laws of motion formulas pyqs

Complete Formula Sheet

A compact all-in-one formula sheet for Laws of Motion, friction, connected bodies, pulleys and circular motion.

Newton's First Law

Inertia

If net external force is zero, velocity remains constant.

ΣF = 0 ⇒ a = 0

Rest and uniform motion are both equilibrium states.

Newton's Second Law

Force Law

Net external force equals rate of change of momentum.

F = dp/dt
F = ma

For constant mass systems.

Newton's Third Law

Interaction Pair

Action and reaction are equal, opposite and act on different bodies.

F_AB = -F_BA

Momentum

Linear Momentum

Vector quantity measuring motion content.

p = mv

SI unit: kg m s^-1.

Impulse

Force-Time Effect

J = ∫F dt
J = Δp

Area under F-t graph gives impulse.

Momentum Conservation

Isolated System

Σp_initial = Σp_final

Valid when net external impulse is zero.

Pseudo Force

Non-Inertial Frame

F_pseudo = -ma_frame

Applied opposite to frame acceleration.

Friction

Contact Resistance

f_s ≤ μ_sN
f_k = μ_kN

Connected Bodies

Common Acceleration

a = F_net / (m₁ + m₂ + ...)

Use system equation first, then individual body equation for tension.

Pulley Systems

Ideal String

T same in one massless string
a_Atwood = (m₂-m₁)g/(m₁+m₂)

Circular Motion

Centripetal Need

v = rω
a_n = v²/r = rω²
F_c = mv²/r

Banking

Road Without Friction

tan θ = v²/(rg)
v = √(rg tan θ)

Conical Pendulum

Horizontal Circle

T cos θ = mg
T sin θ = mv²/r
ω = √(g/(l cos θ))

Vertical Circle

Tension Equations

Bottom: T - mg = mv²/r
Top: T + mg = mv²/r

Loop Condition

Complete Loop

v_top,min = √(gr)
v_bottom,min = √(5gr)

Radius of Curvature

Curved Path

a_n = v²/ρ

ρ is instantaneous radius of curvature.

Newton's Laws Formulae

Definitions, units, dimensions and exam-ready notes for the foundation of mechanics.

Definitions

Force: An external influence that changes or tends to change the state of motion.
Inertia: Natural tendency of a body to resist change in motion.
Momentum: Product of mass and velocity, p = mv.
Impulse: Change in momentum due to a force acting for a time interval.

Units And Dimensions

Quantity	SI Unit	Dimension
Force	newton, N	[M L T^-2]
Momentum	kg m s^-1	[M L T^-1]
Impulse	N s	[M L T^-1]
Coefficient of friction	unitless	[M⁰L⁰T⁰]

Formula Table

Concept	Formula	Important Note
Net force	ΣF = ma	Use vector sum along chosen axes.
Variable force impulse	J = ∫F dt	Area under force-time graph.
Average force	F_avg = Δp/Δt	Useful in collision and impact problems.
Equilibrium	ΣF_x = 0, ΣF_y = 0	Body may be at rest or moving uniformly.
Pseudo force	F_p = -ma₀	Only in accelerating reference frame.

Friction Formulae

Static, limiting, kinetic and rolling friction with angle relations and graph behavior.

Static Friction

Self-adjusting friction before sliding starts.

0 ≤ f_s ≤ μ_sN

Limiting Friction

Maximum possible static friction.

f_L = μ_sN

Kinetic Friction

Opposes relative sliding motion.

f_k = μ_kN

Usually μ_k < μ_s.

Coefficient Of Friction

Ratio of friction to normal reaction at limiting or sliding state.

μ = f/N

Angle Of Friction

Angle made by resultant contact force with normal at limiting friction.

tan λ = μ

Angle Of Repose

Incline angle at which sliding just begins.

tan α = μ_s

Rolling Friction

Rolling friction is generally much smaller than sliding friction. It depends on deformation, surface nature and wheel radius.

f_rolling ≪ f_k

Important Graph Summary

Static friction rises with applied force until the limiting value.
At impending motion, friction equals μ_sN.
Once motion starts, friction drops to μ_kN and becomes nearly constant.

Circular Motion Formulae

Uniform circular motion, banking, conical pendulum, vertical circle and radius of curvature formulas.

Speed And Angular Speed

v = rω
ω = 2π/T = 2πf

Normal Acceleration

a_n = v²/r
a_n = rω²

Centripetal Force

F = mv²/r
F = mrω²

Radius Of Curvature

a_n = v²/ρ

ρ changes along non-circular curved paths.

Banking And Vertical Circle Table

Topic	Formula	Use
Frictionless banking	tan θ = v²/(rg)	Safe speed on banked road without friction.
Conical pendulum	tan θ = v²/(rg), r = l sin θ	Tension components provide mg and centripetal force.
Vertical circle top	T + mg = mv²/r	Minimum top speed occurs when T = 0.
Vertical circle bottom	T - mg = mv²/r	Tension is maximum at bottom.
Complete loop	v_bottom,min = √(5gr)	For a string or bead completing a smooth loop.

Free Body Diagram Summary

A quick revision section for FBD rules, tension, normal reaction, constraints and pulley systems.

FBD Rules

Isolate one body at a time.
Draw only external forces acting on that body.
Choose axes along acceleration or along the plane.
Resolve vectors after drawing the FBD.

Tension Rules

Tension pulls away from the body along the string.
In a massless, frictionless string over an ideal pulley, tension is same throughout.
Different strings may have different tensions.

Normal Reaction Rules

Normal is perpendicular to the contact surface.
Normal is not always equal to weight.
On an accelerating lift, use vertical force balance with acceleration.

Constraint Motion Rules

Length of an inextensible string remains constant.
Differentiate length relation to connect velocities and accelerations.
For a movable pulley, displacement relation often gives a factor of 2.

Pulley Rules

Ideal pulley changes direction of tension only.
For Atwood machine, heavier mass moves downward.
Write equations in the direction of acceleration.

Exam Shortcut

For connected bodies, first solve acceleration using the whole system. Then isolate one body to find internal tension or contact force.

NCERT Examples

Solved NCERT-style examples for conceptual clarity and board-level preparation.

NCERT Example

Force From Momentum Change

A ball of mass 0.15 kg moving at 20 m/s is stopped in 0.05 s. Find average force.

Solution: Δp = 0 - 0.15 × 20 = -3 kg m/s. F_avg = Δp/Δt = -3/0.05 = -60 N. Magnitude = 60 N opposite to motion.

NCERT Example

Elevator Apparent Weight

A 50 kg student is in a lift accelerating upward at 2 m/s². Find normal reaction. Take g = 10 m/s².

Solution: N - mg = ma. N = m(g + a) = 50(10 + 2) = 600 N.

NCERT Example

Block On Smooth Table

A 4 kg block is pulled by a horizontal force of 20 N. Find acceleration.

Solution: F = ma ⇒ a = 20/4 = 5 m/s².

NCERT Example

Impulse From Force-Time Graph

A force increases linearly from 0 to 40 N in 2 s. Find impulse.

Solution: Impulse = area of triangle = 1/2 × 2 × 40 = 40 N s.

NCERT Exercises

Solved exercise-style questions covering force, friction and connected body equations.

NCERT Exercise

Two Blocks Connected By String

Blocks of 3 kg and 2 kg are pulled by 20 N on a smooth table. Find acceleration and tension.

Solution: System mass = 5 kg, a = 20/5 = 4 m/s². For 2 kg block, T = ma = 2 × 4 = 8 N.

NCERT Exercise

Friction On Horizontal Surface

A 10 kg block has μ_k = 0.2. Find kinetic friction. Take g = 10 m/s².

Solution: N = mg = 100 N. f_k = μ_kN = 0.2 × 100 = 20 N.

NCERT Exercise

Atwood Machine

Masses 6 kg and 4 kg hang over a smooth pulley. Find acceleration.

Solution: a = (6 - 4)g/(6 + 4) = 2g/10 = 2 m/s² for g = 10 m/s².

NCERT Exercise

Equilibrium On Incline

A block rests on a rough incline. What friction direction is possible?

Solution: Gravity component mg sin θ tends to pull the block down the incline, so static friction acts up the incline.

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CBSE PYQs

Solved CBSE-style previous year questions focused on definitions, reasoning and numericals.

CBSE PYQ Style

Why Does A Passenger Fall Forward?

A bus suddenly stops. Explain why passengers fall forward.

Answer: Due to inertia of motion, the lower part of the body stops with the bus but the upper part continues moving forward.

CBSE PYQ Style

Impulse And Momentum

State impulse-momentum theorem and give its SI unit.

Answer: Impulse equals change in momentum: J = Δp. SI unit is N s or kg m s^-1.

CBSE PYQ Style

Action-Reaction Pair

If action and reaction are equal and opposite, why do they not cancel?

Answer: They act on different bodies, so they cannot cancel while drawing FBD of one body.

CBSE PYQ Style

Friction Advantage

Give two cases where friction is useful.

Answer: Walking and writing require friction. Vehicle tyres also need friction for acceleration, braking and turning.

NEET PYQs

High-quality solved NEET questions. Authentic year is mentioned whenever known; otherwise marked as NEET exam-style.

NEET Exam-style Question

Minimum Speed At Top

A stone tied to a string moves in a vertical circle of radius r. Minimum speed at top for just completing the circle is:

Solution: At top, for just taut string T = 0, so mg = mv²/r. Therefore v = √(gr).

NEET Exam-style Question

Angle Of Repose

If coefficient of static friction is 1/√3, find angle of repose.

Solution: tan α = μ = 1/√3, so α = 30°.

NEET Exam-style Question

Banked Road

A road is banked at angle θ. Find speed for which no friction is needed.

Solution: tan θ = v²/(rg), hence v = √(rg tan θ).

NEET Exam-style Question

Force-Time Graph

A constant force of 25 N acts for 0.4 s. Find impulse.

Solution: J = FΔt = 25 × 0.4 = 10 N s.

JEE Main PYQs

Solved JEE Main-level problems emphasizing equations, constraints and force balance.

JEE Main Exam-style Question

Block Pulled On Rough Surface

A 5 kg block on a horizontal rough surface has μ = 0.2. A force of 30 N pulls it horizontally. Find acceleration. Take g = 10 m/s².

Solution: Friction = μmg = 0.2 × 5 × 10 = 10 N. Net force = 30 - 10 = 20 N. a = 20/5 = 4 m/s².

JEE Main Exam-style Question

Two-Body Contact Force

Two blocks 2 kg and 3 kg are pushed by 25 N on a smooth surface. Find contact force on 3 kg block if force is applied on 2 kg block.

Solution: a = 25/(2 + 3) = 5 m/s². Contact force on 3 kg block = 3 × 5 = 15 N.

JEE Main Exam-style Question

Conical Pendulum

A bob moves as a conical pendulum with string length l and angle θ from vertical. Find angular speed.

Solution: T cos θ = mg and T sin θ = mω²r with r = l sin θ. Dividing gives tan θ = ω²l sin θ/g, so ω = √(g/(l cos θ)).

JEE Main Exam-style Question

Pseudo Force

A frame accelerates right with acceleration a. What pseudo force acts on mass m in the frame?

Solution: Pseudo force has magnitude ma and acts left, opposite to acceleration of frame.

JEE Advanced PYQs

High-quality solved JEE Advanced-style questions for multi-step reasoning.

JEE Advanced Exam-style Question

Block On Wedge Idea

A block remains at rest relative to a wedge accelerating horizontally. What method should be used?

Solution: Work in the wedge frame and apply pseudo force opposite wedge acceleration. Then use equilibrium along and perpendicular to the inclined plane.

JEE Advanced Exam-style Question

Variable Force Impulse

A force varies as F = kt from t = 0 to T. Find impulse.

Solution: J = ∫₀^T kt dt = kT²/2.

JEE Advanced Exam-style Question

Loop Normal At Bottom

A bead enters a smooth vertical loop of radius R with speed √(6gR) at bottom. Find normal reaction at bottom.

Solution: N - mg = mv²/R = m(6gR)/R = 6mg. Hence N = 7mg.

JEE Advanced Exam-style Question

Constraint Relation

A movable pulley is supported by two segments of the same string. If free end moves down by x, how much does pulley move?

Solution: Length relation gives x = 2y, so pulley moves up by y = x/2.

IB Questions

Solved IB Physics questions with clear command-term style answers.

IB Physics

Explain Net Force

Explain why an object moving at constant velocity has zero resultant force.

Answer: Constant velocity means acceleration is zero. Since F_net = ma, resultant force must be zero.

IB Physics

Determine Friction

A 12 kg box is pulled at constant speed by 18 N horizontally. Determine friction.

Answer: Constant speed means net force is zero. Friction balances applied force, so friction = 18 N opposite motion.

IB Physics

Momentum Conservation

State one condition for conservation of linear momentum.

Answer: Total linear momentum is conserved when no resultant external force or external impulse acts on the system.

IB Physics

Circular Motion Direction

State the direction of centripetal acceleration.

Answer: It is directed toward the center of the circular path at every instant.

IGCSE Questions

Solved IGCSE Physics questions with simple language and clean working.

IGCSE Physics

Calculate Acceleration

A resultant force of 15 N acts on a 3 kg object. Calculate acceleration.

Solution: a = F/m = 15/3 = 5 m/s².

IGCSE Physics

Friction In Walking

Why is friction needed for walking?

Answer: The foot pushes the ground backward. Friction from the ground acts forward and helps the person move.

IGCSE Physics

Mass And Weight

State the difference between mass and weight.

Answer: Mass is amount of matter and is measured in kg. Weight is gravitational force, W = mg, and is measured in newtons.

IGCSE Physics

Balanced Forces

What happens when forces on a moving object are balanced?

Answer: Resultant force is zero, so the object continues with constant velocity.

A-Level Questions

Solved A-Level mechanics questions for force resolution, momentum and circular motion.

A-Level Physics

Inclined Plane Acceleration

A block slides down a smooth plane inclined at θ. Find acceleration.

Solution: Component of weight along plane is mg sin θ. Hence a = g sin θ.

A-Level Physics

Impulse In Collision

A 0.2 kg ball reverses velocity from +15 m/s to -10 m/s. Find impulse.

Solution: J = m(v - u) = 0.2(-10 - 15) = -5 N s. Magnitude = 5 N s.

A-Level Physics

Normal On Banked Track

For a frictionless banked track, what provides centripetal force?

Answer: The horizontal component of normal reaction provides centripetal force; the vertical component balances weight.

A-Level Physics

Terminal Equilibrium

If a body moves with constant speed under several forces, what is the resultant force?

Answer: Resultant force is zero because acceleration is zero.

Assertion Reason

Question bank for assertion-reason practice in board, NEET and competitive exam format.

Question 1

Assertion: A body can be in motion even when net force is zero.
Reason: Zero net force means zero acceleration, not necessarily zero velocity.

Answer: Both assertion and reason are true, and reason correctly explains assertion.

Question 2

Assertion: Static friction is always equal to μ_sN.
Reason: Static friction is self-adjusting.

Answer: Assertion is false, reason is true. Static friction can take any value up to μ_sN.

Question 3

Assertion: Centripetal force is not a new type of force.
Reason: It is the name of the net inward force required for circular motion.

Answer: Both are true, and reason correctly explains assertion.

Question 4

Assertion: Action and reaction cancel each other.
Reason: They are equal and opposite.

Answer: Assertion is false. They act on different bodies, so they do not cancel for one body's FBD.

Case Study Questions

Case-based question bank for applied reasoning in Laws of Motion.

Case Study 1: Car On A Banked Road

A car takes a circular turn on a banked road of radius r. The road is designed so that at speed v no friction is required.

Which force has a horizontal component toward the center?
Write the safe speed formula.
What happens if the car speed is much higher?

Answers: Normal reaction provides the inward component. v = √(rg tan θ). At very high speed, friction acts down the slope to prevent outward slipping.

Case Study 2: Box In An Accelerating Lift

A box of mass m rests on the floor of a lift accelerating upward with acceleration a.

Draw the forces on the box.
Find normal reaction.
What is apparent weight?

Answers: Forces are N upward and mg downward. N - mg = ma, so N = m(g + a). Apparent weight is N.

Case Study 3: Two Blocks And A String

Two blocks are connected by a light string and pulled on a smooth horizontal table by an external force.

Do both blocks have same acceleration?
How is system acceleration found?
How is tension found?

Answers: Yes, because the string is inextensible. a = F/(m₁ + m₂). Tension is found using the equation of one block.

Case Study 4: Football Impact

A football changes velocity when kicked. The contact time is small, so the force is large.

Which theorem is used?
How can force be reduced for the same momentum change?
What is area under force-time graph?

Answers: Impulse-momentum theorem. Increase contact time to reduce average force. Area under F-t graph is impulse.

Quick Revision Notes

One-page revision for final exam day: formulas, concepts, mistakes, NEET and JEE focus points.

ΣF = maAlways use net external force.
J = ΔpImpulse is area under F-t graph.
fs ≤ μsNStatic friction self-adjusts.
mv²/rRequired inward force for circular motion.

Most Important Formulas

F = ma
p = mv
J = Δp
f_k = μ_kN
tan α = μ
v = rω
a_n = v²/r
tan θ = v²/(rg)

Most Important Concepts

Action and reaction act on different bodies.
Normal reaction depends on direction of acceleration and contact geometry.
Friction opposes relative motion or tendency of relative motion.
Centripetal force is net inward force, not a separate force.

Most Common Mistakes

Writing f_s = μ_sN in every static case.
Equating normal reaction to mg blindly.
Putting action-reaction forces on the same FBD.
Forgetting pseudo force in accelerating frame.

NEET Quick Revision

Memorize angle of repose and banking formulas.
Practice one-step friction and vertical circle questions.
Use dimensional checks for force, impulse and momentum.
Watch signs in lift and pulley problems.

JEE Quick Revision

Master FBD before writing equations.
Use system equations for acceleration.
Use constraints for pulley problems.
For non-inertial frames, add pseudo force and apply equilibrium or Newton's law.

Last-Minute Method

Draw FBD.
Choose axes smartly.
Write ΣF = ma along each axis.
Apply constraint or friction limit.
Check units and limiting cases.