Question

In the List-1 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; $\overrightarrow{p}$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. $\overrightarrow{r}\left( t \right)=\alpha t\widehat{i}+\beta t\widehat{j}$	1. $\overrightarrow{p}$
Q.$\overrightarrow{r}\left( t \right)=\alpha \cos \left( \omega t \right)\widehat{i}+\beta \sin \left( \omega t \right)\widehat{j}$	2.$\overrightarrow{L}$
R. $\overrightarrow{r}\left( t \right)=\alpha \cos \left( \omega t \right)\widehat{i}+\sin \left( \omega t \right)\widehat{j}$	3.K
S. $\overrightarrow{r}\left( t \right)=\alpha t\widehat{i}+\dfrac{\beta }{2}{{t}^{2}}\widehat{j}$	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. $P\to 2,3,4;Q\to 5;R\to 1,2,4,5;S\to 2,5$
D. $P\to 1,2,3,5;Q\to 2,5;R\to 2,3,4,5;S\to 2,5$

Answer 1

Hint: You could consider each of the paths given in the list I and find the acceleration, velocity, force and thereby the physical quantities on the list II. So, in each case you could look for quantities independent of time and hence conserved along the given path. Doing the matching correctly would give you the answer.

Complete answer:
Let us consider each path one by one.
P. $\overrightarrow{r}\left( t \right)=\alpha t\widehat{i}+\beta t\widehat{j}$
Differentiating this with respect to time we get velocity,
$\overrightarrow{v}=\dfrac{d\overrightarrow{r}(t)}{dt}=\alpha \widehat{i}+\beta \widehat{j}$ which is constant and hence acceleration would be zero.
Linear momentum, $\overrightarrow{P}=m\overrightarrow{v}$would also be constant and so will be the kinetic energy $K=\dfrac{1}{2}m{{v}^{2}}$.
$\overrightarrow{F}=-\left( \dfrac{\partial U}{\partial x}\widehat{i}+\dfrac{\partial U}{\partial y}\widehat{i} \right)=0$
$\Rightarrow U=cons\tan t$
Total energy E=U+K would also be a constant.
Also, we have time rate of change of angular momentum given by,
$\dfrac{d\overrightarrow{L}}{dt}=\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=0$
Now, we could conclude that $\overrightarrow{L}$is a constant.
So, for P all the quantities in list II remain conserved along the path.
Q. $\overrightarrow{r}\left( t \right)=\alpha \cos \left( \omega t \right)\widehat{i}+\beta \sin \left( \omega t \right)\widehat{j}$
$\Rightarrow \overrightarrow{v}=-\alpha \sin \left( \omega t \right)\widehat{i}+\beta \omega \cos \left( \omega t \right)\widehat{j}$
$\Rightarrow \overrightarrow{a}=-{{\omega }^{2}}\left( \alpha \cos \left( \omega t \right)\widehat{i}+\beta \sin \left( \omega t \right)\widehat{j} \right)=-{{\omega }^{2}}\overrightarrow{r}$
Now the torque, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=0$, $\overrightarrow{r}$and$\overrightarrow{F}$are parallel.
Now, the potential energy,
$\Delta U=-\int{\overrightarrow{F}}\centerdot dr=\int\limits_{0}^{r}{m{{\omega }^{2}}}rdr=\dfrac{m{{\omega }^{2}}{{r}^{2}}}{2}$
$\Rightarrow U\propto {{r}^{2}}$
We see that r is a function of time(t) and since U depends on r it would further depend on time. Also, since the force is central, the total energy would remain constant. Here, we find that angular momentum and total energy is conserved along the path.
R. $\overrightarrow{r}\left( t \right)=\alpha \cos \left( \omega t \right)\widehat{i}+\sin \left( \omega t \right)\widehat{j}$
$\Rightarrow \overrightarrow{v}\left( t \right)=\alpha \left( -\omega \sin \left( \omega t \right)\widehat{i}+\omega \cos \left( \omega t \right)\widehat{j} \right)$
$\Rightarrow \left| \overrightarrow{v} \right|=\alpha \omega $
So, speed remains constant.
$\overrightarrow{a}\left( t \right)=\alpha \left( -{{\omega }^{2}}\cos \left( \omega t \right)\widehat{i}-{{\omega }^{2}}\sin \left( \omega t \right)\widehat{j} \right)=-{{\omega }^{2}}\overrightarrow{r}$
Also, $\overrightarrow{\tau }=\overrightarrow{F}\times \overrightarrow{r}=0$
Since the force is central in nature and also the distance from the fixed point is also central. Then we have constant potential energy and kinetic energy. Here, whole list II with the exception of linear momentum is conserved along the given path.
S. $\overrightarrow{r}\left( t \right)=\alpha t\widehat{i}+\dfrac{\beta }{2}{{t}^{2}}\widehat{j}$
$\Rightarrow \overrightarrow{v}=\alpha t\widehat{i}+\beta t\widehat{j}$
$\Rightarrow \overrightarrow{a}=\beta \widehat{j}$
So, force would also be constant.
Potential energy, $\Delta U=-\int{\overrightarrow{F}}\centerdot \overrightarrow{dr}=-m\int\limits_{0}^{t}{\beta \widehat{j}}\left( \alpha \widehat{i}+\beta t\widehat{j} \right)dt$
$\Rightarrow U=\dfrac{-m{{\beta }^{2}}{{t}^{2}}}{2}$
Kinetic energy, $K=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}m\left( {{\alpha }^{2}}+{{\beta }^{2}}{{t}^{2}} \right)$
Total energy, $E=K+U=\dfrac{1}{2}m{{\alpha }^{2}}$
Here, we have only the total energy conserved along the given path.
Therefore, the matching would be as follows:
$P\to 1,2,3,4,5;Q\to 2,5;R\to 2,3,4,5;S\to 5$

Hence, option A is correct.

Note:
As we know that matching is actually a messy stuff to do, we should deal with the given question as organized as possible. See for the above example, we have dealt with each path separately and have looked for quantities conserved along each of the paths on the list I in list II.