JEE Advanced 2020 · Paper 2 · Physics Q12

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PhysicsMultiple correct+4 / -2Question 12 of 54

Starting at time $t = 0$ from the origin with speed $1\ \mathrm{m\,s^{-1}}$ , a particle follows a two-dimensional trajectory in the $x$ - $y$ plane so that its coordinates are related by the equation $y = \dfrac{x^2}{2}$ . The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$ , respectively. Then

a_x = 1\ \mathrm{m\,s^{-2}}

implies that when the particle is at the origin,

a_y = 1\ \mathrm{m\,s^{-2}}

a_x = 0

implies

a_y = 1\ \mathrm{m\,s^{-2}}

at all times

t = 0

, the particle's velocity points in the

x

-direction

a_x = 0

implies that at

t = 1

s, the angle between the particle's velocity and the

x

axis is

45^\circ

Answer and worked solution

Official answer: A, B, C, D

Relations along the path. From $y=\frac{x^2}2$ :
$v_y=xv_x,\qquad a_y=v_x^2+xa_x.$

(C) At the origin $v_y=0$ , so the velocity points along $x$ . True.

(A) At the origin $v_x=1$ (all of the speed) and $x=0$ , so $a_y=v_x^2=1\ \mathrm{m\,s^{-2}}$ . True.

(B) If $a_x=0$ , $v_x$ stays 1, so $a_y=1\ \mathrm{m\,s^{-2}}$ at all times. True.

(D) With $a_x=0$ , $x=t$ . At $t=1$ s, $x=1$ , so $\frac{v_y}{v_x}=x=1$ and the velocity is at $45^\circ$ . True.

(A), (B), (C), (D)

Topics:Motion in a plane

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JEE Advanced 2020 · Paper 2, Physics question 12