1}$.Then,

• A. The velocity of the particle at $t=4s$, is $5m{s}^{-1}$
• B. The acceleration of the particle is $5m{s}^{-2}$
• C. The direction of motion of the particle initially (at $t=0$) is at right angles to the direction of acceleration
• D. The path of the particle is circle

Answer 1

Answer: (A), (B)

The velocity of the particle at $t=4s$, is $5m{s}^{-1}$
The acceleration of the particle is $5m{s}^{-2}$

$\overrightarrow{F}=-\left[\frac{\delta U}{\delta x}\hat{i}+\frac{\delta U}{\delta y}\hat{j}\right]=(-7\hat{i}-24\hat{j})N$

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=(-\frac{7}{5}\hat{i}-\frac{24}{5}\hat{j})m/s$

$|\overrightarrow{a}|=\sqrt{{\left(\frac{7}{5}\right)}^2+{\left(\frac{24}{5}\right)}^2}=5m/s^2$
$\text{Since,}\overrightarrow{a}=\text{constant, we can apply,}$

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$

$=(8.6\hat{i}+23.2\hat{j})+(-\frac{7}{5}\hat{i}-\frac{24}{5}\hat{j})\left(4\right)$

$=(3\hat{i}+4\hat{j})m/s$

$|\overrightarrow{v}|=\sqrt{(3{)}^2+(4{)}^2}=5m/s$