PREPWISER.
Practise online

Newton's Laws of Motion revision notes

Every idea in the Newton's Laws of Motion chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 28 and download the PDF.

Sign in to read all 28Download PDF

Read the first part free. Sign in free with Google to read the rest and download the PDF: 11 A4 pages.11 A4 pages, ready to print.

How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Newton's laws, momentum and impulse

First law and inertia
3× asked, last 2023
F⃗net=0  ⟺  v⃗=constant\vec F_{\text{net}}=0\;\Longleftrightarrow\;\vec v=\text{constant}
Holds in inertial frames only. Mass measures inertia: a larger mass resists a change in velocity more.
  • Inertia of rest (passengers jerk back when a bus starts), of motion (a fan keeps turning after switch-off) and of direction (mud flies off a wheel along the tangent).
Second law
7× asked, last 2025
F⃗net=dp⃗dt,F⃗net=ma⃗    (m constant)\vec F_{\text{net}}=\frac{d\vec p}{dt},\qquad \vec F_{\text{net}}=m\vec a\;\;(m\text{ constant})
Net external force. It is a vector law, so apply it to each component. 1 N = 1 kg m s⁻².
  • Zero net force on a fixed mass means constant momentum, not necessarily rest.
  • F=maF=ma is a special case. Mass gained with no velocity along the motion (sand dropped on a belt): F=dp/dt=m dv/dt+v dm/dtF=dp/dt=m\,dv/dt+v\,dm/dt. Mass lost at the body's own velocity (sand leaking from a cart): F=m dv/dtF=m\,dv/dt.
  • Momentum of fixed size that only turns (like i^cos⁡kt−j^sin⁡kt\hat i\cos kt-\hat j\sin kt): F⃗=dp⃗/dt\vec F=d\vec p/dt is always perpendicular to p⃗\vec p, as in uniform circular motion.
  • Not the full balloon mass MM after ballast Δm\Delta m is dropped. Upthrust FF stays fixed, so F=M(g−a)F=M(g-a) falling at aa and (M−Δm)(g+a)(M-\Delta m)(g+a) rising at aa: Δm=2Ma/(g+a)\Delta m=2Ma/(g+a).
  • Quick method. Forces in equilibrium: remove one and the rest add up to minus that force, so a=Fremoved/ma=F_{\text{removed}}/m. To stop a body accelerating, add a force equal and opposite to the resultant.
Constant force: vector motion
8× asked, last 2025
a⃗=F⃗m,v⃗=u⃗+a⃗t,r⃗=u⃗t+12a⃗t2\vec a=\frac{\vec F}{m},\qquad \vec v=\vec u+\vec a t,\qquad \vec r=\vec u t+\tfrac12\vec a t^2
Constant F⃗\vec F only. Work each axis separately; r⃗\vec r is measured from the start point.
  • From rest under a constant force the path is a straight line along F⃗\vec F: displacements are in the ratio x:y:z=Fx:Fy:Fzx:y:z=F_x:F_y:F_z.
Variable force
6× asked, last 2026
F⃗=mdv⃗dt=md2r⃗dt2,F=mvdvdx\vec F=m\frac{d\vec v}{dt}=m\frac{d^2\vec r}{dt^2},\qquad F=mv\frac{dv}{dx}
Use v dv/dxv\,dv/dx when vv is given in terms of xx. Integrate F(t)/mF(t)/m once for v⃗\vec v, twice for r⃗\vec r.
ttvvOmg/kmg/kslope ggterminal speed
Drag −kv-kv: starts at slope gg, levels off at vt=mg/kv_t=mg/k
  • Direction of the net force = direction of d2r⃗/dt2d^2\vec r/dt^2. Terms in r⃗\vec r that are constant or linear in tt give no force.
  • v=kxv=k\sqrt x means v2=k2xv^2=k^2x: uniform acceleration a=k2/2a=k^2/2, so the force is constant.
  • Falling from rest against drag −kv-kv: v=(mg/k)(1−e−kt/m)v=(mg/k)(1-e^{-kt/m}). The vv–tt graph starts with slope gg and levels off at terminal speed mg/kmg/k.
  • Not r⃗=a⃗t2/2\vec r=\vec a t^2/2 when FF depends on tt. That needs constant a⃗\vec a. Integrate step by step, adding the starting values v⃗(0)\vec v(0) and r⃗(0)\vec r(0).
  • Quick method. Velocity-dependent force: write m dvx/dtm\,dv_x/dt and m dvy/dtm\,dv_y/dt separately. A quantity claimed to be constant must have zero time derivative; test each option that way.
Third law
2× asked, last 2024
F⃗AB=−F⃗BA\vec F_{AB}=-\vec F_{BA}
Equal, opposite, same kind of force, acting on different bodies, so the pair never cancels in one FBD.
Impulse
3× asked, last 2025
J⃗=∫F⃗ dt=Δp⃗,Favg=ΔpΔt\vec J=\int\vec F\,dt=\Delta\vec p,\qquad F_{\text{avg}}=\frac{\Delta p}{\Delta t}
Impulse = area under the FF–tt graph. Unit N s = kg m s⁻¹. Δp⃗\Delta\vec p is final minus initial, as vectors.
ttFFOτ\tauF0F_0impulse =23F0τ=\tfrac23F_0\tau
Area under FF–tt is the impulse; a parabolic hump is 23\tfrac23 of its rectangle
  • Quick method. Area under a parabolic hump that is zero at both ends = two-thirds of its bounding rectangle (base × peak). This saves the integral.
Force by a jet or stream
1× asked, last 2022
F=dmdt v=ρAv2F=\frac{dm}{dt}\,v=\rho A v^2
ρ\rho = density, AA = jet cross-section, vv = jet speed relative to surface; stopped dead. Bouncing straight back at vv: 2F2F.

Practise Newton's Laws of Motion questions as you read. Free, and you can start without signing up.

Practise Newton's Laws of Motion

21 more concepts in 4 topics

Free-body diagrams and equilibrium; Connected bodies, pulleys and constraints; Non-inertial frames and pseudo force; and Springs. Read the first part free. Sign in free with Google to read the rest and download the PDF.

Sign in to read the restYou come straight back to this spot.Download the PDF

Free-body diagrams and equilibrium

  • Free-body diagram5×
  • Three forces and Lami's theorem4×
  • Rope pulled sideways at a point4×
  • Smooth incline4×
  • Rope or chain with mass3×
  • Chain hanging between two supports1×

Connected bodies, pulleys and constraints

  • Blocks pulled or pushed together4×
  • Atwood machine6×
  • Block on a table, hanging mass6×
  • Double incline3×
  • Chain over an edge or pulley2×
  • Constraint relations4×
  • Movable pulley

Non-inertial frames and pseudo force

  • Apparent weight in a lift6×
  • Pseudo force5×
  • Bob hanging in an accelerating vehicle1×
  • Block on an accelerating incline2×
  • Block on a free wedge1×

Springs

  • Spring force1×
  • Series, parallel and cut springs
  • Sudden cut: spring vs string

Practise Newton's Laws of Motion next

About 70 past JEE Main questions from this chapter, with instant answer checking and step-by-step solutions where available. Free, and you can start without signing up.

Practise Newton's Laws of Motion

Also free on PrepWiser

Spotted a mistake? Write to support@prepwiser.in and we will fix it in the next version.

Version 1, October 2026. Free to share for study, as it is. Please don't sell it or remove the PrepWiser name.

PractiseSign in to read all 28Download PDF
Free with Google, takes a few seconds11 pages, free