Question

The potential energy of a particle of mass 5kg moving in xy-plane is given as
$U=(7x+24y)$ joule, $x$ and $y$ being in meter. 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 angle to the direction 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 $5m{s}^{-2}$

We know,

$-\frac{dU}{dx}=F$

$\Rightarrow F=\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}$

Given,

$U=(7x+24y)$

$\overrightarrow{F}=-\frac{\partial (7x+24y)}{\partial x}\hat{i}-\frac{\partial (7x+24y)}{\partial y}\hat{j}$

$\overrightarrow{F}=-7\hat{i}-24\hat{j}$

We know $\overrightarrow{F}=m\overrightarrow{a}$

$\overrightarrow{a}=-\frac{7}{5}\hat{i}-\frac{24}{5}\hat{j}$

Evaluating OPTION A,

$u_x=8.6\hat{i}$ and $u_y=23.2\hat{j}$

$a_x=-1.4\hat{i}$ and $a_y=-4.8\hat{j}$

Using $v=u+at$ at $t=4sec$

$v_x=u_x-a_{x}t=8.6-\left(1.4\right)\left(4\right)=3m/sec$

$v_y=u_y-a_{y}t=23.2-\left(4.8\right)\left(4\right)=4m/sec$

$v=\sqrt{{v_x}^2+{v_y}^2}=\sqrt{3^2+4^2}=5m/sec$

Evaluating OPTION B,

$|a|=\sqrt{{a_x}^2+{a_y}^2}$

$|a|=\sqrt{(-1.4{)}^2+(-4.8{)}^2}=\sqrt{25}=5m/se{c}^2$

Evaluating OPTION C,

Initial direction of motion $\overrightarrow{d_1}=8.6\hat{i}+23.2\hat{j}$

Direction of acceleration $\overrightarrow{d_2}=-1.4\hat{i}-4.8\hat{j}$

$\overrightarrow{d_1}.\overrightarrow{d_2}=\left(8.6\right)(-1.4)+\left(23.2\right)(-4.8)=-12.04-111.36=-123.4$

Since, $\overrightarrow{d_1}.\overrightarrow{d_2}\ne 0$

$\therefore$ The direction of motion of the particle initially (at $t=0$) is NOT at right angle to the direction of acceleration.

Evaluating OPTION D

The path of the particle is

$x=u_{x}t-\frac{1}{2}{a}_x{t}^2=8.6t-0.7t^2$

$y=u_{y}t-\frac{1}{2}{a}_y{t}^2=23.2t-2.4t^2$

For path to be circular $x^2+y^2=constant$

But, $(8.6t-0.7t^2{)}^2+(23.2t-2.4t^2{)}^2\ne Constant$. It will be always be in terms of variable $t$.

Hence, the correct answers are OPTIONS A and B.