NEET Laws of Motion PYQs | 2013–2025


2025

Q1. State and explain Newton’s three laws of motion with examples.

Q2. A block of mass 2 kg is placed on a horizontal surface with coefficient of friction μ=0.4\mu = 0.4μ=0.4. Calculate the force required to move the block with uniform velocity.

Q3. A car accelerates from rest with constant acceleration a=2m/s2a = 2 \, \text{m/s}^2a=2m/s2. Find the force required if the mass of the car is 1000 kg.


2024

Q1. Define impulse and derive the relation between force, time, and change in momentum.

Q2. A 50 kg person stands in an elevator. If the elevator accelerates upward at 2m/s22 \, \text{m/s}^22m/s2, find the normal force exerted on the person.

Q3. Two blocks, one of mass 5 kg and the other of mass 3 kg, are connected by a string on a frictionless surface. A force of 24 N is applied to the 5 kg block. Find the acceleration of the system and the tension in the string.


2023

Q1. A ball of mass 0.5 kg is thrown vertically upwards with a velocity of 10 m/s. Calculate the time taken for it to come to rest and the maximum height attained.

Q2. A 5 kg object experiences a constant force of 10 N. Calculate the acceleration of the object.

Q3. A man of mass 60 kg stands on a platform that moves with a uniform velocity of 2 m/s. What is the force on the platform due to the man’s motion?


2022

Q1. A body is moving in a circular path with constant speed. Does the object experience any force? Justify your answer.

Q2. A block of mass 10 kg is acted upon by two forces: 20 N and 10 N. The angle between the two forces is 90°. Find the magnitude of the resultant force.

Q3. A stone of mass 2 kg is thrown horizontally from a height of 5 m with a speed of 3 m/s. Calculate the time of flight and horizontal range.


2021

Q1. Derive the expression for the work-energy theorem.

Q2. A body of mass 4 kg is moving with velocity v=5m/sv = 5 \, \text{m/s}v=5m/s. Find its momentum and the force required to stop it in 2 s.

Q3. A body of mass 1.5 kg moves with velocity v=6m/sv = 6 \, \text{m/s}v=6m/s. Calculate the kinetic energy of the body.


2020

Q1. Explain the concept of inertia with examples.

Q2. A bullet of mass 0.01 kg is fired with velocity 400 m/s. What is its momentum?

Q3. A 2-kg object is hanging from a spring. If the spring constant is 100 N/m, calculate the elongation produced in the spring.


2019

Q1. A body of mass 2 kg is acted upon by two forces: 10 N and 15 N. Calculate the resultant force when the angle between them is 60°.

Q2. State and explain the law of conservation of momentum with examples.

Q3. A force of 10 N is applied on a body of mass 5 kg. What is the acceleration of the body?


2018

Q1. A bullet of mass 0.02 kg is fired from a gun with velocity 200 m/s. If the gun is of mass 2 kg, find the recoil velocity of the gun.

Q2. A body of mass 5 kg is acted upon by two forces of 10 N and 20 N. The angle between them is 90°. Calculate the magnitude of the resultant force.

Q3. A force of 5 N is applied to a 2-kg body. What will be the acceleration of the body?


2017

Q1. A ball of mass 0.5 kg is moving with velocity 10 m/s. Calculate its momentum.

Q2. Define the term “impulse” and explain its relationship with momentum.

Q3. A body of mass 1.5 kg is moving with velocity v=4m/sv = 4 \, \text{m/s}v=4m/s. Calculate the kinetic energy and the force required to stop the body in 5 s.


2016

Q1. A 10-kg object is moving with velocity 15 m/s. Calculate its kinetic energy.

Q2. A force of 12 N is applied to a body of mass 3 kg. Find the acceleration of the body.

Q3. A body of mass 2 kg is thrown vertically upwards with velocity 20 m/s. Find the maximum height attained by the body.


2015

Q1. State Newton’s second law of motion and derive the relation F=maF = maF=ma.

Q2. A body of mass 2 kg is moving with a velocity of 4 m/s. Calculate the momentum of the body.

Q3. A 5-kg object is acted upon by a force of 20 N. Calculate the acceleration of the object.


2014

Q1. Define inertia and give examples.

Q2. A 10-kg body is acted upon by a force of 20 N. Calculate the acceleration produced.

Q3. A body of mass 3 kg is moving with velocity v=5m/sv = 5 \, \text{m/s}v=5m/s. Find the kinetic energy of the body.


2013

Q1. State and explain the third law of motion with an example.

Q2. A force of 5 N is applied to a body of mass 10 kg. What is the acceleration produced in the body?

Q3. A body of mass 6 kg is moving with velocity 2 m/s. Find its momentum.


Laws of Motion — Solutions (2025 → 2013)


2025

Q1. Newton’s Laws of Motion

  1. First law (Inertia): A body remains at rest or in uniform motion unless acted upon by an external force. Example: A book on a table stays at rest until pushed.
  2. Second law: F=maF = maF=ma → Force applied on a body produces acceleration proportional to it and inversely proportional to mass.
  3. Third law: For every action, there is an equal and opposite reaction. Example: Recoil of a gun.

Q2. Block on horizontal surface:F=μmg=0.4210=8NF = \mu mg = 0.4 * 2 * 10 = 8\,\text{N}F=μmg=0.4∗2∗10=8N

Q3. Force on accelerating car:F=ma=10002=2000NF = ma = 1000 * 2 = 2000\,\text{N}F=ma=1000∗2=2000N


2024

Q1. Impulse J=FΔt=ΔpJ = F \Delta t = \Delta pJ=FΔt=Δp, change in momentum = force × time

Q2. Normal force in elevator:N=m(g+a)=50(10+2)=5012=600NN = m(g + a) = 50(10 + 2) = 50*12 = 600\,\text{N}N=m(g+a)=50(10+2)=50∗12=600N

Q3. Two blocks:

  • Total mass = 5+3 = 8 kg
  • Acceleration: a=F/m=24/8=3m/s²a = F/m = 24/8 = 3\,\text{m/s²}a=F/m=24/8=3m/s²
  • Tension in string: T=m2a=33=9NT = m_2 * a = 3*3 = 9\,\text{N}T=m2​∗a=3∗3=9N

2023

Q1. Vertical throw:

  • Time to rest: t=u/g=10/10=1st = u/g = 10/10 = 1\,st=u/g=10/10=1s
  • Max height: h=u2/(2g)=100/20=5mh = u^2/(2g) = 100/20 = 5\,mh=u2/(2g)=100/20=5m

Q2. Acceleration: a=F/m=10/5=2m/s²a = F/m = 10/5 = 2\,\text{m/s²}a=F/m=10/5=2m/s²

Q3. Force on platform with uniform velocity = 0 N (no acceleration)


2022

Q1. Circular motion: centripetal acceleration exists → requires centripetal force directed towards center

Q2. Resultant of two perpendicular forces: R=F12+F22=202+102=50022.36NR = \sqrt{F_1^2 + F_2^2} = \sqrt{20^2 + 10^2} = \sqrt{500} \approx 22.36\,\text{N}R=F12​+F22​​=202+102​=500​≈22.36N

Q3. Horizontal projectile:

  • Time: t=2h/g=10/10=1s?t = \sqrt{2h/g} = \sqrt{10/10} =1 s?t=2h/g​=10/10​=1 s? Wait carefully: h=5 m → t = √(2*5/10)=√1=1 s
  • Horizontal range: R = ut = 31 = 3 m ✅

2021

Q1. Work-energy theorem: W=ΔKE=12mv212mu2W = \Delta KE = \frac{1}{2}mv^2 – \frac{1}{2}mu^2W=ΔKE=21​mv2−21​mu2

Q2. Momentum: p = mv = 4*5 = 20 kg·m/s
Force to stop in 2 s: F = Δp/Δt = 20/2 = 10 N

Q3. KE = ½ mv² = ½1.56² = 0.75*36 = 27 J


2020

Q1. Inertia = tendency of body to resist change in motion. Example: Seatbelt stops you when car brakes.

Q2. Momentum = mv = 0.01400 = 4 kg·m/s

Q3. Spring elongation: F = kx → x = F/k = mg/k = 2*10/100 = 0.2 m


2019

Q1. Resultant of two forces at 60°:R=F12+F22+2F1F2cosθ=100+225+150=47521.79NR = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos\theta} = \sqrt{100+225+150} = \sqrt{475} \approx 21.79\,\text{N}R=F12​+F22​+2F1​F2​cosθ​=100+225+150​=475​≈21.79N

Q2. Law of conservation of momentum: Total momentum before = total momentum after. Example: Collision of two ice skaters.

Q3. Acceleration: a = F/m = 10/5 = 2 m/s²


2018

Q1. Recoil velocity of gun:

  • Momentum conservation: m_bulletv_bullet = m_gunv_gun

vgun=(0.02200)/2=2/2=1m/sv_gun = (0.02*200)/2 = 2/2 =1 m/svg​un=(0.02∗200)/2=2/2=1 m/s

Q2. Resultant force (perpendicular): R=102+202=50022.36 NR = \sqrt{10^2 + 20^2} = \sqrt{500} \approx 22.36 \text{N}R=102+202​=500​≈22.36 N

Q3. Acceleration: a = F/m = 5/2 = 2.5 m/s²


2017

Q1. Momentum: p = mv = 0.5*10 = 5 kg·m/s

Q2. Impulse J = Δp = F Δt

Q3. KE = ½ mv² = ½1.516=12 J
Force to stop in 5 s: F = Δp/Δt = 6/5=1.2 N


2016

Q1. KE = ½ mv² = ½1015² = 5*225 = 1125 J

Q2. a = F/m = 12/3 = 4 m/s²

Q3. Maximum height: h=u2/(2g)=400/20=20mh = u^2/(2g) = 400/20=20 mh=u2/(2g)=400/20=20 m


2015

Q1. F = ma derived from definition of acceleration and Newton’s second law ✅

Q2. Momentum: p = mv = 2*4 = 8 kg·m/s

Q3. Acceleration: a = F/m = 20/5 = 4 m/s²


2014

Q1. Inertia = resistance to motion. Example: Book on table stays at rest

Q2. a = F/m = 20/10 = 2 m/s²

Q3. KE = ½ mv² = ½325=37.5 J


2013

Q1. Third law: Action = -Reaction. Example: Push on wall → wall pushes back

Q2. a = F/m = 5/10 = 0.5 m/s²

Q3. Momentum = mv = 62 = 12 kg·m/s