JEE Advanced 2018 · Paper 2 · Physics Q18

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PhysicsList match+3 / -1Question 18 of 54

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 $\alpha \ne \beta$ . 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: $\vec{p}$ 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

P
$\vec{r}(t) = \alpha t\,\hat{i} + \beta t\,\hat{j}$
Q
$\vec{r}(t) = \alpha\cos\omega t\,\hat{i} + \beta\sin\omega t\,\hat{j}$
R
$\vec{r}(t) = \alpha(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j})$
S
$\vec{r}(t) = \alpha t\,\hat{i} + \dfrac{\beta}{2}t^2\,\hat{j}$

LIST-II

1
$\vec{p}$
2
$\vec{L}$
3
$K$
4
$U$
5
$E$

P → 1, 2, 3, 4, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 5

P → 1, 2, 3, 4, 5; Q → 3, 5; R → 2, 3, 4, 5; S → 2, 5

P → 2, 3, 4; Q → 5; R → 1, 2, 4; S → 2, 5

P → 1, 2, 3, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 2, 5

Answer and worked solution

Official answer: A

Find the force from $\vec F=m\ddot{\vec r}$ in each case.

P. $\vec r=(\alpha t,\beta t)$ : uniform velocity and $\vec F=0$ . $\vec r\parallel\vec v$ always, so $\vec L=m\vec r\times\vec v=0$ , a constant. $\vec p$ , $\vec L$ , $K$ , $U$ and $E$ are all conserved: 1, 2, 3, 4, 5.
Q. $\vec r=(\alpha\cos\omega t,\beta\sin\omega t)$ with $\alpha\ne\beta$ : $\vec F=-m\omega^2\vec r$ , a central spring-like force. $\vec L$ and $E$ are conserved. The speed and the distance vary, so $K$ and $U$ do not stay constant: 2, 5.
R. Uniform circular motion with $\vec F=-m\omega^2\vec r$ : $\vec L$ , $K$ , $U$ (since $r$ is constant) and $E$ are conserved. $\vec p$ is not: 2, 3, 4, 5.
S. $\vec r=(\alpha t,\tfrac\beta2t^2)$ : a constant force $m\beta\hat j$ , as in a projectile. Only $E$ is conserved. The torque about the origin, $m\alpha\beta t\,\hat k$ , is nonzero, so $\vec L$ changes: 5.

(A)

Topics:Conservation of mechanical energy Angular momentum and its conservation Simple harmonic motion

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JEE Advanced 2018 · Paper 2, Physics question 18