NEET PhysicsUpdated 19 Oct 2026

Laws of Motion: NEET PYQs and weightage

Laws of Motion carries 4.0 of 100 Physics marks in NEET by past-paper share: 30 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 8 of 30 Physics chapters.

Take the 25-question Laws of Motion test Formula sheet ↓All NEET Physics chapters

4.0/100 Weight in Physics

Exam importance (3/5, by past-paper share)

30 NEET PYQs

565 Practice questions

504 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (30 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Laws of Motion: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Newton's second law
$F_{net} = \frac{d p}{d t}$
$F_{net} = m a (body of fixed mass m)$
$\sum F_{x} = m a_{x}$
$\sum F_{y} = m a_{y}$
$1 N = 1 kg m s^{- 2}$
Use it when: Any body or system accelerates. Count only external forces, and never add $ma$ as an extra force.
Static friction
$f_{s} \leq μ_{s} N$
$(f_{s})_{m a x} = μ_{s} N$
Use it when: A body is at rest relative to a surface, e.g. a box on a train floor: $a_{m a x} = μ_{s} g$ .
Kinetic friction
$f_{k} = μ_{k} N (μ_{k} < μ_{s})$
$a = \frac{F - f _{k}}{m}$
Use it when: The surfaces are actually sliding. If $F$ is removed, $a = - f_{k} / m$ .
Block on an incline
$m g cos θ (into the slope)$
$m g sin θ (down the slope)$
$N = m g cos θ (only when no other applied force has a component perpendicular to the slope)$
Use it when: Resolving weight for any body resting or sliding on a slope of angle $θ$ .
Angle of repose
$tan θ_{m a x} = μ_{s} (mass does not matter)$
Use it when: The incline is raised until the block just begins to slide.
Normal force in a lift
$N = m (g + a) (a upward: speeding up while rising, or slowing down while descending)$
$N = m (g - a) (a downward: speeding up while descending, or slowing down while rising)$
$N = 0 (free fall, a = g downward)$
Use it when: Finding the weighing-scale reading or the floor's push on a body in an accelerating lift.
Centripetal force
$f_{c} = \frac{m v ^{2}}{R} (directed toward the centre)$
Use it when: Anything moves in a circle. Identify the real force (tension, friction, gravity) supplying it.
Turn on a level road
$v_{m a x} = μ_{s} R g$
Use it when: Friction alone keeps a car or cyclist on a flat curve. Convert km/h to m/s by dividing by $3.6$ .
Banked road, maximum speed
$v_{m a x} = \frac{R g ( μ _{s} + tan θ )}{1 - μ _{s} tan θ}$
Use it when: The curve is banked at $θ$ and friction also helps, with the car on the point of slipping outward.
Banked road, optimum speed
$v_{0} = R g tan θ$
Use it when: No friction is needed at all. Below $v_{0}$ , friction acts up the slope.
Impulse
$J = F Δ t = Δ p (unit N s)$
$∣ J ∣ = 2 m u cos θ (along the normal)$
For the wall bounce: a ball hits a wall at angle $θ$ from the normal with speed $u$ and bounces back at $u$ .
Use it when: A large force acts for a very short time, like a bat, a catch or a wall bounce.
Conservation of momentum
$p_{A} + p_{B} = p_{A}^{'} + p_{B}^{'}$
$m_{g} v_{g} = - m_{b} v_{b} (gun recoil)$
Use it when: No net external force acts on the system: collisions, explosions, firing a gun.
Equilibrium
$F_{1} + F_{2} + F_{3} = 0$
$\sum F_{x} = 0$
$\sum F_{y} = 0$
Use it when: A particle or a knot point in a string is at rest or moving with constant velocity.
Third law
$F_{A B} = - F_{B A} (equal, opposite, on different bodies)$
Use it when: Drawing the second body's free-body diagram, or finding the force on the floor or wall.

Trap

Static friction adjusts itself. Put f_s = μ_s N only when slipping is just about to begin. Otherwise, find f_s from the equations of motion. It can be less than μ_s N.

How does NEET ask Laws of Motion?

6 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQCircular Motion in Horizontal Plane

Try right here · No sign-in

A car is negotiating a curved road of radius R. The road is banked at an angle θ, the the coefficient of friction between the tyres of the car and the road is μ

_{s}

. The maximum safe velocity on this road is

Answer key (B) & worked solution

Correct answer: Option B

\frac{v ^{2}}{R g} = (\frac{μ _{s} + tan θ}{1 - μ _{s} tan θ})

\Rightarrow v = R g (\frac{μ _{s} + tan θ}{1 - μ _{s} tan θ})

Q2PhysicsNEET UG PYQFriction

Which one of the following statements is incorrect?

AFrictional force opposes the relative motion

BLimiting value of static friction is directly proportional to normal reaction

CRolling friction is smaller than sliding friction

DCoefficient of sliding friction has dimensions of length

Answer key (D) & worked solution

Correct answer: Option D

Coefficient of sliding friction has no dimension

f = μ_{s} N

\Rightarrow μ_{s} = \frac{f}{N}

Q3PhysicsNEET UG PYQConservation of Momentum

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angle to each other. The first part of mass 1 kg moves with a speed of 12 m s

^{- 1}

and the second part of mass 2 kg moves with 8 m s

^{- 1}

speed. If the third part flies off with 4 m s

^{- 1}

speed, then its mass is

A3 Kg

B5 Kg

C7 Kg

D17 Kg

Answer key (B) & worked solution

Correct answer: Option B

From the law of conservation of momentum, we get.

P_{1} + P_{2} + P_{3} = 0

Given that two of the masses of rock go off at right angles to each other. Let

P_{1}

and

P_{2}

be the momentum of the rocks that go off at right angles to each other.

P_{3} = P_{1}^{2} + P_{2}^{2} (∵ V_{1} ⊥ V_{2})

m_{3} V_{3} = (m_{1} V_{1})^{2} + (m_{2} V_{2})^{2}

m_{3} \times 4 = (1)^{2} \times (12)^{2} + (2)^{2} (8)^{2}

m_{3} \times 4 = 20

∴ m_{3} = 5 k g

Q4PhysicsNEET UG PYQCircular Motion in Horizontal Plane

One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v', the net force on the particle (directed towards center) will be (T represents the tension in the string)

T - \frac{m v ^{2}}{l}

BZero

T + \frac{m v ^{2}}{l}

Answer key (C) & worked solution

Correct answer: Option C

Centripetal force

(\frac{m v ^{2}}{l})

is provided by tension in string, So the net force will be equal to tension i.e T.

Q5PhysicsNEET UG PYQCircular Motion in Horizontal Plane

A car travels on a circular racetrack of radius 50 m , which is banked at an angle

θ

. If the car travels at a speed

10 ms^{- 1}

, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be

10 ms^{- 2}

, the value of

θ

is:

tan^{- 1} (\frac{2}{5})

tan^{- 1} (3 /2)

tan^{- 1} (23)

tan^{- 1} (\frac{1}{5})

Answer key (D) & worked solution

Correct answer: Option D

tan θ = \frac{v ^{2}}{r g} = \frac{10 \times 10}{50 \times 10} = \frac{1}{5}

θ = tan^{- 1} (\frac{1}{5})

Q6PhysicsNEET UG PYQFriction

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15. (g = 10 ms

^{- 2}

)

150 m s^{- 2}

1.5 m s^{- 2}

50 m s^{- 2}

1.2 m s^{- 2}

Answer key (B) & worked solution

Correct answer: Option B

F_{S} = ma

f_{L} = m a_{m a x}

μ m g = m a_{m a x}

a_{m a x} = μg

$\begin{gathered} = 0.15\left( {10} \right) \\ = 1.5\,\,m/{s^2} \\ \end{gathered} $

Which topics in Laws of Motion carry the most questions?

All 7 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Newton's II Law of Motion	13	141
Friction	6	149
Circular Motion in Horizontal Plane	5	74
Conservation of Momentum	4	70
Equilibrium of Particle	2	114
Newton's I Law of Motion	0	10
Newton's III Law of Motion	0	7

Why does Laws of Motion show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Laws of Motion, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Laws of Motion?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Laws of Motion test

Chapters either side of Laws of Motion by weight

Moving Charges and Magnetism Weight 4.6/100 Electrostatic Potential and Capacitance Weight 4.2/100 Alternating Current Weight 3.8/100 Gravitation Weight 3.8/100

Laws of Motion: NEET PYQs and weightage

Laws of Motion: formula sheet for NEET

Newton's second law

Static friction

Kinetic friction

Block on an incline

Angle of repose

Normal force in a lift

Centripetal force

Turn on a level road

Banked road, maximum speed

Banked road, optimum speed

Impulse

Conservation of momentum

Equilibrium

Third law

How does NEET ask Laws of Motion?

Which topics in Laws of Motion carry the most questions?

Why does Laws of Motion show "–" until you answer 3 questions?

What happens when you test Laws of Motion?

Chapters either side of Laws of Motion by weight