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 $\alpha \ne \beta$. 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: $\vec{L}$ is the linear momentum, $\vec{L}$ 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 that path.

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

Answer: (A)

(P) $\vec{r}\left(t\right)=\alpha t\hat{i}+\beta t\hat{j}$
$\vec{v}=\frac{d\vec{r}\left(t\right)}{dt}=\alpha \hat{i}+\beta \hat{j}$ {constant}
$\vec{a}=\frac{\overrightarrow{dv}}{dt}=0$
$\vec{P}=m\vec{v}$ (remain constant)
$k=\frac{1}{2}m{v}^2$ {remain constant}
$\vec{F}=-\left[\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{i}\right]=0$
$\Rightarrow U\to$ constant
$E=K+U$
$\frac{d\vec{L}}{dt}=\vec{\tau }=\vec{r}\times \vec{F}=0$
$\vec{L}$ = constant
(Q) $\vec{r}=\alpha \cos \left(\omega t\right)\hat{i}+\beta \sin \left(\omega t\right)\hat{j}$
$\vec{v}=\frac{d\vec{r}}{dt}=-\alpha \omega \sin \left(\omega t\right)\hat{i}+\beta \omega \cos \left(\omega t\right)\hat{j}$
$\vec{a}=\frac{d\vec{v}}{dt}=-\alpha {\omega }^2\cos \left(\omega t\right)\hat{i}-\beta {\omega }^2\sin \left(\omega t\right)\hat{j}$
$=-{\omega }^2\left[\alpha \cos \left(\omega t\right)\hat{i}+\beta \sin \left(\omega t\right)\hat{j}\right]$
$\vec{a}=-{\omega }^2\vec{r}$
$\vec{\tau }=\vec{r}\times \vec{F}=0$ {$\vec{r}$ and $\vec{F}$ are parallel}
$\Delta U=-\int \vec{F}.dr=+{\int }_0^{r}m{\omega }^2.r.dr$
$\Delta U=m{\omega }^2\left[\frac{r^2}{2}\right]$
$U\propto r^2$
$r=\sqrt{{\alpha }^2{\cos }^2\left(\omega t\right)+{\beta }^2{\sin }^2\left(\omega t\right)}$
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.
(R) $\vec{r}\left(t\right)=\alpha \left(\cos \omega t\hat{i}+\sin \left(\omega t\right)\hat{j}\right)$
$\vec{v}\left(t\right)=\frac{d\vec{r}\left(t\right)}{dt}=\alpha \left[-\omega \sin \left(\omega t\right)\hat{i}+\omega \cos \left(\omega t\right)\hat{j}\right]$
$\left|\vec{v}\right|=\alpha \omega$ (Speed remains constant)
$\vec{a}\left(t\right)=\frac{d\vec{v}\left(t\right)}{dt}=\alpha \left[-{\omega }^2\cos \left(\omega t\right)\hat{i}-{\omega }^2\sin \left(\omega t\right)\hat{j}\right]$
$=-\alpha {\omega }^2\left[\cos \left(\omega t\right)\hat{i}+\sin \left(\omega t\right)\hat{j}\right]$
$\vec{a}\left(t\right)=-{\omega }^2\left(\vec{r}\right)$
$\vec{\tau }=\vec{F}\times \vec{r}=0$
$|\vec{r}|=\alpha$ (remain constant)
Force is central in nature and distance from fixed point is constant.
Potential energy remains constant
Kinetic energy is also constant (speed is constant)
(S) $\vec{r}=\alpha t\hat{i}+\frac{\beta }{2}{t}^2\hat{j}$
$\vec{v}=\frac{d\vec{r}}{dt}=\alpha t\hat{i}+\beta t\hat{j}$ (speed of particle depends on 't')
$\vec{a}=\frac{d\vec{v}}{dt}=\beta \hat{j}$ {constant}
$\vec{F}=m\vec{a}$ {constant}
$\Delta U=-\int \vec{F}.d\vec{r}=-m{\int }_0^t\beta \hat{j}.\left(\alpha \hat{i}+\beta t\hat{j}\right)dt$
$U=\frac{-m{\beta }^2{t}^2}{2}$
$k=\frac{1}{2}m{v}^2=\frac{1}{2}m\left({\alpha }^2+{\beta }^2+t^2\right)$
$E=k+U=\frac{1}{2}m{\alpha }^2$ [remain constant]