JEE Advanced Previous Year Questions (2018 - 2024): Laws of Motion | Physics for JEE Main & Advanced PDF Download

2024

Q1: A ball is thrown from the location (x₀, y₀) = (0, 0) of a horizontal playground with an initial speed v₀ at an angle θ₀ from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x₁, y₁) = (L, 0). The stone is thrown at an angle (180° - θ₁) from the +x-direction with a suitable initial speed. For a fixed v₀, when (θ₀, θ₁) = (45°, 45°), the stone hits the ball after time T₁, and when (θ₀, θ₁) = (60°, 30°), it hits the ball after time T₂. In such a case, (T₁/T₂)² is ____.
Ans: 2
For Case I :
For case II,

2023

Q1: A particle of mass m is moving in the xy-plane such that its velocity at a point (x,y) is given as  where α is a non-zero constant. What is the force   acting on the particle?
(a)

(b)

(c)

(d) [JEE Advanced 2023 Paper 2]
Ans: 

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2021

Both particle have no vertical velocity after splitting so both will take same time to reach the ground.
∵ Time of motion of one part falling vertically downwards is 0.5 s
⇒ Time of motion of another part, t = 0.5 s

Q2: A projectile is thrown from a point O on the ground at an angle 45^∘ from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g = 10 m/s².
The value of x is _____________.     [JEE Advanced 2021 Paper 1 ]
Ans: 7.5

Range =

Time of flight =

Both particle have no vertical velocity after splitting so both will take same time to reach the ground.
∵ Time of motion of one part falling vertically downwards is 0.5 s
⇒ Time of motion of another part, t = 0.5 s
From momentum conservation, p_i = p_f
2m × 5 = m × v
v = 10 m/s
Displacement of other part in 0.5 s in horizontal direction,
= v(T/2) = 10 × 0.5 = 5 m = R
∴ Total distance of second part from point O is x =3R/2 = 3 × 5/2
⇒ x = 7.5 m 

2020

Here, μ_s = 0.40, μ_k = 0.32
∵ μ_kN₃ = μ_sN₄
and x₁N₃ = 40 × N₄
So, 
∴ x₁ = 32 cm

⇒ x_g = 25.6

Q2: A football of radius R is kept on a hole of radius r (r < R) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle θ from the horizontal as shown in the figure below. The maximum value of θ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale)      [JEE Advanced 2020 Paper 1]

(a) sin θ = r / R
(b) tan θ = r / R
(c) sin θ = r / 2R
(d) cos θ = r / 2R
Ans: (a)

For θ_max, the football is about to roll, then N₂ = 0 and all the forces (mg and N₁) must pass through contact point.
∴
⇒ sin θ_max = r / R

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2019

Q1: A block of mass 2M is attached to a massless spring with spring-constant k.This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are a₁, a₂ and a₃ as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x₀. Which of the following option(s) is/are correct? [g is the acceleration due to gravity. Neglect friction] 
(a) a₂ −  a₁= a₁ − a₃
(b) At an extension of x₀ / 4 of the spring, the magnitude of acceleration of the block connected to the spring is 3g / 10.
(c) x0 = 4Mg / k
(d) When spring achieves an extension of x₀ / 2 for the first time, the speed of the block connected to the spring is      [JEE Advanced 2019 Paper 2]
Ans: 

In the frame of pulley B,
the hanging masses have accelerations :
M → (a₂ − a₁)
2M → (a₃ − a₁) : downward.
∴ (a₂ − a₁) = (−a₃ − a₁) [constant]
Assuming that the extension of the spring is x
We consider the FBD of A : 

where, a1 = d²x / dt²  .... (i)
and the FBD of the rest of the system in the frame of pulley B :

Upward acceleration of block M w.r.t. the pulley B = Downward acceleration of block 2M w.r.t. the pulley.

Substituting in Eq. (i), we get

This is equation of SHM
Maximum extension = 2 × amplitude
i.e.,
Amplitude = x₀ / 2 =  and ω =
At x₀ / 4, acceleration is easily found from Eq. (iii),

At x₀/2, speed of the block
(2M) = ω x Amplitude
=

∴  option (a) is correct.

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2018

Q1: A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of 1.0N is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = v₀e^−t/τ, where �0 is a constant and τ = 4s. The displacement of the block, in metres, at t = τ is ______________ Take e⁻¹= 0.37.    [JEE Advanced 2018 Paper 2]
Ans: 6.30
The block was initially at rest and its velocity just after the application of impulse is v(0) = v₀e^−0/τ  = v₀. The applied impulse is equal to the change in linear momentum of the block i.e., J = mv₀, which give
v₀ = J/m = 1/0.4 = 2.5 m/s.

The velocity of the particle is given as
v(t) = v₀e^−t/τ
Integrate to get the displacement 

Substitute t = τ = 4 s and 
v₀ = 2.5 m/s to get x(τ) 
= (2.5) (4) (1 − e⁻¹) 
= 6.3 m.