Question

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: $\vec{L}$ is the linear momentum, $\vec{L}$ 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.	$\vec{r} (t) = \alpha t \hat{i} + \beta t \hat{j}$	1.	$\vec{p}$
Q.	$\vec{r}(t) = \alpha \, \cos \omega t \, \hat{i} + \beta \, \sin \, \omega t \hat{j}$	2	$\vec{L}$
R.	$\vec{r} (t) = \alpha (\cos \omega t \hat{i} + \sin \, \omega t \hat{j}$	3	K
S.	$\vec{r} (t) = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}$	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; 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

Answer: (A)

(P) $\vec{r} \left(t\right) = \alpha t\hat{i} + \beta t \hat{j} $
$\vec{v} = \frac{d\vec{r}\left(t\right)}{dt} = \alpha\hat{i} + \beta \hat{j} $ {constant}
$\vec{a} = \frac{\vec{dv}}{dt} = 0 $
$\vec{P} = m\vec{v} $ (remain constant)
$k = \frac{1}{2} mv^{2}$ {remain constant}
$ \vec{F} = - \left[\frac{\partial U}{\partial x} \hat{i } + \frac{\partial U}{\partial y} \hat{i}\right] = 0 $
$ \Rightarrow U \to $ constant
$E = K + U $
$\frac{d \vec{L}}{dt} = \vec{\tau} = \vec{r} \times\vec{F} = 0 $
$\vec{L} $ = constant
(Q) $\vec{r} = \alpha \cos\left(\omega t\right) \hat{i} + \beta \sin\left(\omega t\right)\hat{j} $
$\vec{v} = \frac{d\vec{r}}{dt} = - \alpha \omega\sin\left(\omega t\right) \hat{i} + \beta \omega\cos\left(\omega t\right)\hat{j} $
$\vec{a} = \frac{d\vec{v}}{dt} = -\alpha \omega^{2} \cos\left(\omega t\right) \hat{i} - \beta\omega^{2} \sin\left(\omega t\right)\hat{j}$
$ = - \omega^{2} \left[\alpha \cos\left(\omega t\right) \hat{i} + \beta \sin\left(\omega t\right)\hat{j}\right]$
$ \vec{a} = - \omega^{2} \vec{r}$
$ \vec{\tau} = \vec{r } \times\vec{F} = 0$ {$\vec{r}$ and $\vec{F}$ are parallel}
$ \Delta U = - \int\vec{F}.dr = + \int^{r}_{0} m \omega^{2} .r.dr $
$\Delta U = m\omega^{2} \left[ \frac{r^{2}}{2}\right] $
$U \propto r^{2} $
$r = \sqrt{\alpha^{2} \cos^{2} \left(\omega t\right) + \beta^{2} \sin^{2}\left(\omega t\right)} $
r is a function of time (t)
U depends on r hence it will change with time
Total energy remain constant because force is central.
(R) $\vec{r} \left(t\right) = \alpha\left(\cos\omega t \hat{i} + \sin\left(\omega t\right)\hat{j}\right) $
$\vec{v } \left(t\right) = \frac{d\vec{r}\left(t\right)}{dt} = \alpha \left[-\omega\sin\left(\omega t\right)\hat{i} + \omega\cos\left(\omega t\right)\hat{j}\right] $
$\left|\vec{v} \right| = \alpha\omega $ (Speed remains constant)
$\vec{a} \left(t\right) = \frac{d\vec{v}\left(t\right)}{dt}=\alpha \left[-\omega^{2} \cos\left(\omega t\right)\hat{i} - \omega^{2} \sin \left(\omega t\right) \hat{j}\right]$
$ = - \alpha\omega^{2} \left[\cos\left(\omega t\right)\hat{i} + \sin \left(\omega t\right)\hat{j}\right] $
$\vec{a}\left(t\right) = - \omega^{2}\left(\vec{r}\right) $
$\vec{\tau} = \vec{F} \times\vec{r} = 0 $
$|\vec{r} | = \alpha$ (remain constant)
Force is central in nature and distance from fixed point is constant.
Potential energy remains constant
Kinetic energy is also constant (speed is constant)
(S) $\vec{r} = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}$
$\vec{v} = \frac{d\vec{r}}{dt} = \alpha t \hat{i} + \beta t \hat{j} $ (speed of particle depends on 't')
$ \vec{a} = \frac{d\vec{v}}{dt} = \beta\hat{j} $ {constant}
$\vec{F} = m \vec{a}$ {constant}
$ \Delta U = - \int\vec{F} . d \vec{r} = - m \int^{t}_{0} \beta\hat{j} . \left(\alpha \hat{i} + \beta t \hat{j}\right)dt $
$U = \frac{-m \beta^{2}t^{2}}{2}$
$ k = \frac{1}{2} mv^{2} = \frac{1}{2} m \left(\alpha^{2} + \beta^{2} + t^{2}\right) $
$E = k + U = \frac{1}{2} m\alpha^{2} $ [remain constant]