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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 $\alpha \neq \beta$. 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, $\bar{L}$ 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-$II$ |
| $P$ $\dot{r}(t)=\alpha t \hat{t}+\beta t \hat{j}$ | $1$ $\overrightarrow{ p }$ |
| $Q$ $\dot{r}(t)=\alpha \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$ | $2$ $\overrightarrow{ L }$ |
| $R$ $\dot{r}(t)=\alpha(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ | $3$ $K$ |
| $S$ $\dot{r}(t)=\alpha t \hat{i}+\frac{\beta}{2} t^2 \hat{j}$ | $4$ $U$ |
| $5$ $E$ |
$P \rightarrow 1,2,3,4,5 ; \quad Q \rightarrow 2,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 5$
$P \rightarrow 1,2,3,4,5 ; \quad Q \rightarrow 3,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 2,5$
$P \rightarrow 2,3,4 ; \quad Q \rightarrow 5 ; \quad R \rightarrow 1,2,4 ; \quad S \rightarrow 2,5$
$P \rightarrow 1,2,3,5 ; \quad Q \rightarrow 2,5 ; \quad R \rightarrow 2,3,4,5 ; \quad S \rightarrow 2,5$
Solution
$( P ) \rightarrow(1),(2),(3),(4),(5) ;(Q) \rightarrow(2),(5) ;(R) \rightarrow(2),(3),(4),(5) ;(S) \rightarrow(5)$
$(P)\overrightarrow{ r }=\alpha t \hat{i}+\beta t \hat{j}$
$\dot{ v }=\frac{ dr }{ dt }=\alpha \hat{ i }+\beta \hat{ j } \text { (constant) }$
$a=0$
$\overrightarrow{ F }=0$
$\text { Angular momentum } \quad \dot{ L }= m (\dot{ r } \times \dot{ v })$
$\overrightarrow{ L }= m [\alpha \beta t (\hat{ k })+\alpha \beta t (-\hat{ k })]=0 \text { (conserved) }$
$(Q) \vec{r}=a \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$
Particle is moving on an elliptical path.
Angular momentum about origin and total energy remains conserved.
$(R)$ $\overrightarrow{ r }= a (\cos \omega t \hat{ i }+\sin \omega t \hat{j})$
Particle moves on a circular path
radius $= a$
angular velocity $=\varrho$
speed $v = a$
but $\dot{ v } \neq$ constant
hence $\overrightarrow{ P } \neq$ constant
Angular momentum $\overrightarrow{ L }=\left( moa ^2\right) \hat{ k }$ is constant
Kinetic energy $E=\frac{1}{2} m v^2$ (constant)
P.E. $U =$ constant
$E =$ constant
$(S)$ $\overrightarrow{ r }=\alpha t \hat{ i }+\frac{\beta t ^2}{2} \hat{j}$
$\vec{v}=\frac{ d \vec{r}}{d t}=\alpha \hat{i}+\beta t \hat{j}$ (depends on $t$ )
hence $\overrightarrow{ P }= m \overrightarrow{ v }$ also depends on $t$.
$\dot{ a }=\frac{ d \dot{v}}{ dt }=\beta \hat{j}$
$\overrightarrow{ F }= ma = m \beta \hat{ j }$ (constant)
$\dot{ L }= m (\dot{ r } \times \dot{ v })= m \left[\alpha \beta t ^2 \hat{ k }+\frac{\alpha \beta t ^2}{2}(-\hat{ k })\right]$
$\overrightarrow{ L }=\frac{ m \alpha \beta t ^2}{2} \cdot \hat{ k }$
(depends on $t$ )
Only total energy remains conserved.
