JEE Advanced Previous Year Questions (2018 - 2023): Motion | Physics for JEE Main & Advanced PDF Download

2023

Also dx₂/dt = speed of tip of person's shadow
Applying similar triangle rule in ΔABE and ΔDCE

Differentiate both sides w.r.t. 

= 100 cm/s
This is the speed of the tip of the person's shadow with respect to the lamp post. But, we need to find the speed of the shadow's tip with respect to the person, which is the relative speed :

= 40 cms^-1

2022

So, horizontal displacement done by the projectile in new region is

So, n = 0.95d

2020

Q1: Starting at time t = 0 from the origin with speed 1 ms^-1, a particle follows a two-dimensional trajectory in the x-y plane so that its coordinates are related by the equation y = x² / 2. The x and y components of its acceleration are denoted by a_x and a_y, respectively. Then    [JEE Advanced 2020 Paper 2]
(a) a_x = 1 ms^-2 implies that when the particle is at the origin, a_y = 1 ms^-2
(b) a_x = 0 implies a_y = 1 ms^-2 at all times
(c) at t = 0, the particle's velocity points in the x-direction
(d) a_x = 0 implies that at t = 1s, the angle between the particle's velocity and the x axis is 45^o
Ans: (b), (c) & (d)
Given, equation
y = x² / 2

For option A :
when particle is at origin then x = 0.

Given that at origin, v_x = 1 m/s
So, a_y= (1)² = 1 ms^-2
At origin, 0 × a_x = 0 means a_y does not depend on the value of a_x. 

For option B :
When a_x = 0 ⇒ v_x = constant
So v_x will stay 1 ms^-1.
when a_x = 0,
 
= 1 ms^-2 

For option C :
Velocity is always tangential to the path of motion of the particle. So at origin path is along x axis as tangent at origin along x axis.

For option D :
When a_x = 0 ⇒ v_x = constant
So v_x will stay 1 ms^-1.
At t = 1, x = 1 m
∴v_y = xv_x = 1 × 1 = 1
Now, slope of velocity at t = 1 is,
tanθ = v_y / v_x = 11 = 1
⇒ θ = 45^o 

2019

Ans: 4.0
For first projectile,

For journey,

⇒ α = 4

2018

If speed is v after the first collision, then speed should remain  1/√2 times, because kinetic energy has reduced to half.

⇒ h_max = 30 m