$( P ) \rightarrow(1),(2),(3),(4),(5) ;(Q) \rightarrow(2),(5) ;(R) \rightarrow(2),(3),(4),(5) ;(S) \rightarrow(5)$

$(P)\overrightarrow{ r }=\alpha t \hat{i}+\beta t \hat{j}$

$\dot{ v }=\frac{ dr }{ dt }=\alpha \hat{ i }+\beta \hat{ j } \text { (constant) }$

$a=0$

$\overrightarrow{ F }=0$

$\text { Angular momentum } \quad \dot{ L }= m (\dot{ r } \times \dot{ v })$

$\overrightarrow{ L }= m [\alpha \beta t (\hat{ k })+\alpha \beta t (-\hat{ k })]=0 \text { (conserved) }$

$(Q) \vec{r}=a \cos \omega t \hat{i}+\beta \sin \omega t \hat{j}$

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.