Question

In the List-I below, four different paths of a particle are given as functions of time. In these functions, $\alpha$ and $\beta$ are positive constants of appropriate dimensions and $\alpha \ne B$. In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned; $\overrightarrow{p}$ is the linear momentum $\overrightarrow{L}$ is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for the path.

List - I	List - II
P. $\overrightarrow{r}\left(t\right)=\alpha t\hat{i}+\beta t\hat{j}$	1. $\overrightarrow{p}$
Q. $\overrightarrow{r}\left(t\right)=\alpha cos\left(\omega t\right)\hat{i}+\beta sin\left(\omega t\right)\hat{j}$	2. $\overrightarrow{L}$
R. $\overrightarrow{r}\left(t\right)=\alpha \left(cos\right(\omega t)\hat{i}+sin(\omega t\left)\hat{j}\right)$	3. K
S. $\overrightarrow{r}\left(t\right)=\alpha t\hat{i}+\frac{\beta }{2}{t}^2\hat{j}$	4. U 5. E

• A. $P\to 1,2,3,4,5;Q\to 2,5;R\to 2,3,4,5;S\to 5$
• B. $P\to 1,2,3,4,5;Q\to 3,5;R\to 2,3,4,5;S\to 2,5$
• C. $P\to 2,3,4;Q\to 5;R\to 1,2,4;S\to 2,5$
• D. $P\to 1,2,3,5;Q\to 2,5;R\to 2,3,4,5;S\to 2,5$

Answer 1

The correct option is A $P\to 1,2,3,4,5;Q\to 2,5;R\to 2,3,4,5;S\to 5$

$\left(P\right)$ $\overrightarrow{r}\left(t\right)=\alpha t\hat{i}+\beta \widehat{tj}$

$\overrightarrow{v}=\frac{d\overrightarrow{r}\left(t\right)}{dt}=\alpha \hat{i}+\beta \hat{j}${constant}

$\overrightarrow{a}=\frac{\overrightarrow{dv}}{dt}=0$

$\overrightarrow{P}=m\overrightarrow{v}$ (remain constant)

$k=\frac{1}{2}m{v}^2${remain constant}

$\overrightarrow{F}=-[\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial Y}\hat{i}]=0$

$\Rightarrow U\to constant$

E = K + U

$\frac{d\overrightarrow{L}}{dt}=\overrightarrow{T}=\overrightarrow{r}\times \overrightarrow{F}=0$

$\overrightarrow{L}=constant$

$\left(Q\right)$ $\overrightarrow{r}=-\alpha cos\left(\omega t\right)\hat{i}+\beta sin\left(\omega t\right)\hat{j}$

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=-\alpha sin\left(\omega t\right)\hat{i}+\beta \omega cos\left(\omega t\right)\hat{j}$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=-{\alpha }^{2}cos\left(\omega t\right)\hat{i}-\beta {\omega }^{2}sin\left(\omega t\right)\hat{j}$

$=-{\omega }^2\left[\alpha cos\right(\omega t)\hat{i}+\beta sin(\omega t\left)\hat{j}\right]$

$\overrightarrow{a}=-{\omega }^2\overrightarrow{r}$

$\overrightarrow{T}=\overrightarrow{r}\times \overrightarrow{F}=0\overrightarrow{r}and\overrightarrow{F}areparallel$

$△U=-\int \overrightarrow{F}.dr=+{\int }_0^{r}m{\omega }^2.r.dr$

$△U=m{\omega }^2\left[\frac{r^2}{2}\right]$

$U\propto r^2$

$r=\sqrt{{\alpha }^{2}co{s}^2\left(\omega t\right)+{\beta }^{2}si{n}^2\left(\omega t\right)}$

r is a function of time (t)

U depends on r hence it will change with time

Total energy remain constant because force is central.

$\left(R\right)\overrightarrow{r}\left(t\right)=\alpha (cos\omega t\hat{i}+sin(\omega t\left)\hat{j}\right)$

$\overrightarrow{v}\left(t\right)=\frac{d\overrightarrow{r}\left(t\right)}{dt}=\alpha [-\omega sin(\omega t)\hat{i}+\omega cos(\omega t\left)\hat{j}\right]$

$|\overrightarrow{v}|=\alpha \omega$ (Speed remains constant)

$\overrightarrow{a}\left(t\right)=\frac{d\overrightarrow{v}\left(t\right)}{dt}=\alpha [-{\omega }^{2}cos(\omega t)\hat{i}-{\omega }^{2}sin(\omega t\left)\hat{j}\right]$

= $-\alpha {\omega }^2\left[cos\right(\omega t)\hat{i}+sin(\omega t\left)\hat{j}\right]$

$\overrightarrow{a}\left(t\right)=-{\omega }^2\left(\overrightarrow{r}\right)$

$\overrightarrow{T}=\overrightarrow{F}\times \overrightarrow{r}=0$

$|\overrightarrow{r}|=\alpha$(remain constant)

Force is central in nature and distance from fixed point is central.

Potential energy remains constant

Kinetic energy is also constant (speed is constant)

$\left(S\right)\overrightarrow{r}=\alpha t\hat{i}+\frac{\beta }{2}{t}^2\hat{j}$

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\alpha t\hat{i}+\beta t\hat{j}$ (speed of particle depends on ‘t’)

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\beta \hat{j}$ {constant}

$\overrightarrow{F}=m\overrightarrow{a}$ {constant}

$△U=-\int \overrightarrow{F}.d\overrightarrow{r}=-m{\int }_0^t\beta \hat{j}(\alpha \hat{i}+\beta \widehat{tj})dt$

$U=\frac{-m{\beta }^2{t}^2}{2}$

$k=\frac{1}{2}m{v}^2=\frac{1}{2}m({\alpha }^2+{\beta }^2{t}^2)$

$E=k+U=\frac{1}{2}m{\alpha }^2$ [remain constant].