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Work, Power & Energy

Section: Physics

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Question : 5 of 22

Marks: +1, -0

In the List-I below, four different paths of a particle are given as functions of time. In these functions,

\alpha

\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:

\overset{-}{p}

is the linear momentum

\overset{-}{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. $\overset{-}{r}(t)=\alpha \overset{-}{ti}+\beta \overset{-}{tj}$	1. $\overset{-}{p}$
Q. $\overset{-}{r}(t)=\alpha \cos \omega \overset{-}{ti}+\beta \sin \omega \overset{-}{tj}$	2. $\overset{-}{L}$
R. $\overset{-}{r}(t)=\alpha\left(\cos \omega \overset{-}{ti}+\sin \omega \overset{-}{tj}\right)$	3. K
S. $\overset{-}{r}(t)=\alpha \overset{-}{ti}+\frac{\beta}{2} t^{\overset{-}{2j}}$	4. U
	5. E

	P	Q	R	S
A)	1,2,3,4,5	2,5	2,3,4,5	5
B)	1,2,3,4,5	3,5	2,3,4,5	2,5
C)	2,3,4	5	1,2,4	2,5
D)	1,2,3,5	2,5	2,3,4,5	2,5

[JEE Adv 2018 P2]

A
B
C
D

Solution:

P)

\vec{r}(t)=\alpha \hat{ti}+\beta \hat{tj}

\vec{V}=\frac{\vec{dr}(t)}{dt}=\alpha \hat{i}+\beta \hat{j}

{Constant}

\vec{a}=\frac{\vec{dv}}{dt}=0

\vec{P}=\vec{mv}

(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 \rightarrow

Constant

E=K+U

\frac{\vec{dL}}{dt}=\tau^{\rightarrow}=\vec{r}\times\vec{F}=\vec{0L}=

constant

(Q)\vec{r}=\alpha os(\omega t)\hat{i}+\beta \sin(\omega t)\hat{j}

\vec{a}=\frac{\vec{dv}}{dt}=-\alpha \omega^2 \cos(\omega t)\hat{i}-\beta \omega \sin(\omega t)\hat{j}

\vec{a}=-\omega^{\vec{2r}}

U \propto r^2

U depends on r hence it will change with time

Total energy remain constant because force is central

(R)

\vec{r}(t)=\alpha(\cos \omega \hat{ti}+\sin(\omega t)\hat{j})

\vec{v}(t)=\frac{\vec{dr}(t)}{dt}=\alpha(-\omega \sin(\omega t)\hat{i}+\omega \cos(\omega t)\hat{j})

\vec{a}(t)=-\alpha \omega^2[\cos(\omega t)\hat{i}+\sin(\omega t)\hat{j}]

(S)

\vec{r}=\alpha \hat{ti}+\frac{\beta}{2}t^{\hat{2j}}\vec{a}=\frac{\vec{dv}}{dt}=\beta \hat{j}

Constant

\vec{F}=\vec{ma}

Constant

\Delta U = -\int \vec{F} \cdot \vec{dr} = -m \int\limits_{0}^{t} \beta \hat{j} \cdot (\alpha \hat{i} + \beta t \hat{j}) \, dt \;\;\; E = k + U = \frac{1}{2} m \alpha^2

[Remain constant]

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