The linear momentum $$\overrightarrow p $$ of a particle is conserved if force on it is zero i.e., $$\overrightarrow F = m\overrightarrow a = \overrightarrow 0 $$. If $$\overrightarrow p $$ is conserved then kinetic energy $$K = {p^2}/2m$$ is also conserved. The angular momentum $$\overrightarrow L $$ of a particle is conserved if torque on it is zero i.e., $$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \overrightarrow 0 $$. The total energy E is conserved for conservative forces.

For P : $$\overrightarrow r (t) = \alpha t\widehat i + \beta t\widehat j$$

$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta \widehat j$$ = constant

$$ \Rightarrow \overrightarrow p $$ is constant (conserved)

Now, $$\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}} $$ = constant

Therefore, $$K = {1 \over 2}m{v^2}$$ is constant (conserved)

Now, $$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = 0 \Rightarrow \overrightarrow F = m\overrightarrow a = 0$$

Thus, $$F = \nabla U \Rightarrow U$$ is constant (conserved)

Now, $$E = U + K \Rightarrow E$$ is constant, since U and K are constant

Now, $$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = 0 \Rightarrow \overrightarrow L $$ is constant (conserved)

Thus, the correct mapping is $$P \to 1,2,3,4,5$$.

For Q : $$\overrightarrow r (t) = \alpha \cos \omega t\,\widehat i + \beta \sin \omega t\,\widehat j$$

$$ \Rightarrow v = {{d\overrightarrow r } \over {dt}} = - \alpha \omega \sin \omega t\,\widehat i + \beta \omega \cos \omega t\,\widehat j$$

So v is not a constant $$ \Rightarrow \overrightarrow p $$ is not conserved for the path.

Now, $$\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}{{\sin }^2}\omega t + {\beta ^2}{\omega ^2}\sin \omega t} $$ which is not constant. Therefore,

$$K = {1 \over 2}m{v^2}$$ is not constant (not conserved)

Now, $$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = - \alpha {\omega ^2}\cos \omega t\,\widehat i + \beta {\omega ^2}\sin \omega t\,\widehat j$$

$$ \Rightarrow \overrightarrow a $$ is not constant

$$ \Rightarrow \overrightarrow F $$ is not constant

$$ \Rightarrow \overrightarrow U $$ is not constant (not conserved)

E = U + K is constant (conserved)

Now, $$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega \alpha \beta \widehat k$$

$$ \Rightarrow \overrightarrow L $$ is constant (conserved)

Thus, the correct mapping is $$Q \to 2,5$$.

For R : $$\overrightarrow r (t) = \alpha (\cos \omega t\,\widehat i + \sin \omega t\,\widehat j)$$

$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \omega ( - \sin \omega t\,\widehat i + \cos \omega t\,\widehat j)$$

$$ \Rightarrow \overrightarrow v $$ is not a constant

$$ \Rightarrow \overrightarrow p $$ is not a constant (not conserved)

Now, $$\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}({{\sin }^2}\omega t + {{\cos }^2}\omega t)} = \alpha \omega $$

$$ \Rightarrow \left| {\overrightarrow v } \right|$$ is constant

$$ \Rightarrow K = {1 \over 2}m{v^2}$$ is a constant (conserved)

Now, $$\overrightarrow a = {{dv} \over {dt}} = \alpha {\omega ^2}( - \sin \omega t\,\widehat i - \cos \omega t\,\widehat j)$$

$$ \Rightarrow \overrightarrow a $$ is not constant

Since $$ E = U + K$$ is constant (conserved)

$$\Rightarrow U$$ is constant (conserved)

Now, $$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega {\alpha ^2}\widehat k$$

$$ \Rightarrow \overrightarrow L $$ is constant (conserved)

Thus, the correct mapping is $$R \to 2,3,4,5.$$

For S : $$\overrightarrow r (t) = \alpha t\,\widehat i + {\beta \over 2}{t^2}\,\widehat j$$

$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta t\,\widehat j$$

$$ \Rightarrow \overrightarrow v $$ is not conserved

$$ \Rightarrow \overrightarrow p $$ is not conserved

Now, $$\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}\,{t^2}} $$

$$ \Rightarrow \left| {\overrightarrow v } \right|$$ is not conserved

$$ \Rightarrow K = {1 \over 2}m{v^2}$$ is not conserved

$$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \beta \widehat j$$

$$ \Rightarrow E = U + K$$ is conserved

But, U is not conserved.

Now, $$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = {1 \over 2}\alpha \beta {t^2}\widehat k$$

$$ \Rightarrow \overrightarrow L $$ is not conserved.

Thus, the correct mapping is $$S \to 5$$.

Therefore, option (A) is correct.