QJEEAdv18Ph7e97: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph7e97 Solution: In the List- below, four different paths of a particle are…

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.