Question

The potential energy of a particle of mass $5kg$ moving in x-y plane is given as $U=7x+24y$ joule, $x$ and $y$ being in metre. Initially at $t=0$, the particle is at the 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 distance of acceleration
• D. the path of the particle is circle

Answer 1

Answer: (A), (B)

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

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

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=(-\frac{7}{5}\hat{i}-\frac{24}{5}\hat{j})m/s^2$

$\mid \hat{a}\mid =\sqrt{(\frac{7}{5}{)}^2+(\frac{24}{5}{)}^2}=5m/s^2$

Since, $\overrightarrow{a}=$ constant, we can apply

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

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

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

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