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NDA Physics · Formula sheet

Laws of Motion and Forces formulas

13 formulas, 5 reference tables and 19 common traps for NDA Physics Laws of Motion and Forces, grouped by subtopic.

Full notes with worked examples

Types of Forces — the Vocabulary of the Chapter

Learn this subtopic in the notes

What a force is — the foundation

Definition of the newton (from Newton's second law)

1 N=1 kg⋅1 m s−21\,\text{N} = 1\,\text{kg} \cdot 1\,\text{m s}^{-2}
  • NNnewton, the SI unit of force
  • kgkgkilogram, the SI unit of mass
  • m s⁻²metre per second squared, the SI unit of acceleration

Fundamental forces vs contact forces

ForceTypeRange / note
GravitationalFundamental, non-contactAlways attractive; infinite range; weakest of the four
ElectromagneticFundamental, non-contactSource of friction, tension, normal, contact forces at large scale
Strong nuclearFundamentalBinds protons and neutrons in the nucleus; very short range
Weak nuclearFundamentalResponsible for radioactive (beta) decay; very short range
Friction / Normal / TensionContact (derived)Need physical contact; obey Newton's third law; can act solid-fluid
NDA 2024 — contact forces (1) need contact, (2) obey the third law, (3) can act between a solid and a fluid: all three statements are correct.
The four fundamentals are the only true forces; everyday contact forces are electromagnetic in origin. NDA tests "which are fundamental?" (answer: all of gravity, EM, and the two nuclear forces).

Conservative vs non-conservative forces; central forces

ForceConservative?Central?
GravitationalConservativeCentral
ElectrostaticConservativeCentral
Spring (elastic restoring)ConservativeCentral (along the spring)
FrictionNon-conservativeNon-central
NDA 2019 — the force that is BOTH non-central AND non-conservative is friction (electric and gravitational are central and conservative).
Air resistance / viscous dragNon-conservativeNon-central
Friction is the only common force that is simultaneously non-central and non-conservative — a frequent NDA distractor target.

Equilibrium and restoring forces

TypeResponse to small pushExample
StableReturns to original positionBall at the bottom of a bowl; pendulum bob
UnstableMoves further awayBall balanced on top of a vertical rod
NDA 2018 — a ball balanced on a vertical rod is in UNSTABLE equilibrium.
NeutralStays in the new positionBall resting on a flat horizontal table
Restoring force is the signature of stable equilibrium; for a pendulum, gravity supplies it (NDA 2018).

Common traps

Force is a vector — direction matters

Two forces of the same magnitude can produce very different results depending on their directions. When combining forces you must use vector addition (the parallelogram law), never simple arithmetic — adding 3 N and 4 N gives anything from 1 N to 7 N depending on the angle between them.

Friction is a contact force; magnetism is non-contact — ALWAYS

NDA 2021 asked whether "friction is a contact force while magnetic force is a non-contact force" is true. It is ALWAYS true: friction requires touching surfaces, while magnetism reaches across a gap. The distractors hedge with "sometimes" or "never" — reject them.

Gravity acts as the RESTORING force for a pendulum

NDA 2018 asked what kind of force gravity provides for a vibrating pendulum bob. The answer is restoring force — not "applied" (no external push) and not "frictional". The component of weight along the swing always points back to the mean position.

Newton's Three Laws of Motion

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First law — inertia

Condition for the first law (equilibrium of motion)

F⃗net=0  ⟺  a⃗=0\vec{F}_{\text{net}} = 0 \iff \vec{a} = 0
  • FnetF_netnet (resultant) external force on the body
  • aaacceleration; zero means rest or constant velocity

Second law — F = ma

Newton's second law

F⃗=dp⃗dt=ma⃗(constant m)\vec{F} = \frac{d\vec{p}}{dt} = m\vec{a} \quad (\text{constant } m)
  • FFnet force (N)
  • p=mvp = mvlinear momentum (kg m/s)
  • mmmass (kg) — the constant of proportionality
  • aaacceleration (m/s²)

Third law — action and reaction

Newton's third law (force pair)

F⃗AB=−F⃗BA\vec{F}_{AB} = -\vec{F}_{BA}
  • FABF_ABforce exerted by A on B (action)
  • FBAF_BAforce exerted by B on A (reaction)

Combining forces — the parallelogram law

Magnitude of the resultant of two forces

R=P2+Q2+2PQcos⁡θR = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}
  • RRmagnitude of the resultant force
  • P,QP, Qmagnitudes of the two forces
  • θθangle between the two forces

Rotational inertia — moment of inertia of common bodies

Moment of inertia of common bodies (mass M, radius R)

Iring=MR2,Idisc=12MR2,Isphere=25MR2I_{\text{ring}} = MR^2, \quad I_{\text{disc}} = \tfrac{1}{2}MR^2, \quad I_{\text{sphere}} = \tfrac{2}{5}MR^2
  • IImoment of inertia (kg m²)
  • MMmass of the body
  • RRradius about the central axis

Mass vs weight

PropertyMassWeight
What it isAmount of matter / inertiaGravitational force on the body
Formula—W = mg
SI unitkilogram (kg)newton (N)
Scalar or vectorScalarVector (downward)
Varies with location?No — same everywhereYes — changes with g
NDA 2018 — mass is "the same everywhere"; NDA 2021 — mass is the constant of proportionality in F = ma.
Mass is constant and is the proportionality constant in F = ma; weight = mg varies with g. NDA tests both halves of this distinction.

Common traps

Constant velocity does NOT mean changing speed

NDA 2018 asked which statement is NOT correct for a body moving at constant velocity. The wrong statement is "its speed changes with time" — at constant velocity the speed is fixed, acceleration is zero, and the net force is zero. Don't confuse constant velocity (vector) with merely constant speed; here both are fixed.

Force is proportional to rate of change of momentum, NOT to momentum itself

NDA 2025 asked which second-law statement is NOT correct. The wrong one says "net force is proportional to the body's momentum" — it should be proportional to the RATE OF CHANGE of momentum (dp/dt). A body can have huge momentum yet zero net force (constant velocity).

Watch the units when computing F = ma

Convert km/h to m/s before using F = ma (divide by 3.6: 72 km/h = 20 m/s). NDA 2024 gave 72 km/h and a 0.2 s stop, yielding a = 100 m/s² and F = 100 kN — the large answer is a sign you converted correctly, not an error.

Action-reaction pairs act on DIFFERENT bodies

The classic error is to think action and reaction cancel and so nothing moves. They never cancel because they act on two separate bodies. When analysing one body's motion, only the forces ON THAT body matter — its reaction on something else is irrelevant to its own free-body diagram.

Two equal forces with a resultant equal to each: θ = 120°

NDA 2026 tested two equal forces whose resultant equals one of them. Solving gives cos θ = -1/2, so θ = 120° between them, and each force makes 60° with the resultant. Both statements (60° to resultant, 120° between forces) are correct.

Don't add force magnitudes arithmetically

5 N and 5 N do NOT give 10 N unless they are parallel. At 60° the resultant is √75 ≈ 8.66 N. Always use the parallelogram formula with the angle; only at θ = 0° is the answer the simple sum.

Mass is the constant of proportionality, NOT weight

NDA 2021 asked which is the constant of proportionality between force and acceleration in F = ma. It is MASS, not weight. Weight = mg varies with g; mass is invariant and is what makes a body resist acceleration.

Same mass + radius, different I — distribution decides

Ring > disc > solid sphere for moment of inertia at equal M and R, because the ring keeps all its mass at the rim while the sphere packs mass near the axis. The body with mass concentrated farther from the axis always has the larger I.

Impulse and Momentum

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Linear momentum, p = mv

Linear momentum

p⃗=mv⃗\vec{p} = m\vec{v}
  • pplinear momentum (kg m/s), a vector
  • mmmass (kg)
  • vvvelocity (m/s), a vector

Impulse = change in momentum; the cushioning principle

Impulse-momentum theorem

J=F Δt=Δp=m(v−u)J = F\,\Delta t = \Delta p = m(v - u)
  • JJimpulse (N·s, equivalently kg m/s)
  • FF(average) force
  • ΔtΔttime over which the force acts
  • Δp=m(v−u)Δp = m(v − u)change in momentum

Common traps

On an elastic bounce, momentum changes but speed and KE don't

NDA 2019 asked what changes suddenly when a ball bounces with no energy loss. Speed and kinetic energy (scalars depending on speed only) are unchanged; MOMENTUM changes because its direction reverses. The change in momentum is 2mv, not zero.

Cushioning increases TIME to reduce FORCE

Jumping onto sand, pulling hands back, crumple zones in cars, bending knees on landing — all work by increasing the impact TIME. The momentum change Δp is fixed; spreading it over a longer Δt lowers the peak force F = Δp/Δt. The reason is reduced force (reduced acceleration), not reduced momentum change.

Net force on the floor in a bounce includes weight

When a ball bounces, the floor's reaction must both reverse the ball's momentum AND support its weight. NDA 2024 (0.1 kg dropped from 0.45 m, rebounds to 0.20 m, contact 0.1 s): impact force = Δp/t = 0.1(3+2)/0.1 = 5 N, then add mg = 1 N to get 6 N net.

Conservation of Momentum and Collisions

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Conservation of linear momentum

Conservation of momentum (two bodies)

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
  • m₁, m₂masses of the two bodies
  • u₁, u₂velocities before
  • v₁, v₂velocities after

Force when mass is added or ejected (variable mass)

Force to maintain speed while loading mass at rate dm/dt

F=v dmdtF = v\,\frac{dm}{dt}
  • FFforce needed to keep speed constant
  • vv(constant) speed of the body
  • dm/dtdm/dtrate at which mass is added

Collisions — elastic and the equal-mass result

Equal-mass elastic head-on collision (m₂ initially at rest)

v1′=0,v2′=u1v_1' = 0, \qquad v_2' = u_1
  • u₁speed of the incoming body (mass m)
  • v₁'speed of body 1 after — it stops
  • v₂'speed of body 2 after — it takes the full speed

Common traps

Internal forces can't move the centre of mass

NDA 2019: an object moving at velocity v has a chemical reaction inside it. The reaction is internal, so it cannot change the velocity of the centre of mass (statement 1 correct). It CAN, however, redistribute kinetic energy among the particles, so the energy-conservation statement is false. Internal forces redistribute, never reset the total.

Vertically-falling mass adds no horizontal momentum

NDA 2023 (rain into a moving wagon): the rain has zero horizontal speed, so horizontal momentum is conserved as M₁v₁ = M₂v₂ — the wagon SLOWS. Don't assume the speed stays the same; the added mass must be dragged up to speed, and with no external horizontal force the wagon pays for it by slowing.

Equal-mass elastic collision: velocities are EXCHANGED

NDA 2024 (bob X hits identical bob Y at rest): X does not bounce back or rise on the other side — it STOPS at the collision point and Y carries off all the speed. This swap only happens for equal masses in an elastic head-on hit; unequal masses share the velocity differently.

Use momentum (not KE) to find the unknown speed

When a problem tells you the post-collision state (e.g. the first sphere stops), use conservation of MOMENTUM to find the other speed. Plugging into kinetic-energy conservation is unnecessary and a common time-sink; momentum alone gives the answer directly.

Friction

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Limiting (maximum) friction, f = μN

Limiting friction

fmax⁡=μNf_{\max} = \mu N
  • fmaxf_maxmaximum (limiting) friction force
  • μμcoefficient of friction (dimensionless)
  • NNnormal force (= mg on flat ground)

Friction as the only horizontal force — stopping a block

Friction from stopping (work-energy form)

f s=12mu2  ⇒  f=mu22sf \, s = \tfrac{1}{2} m u^2 \;\Rightarrow\; f = \frac{m u^2}{2s}
  • fffrictional force
  • ssstopping distance
  • uuinitial speed
  • mmmass of the block

Static, kinetic, and rolling friction

TypeWhen it actsRelative size
StaticBefore sliding startsLargest (up to μ_s N)
Kinetic / slidingWhile the body slidesIntermediate (μ_k N)
RollingWhile the body rollsSmallest
NDA 2024 — the correct ordering is Static friction > Kinetic friction > Rolling friction.
Rolling friction is the smallest, which is why wheels and ball bearings reduce resistance. NDA tests the ordering directly.

Common traps

Use the correct normal force, not always mg

f = μN uses the NORMAL force, which equals mg only on flat ground with no vertical applied force. On an incline N = mg cos θ; for a block on top of another, N is the upper block's weight. Plugging the full weight when the geometry says otherwise is the most common friction error.

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