Forces and momentum is the workhorse topic of every IB Physics paper. After Kinematics gives you the language of motion, Topic A.2 tells you why things accelerate, and gives you the two universal currencies — force and momentum — that connect to almost every later topic, from rigid-body rotation to magnetic fields and orbital motion. The new IB syllabus packs Newton's laws, friction, Hooke's law, drag, buoyancy, electric and magnetic forces, conservation of momentum, all four collision types, and circular motion into one topic.
This cheatsheet condenses every formula, trick and trap from Topic A.2 Forces and Momentum into one page. It covers Newton's three laws (and the most-tested third-law subtlety), contact forces (friction, Hooke, Stokes drag, buoyancy), the three field forces, linear momentum and impulse, all four collision types and explosions, and the full circular-motion toolkit including conical pendulums and vertical loops. Scroll to the bottom for the printable PDF and the gated Photon Academy library.
§1 — Newton's Three Laws A.2
1st Law (Inertia): An object remains at rest or in uniform motion unless a net external force acts on it.
2nd Law: $\vec{F}_{\text{net}} = \dfrac{\Delta\vec{p}}{\Delta t} = m\vec{a}$ (constant mass). Net force equals rate of change of momentum.
3rd Law (Action–Reaction): $\vec{F}_{AB} = -\vec{F}_{BA}$. Forces act on different bodies, are the same type, and have equal magnitude but opposite direction.
TrapN3L pairs act on different bodies. Weight (Earth on block) and normal force (table on block) both act on the same block — they are NOT a Newton's third law pair, even though they often happen to be equal in magnitude.
TrickFor an N3L pair: same magnitude, opposite direction, same force type, act simultaneously on different objects. Always state all four conditions for a "describe" mark.
§2 — Contact Forces A.2
Normal (horizontal):$F_N = mg$
Normal (incline $\theta$):$F_N = mg\cos\theta$
Static friction:$F_f \leq \mu_s F_N$ (inequality — up to a maximum)
Dynamic friction:$F_f = \mu_d F_N$ ($\mu_d < \mu_s$)
Hooke's law:$F_H = -k x$ (restoring, opposes displacement)
Stokes' drag:$F_d = 6\pi\eta r v$
Buoyancy (Archimedes):$F_b = \rho_{\text{fluid}}\,V\,g$
Inclined plane decomposition
On an incline at angle $\theta$:
- Parallel to slope: $F_{\text{net}} = F_{\text{app}} - mg\sin\theta \mp F_f$
- Perpendicular to slope: $F_N = mg\cos\theta$
- Weight component down the slope: $mg\sin\theta$.
TrapStatic friction is not always $\mu_s F_N$. It self-adjusts to whatever value (up to that maximum) is needed to prevent sliding. Only at the point of slipping does it equal $\mu_s F_N$.
NoteTerminal velocity for a small sphere falling through a fluid: balance gravity, buoyancy and Stokes drag — $mg = F_b + 6\pi\eta r v_T$, so $v_T = \dfrac{mg - \rho_f V g}{6\pi\eta r}$.
§3 — Field Forces A.2
Gravitational (weight):$F_g = mg$, $g = 9.81$ m s⁻²
Electric force:$F_e = qE$ (uniform field; direction depends on sign of $q$)
Magnetic force:$F_m = qvB\sin\theta$ (perpendicular to $v$, does no work)
TrickInside parallel plates, $E = V/d$. Balance an oil drop: $qE = mg$ ⇒ $q = mgd/V$ (Millikan-style problem).
§4 — Linear Momentum & Impulse A.2
Linear momentum:$\vec{p} = m\vec{v}$ [kg m s⁻¹]
Newton's 2nd law:$\vec{F} = \dfrac{\Delta\vec{p}}{\Delta t}$
Impulse:$\vec{J} = \vec{F}\,\Delta t = \Delta\vec{p}$ [N s]
Conservation of $p$:$\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}$ (no net external force)
NoteOn a force–time graph, the area under the curve equals the impulse $J$ — and therefore the change in momentum $\Delta p$.
TrapFor a bouncing ball, $\Delta p = m(-v_f) - m(v_i) = -m(v_f + v_i)$ when the ball reverses. The magnitude of the change in momentum is $m(v_f + v_i)$, NOT $m(v_f - v_i)$. Always sign your velocities.
§5 — Collisions & Explosions A.2
| Type | Momentum | Kinetic Energy |
|---|
| Elastic | Conserved | Conserved |
| Inelastic | Conserved | Not conserved (decreases) |
| Perfectly inelastic | Conserved | Maximum loss; objects stick together |
| Explosion | Conserved | Increases (from stored energy) |
Perfectly inelastic:$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)\,v_f$
Explosion (at rest):$0 = m_1 v_1 + m_2 v_2 \;\Rightarrow\; m_1 |v_1| = m_2 |v_2|$
KE loss (perf. inelastic):$\Delta E_k = \dfrac{m_1 m_2}{2(m_1 + m_2)}\,(v_{1i} - v_{2i})^2$
TrickIn an explosion, both fragments have equal-and-opposite momenta. Using $E_k = p^2/(2m)$, the lighter fragment always carries more kinetic energy.
§6 — Circular Motion A.2
Angular velocity:$\omega = \dfrac{2\pi}{T} = 2\pi f$ [rad s⁻¹]
Linear speed:$v = \omega r = \dfrac{2\pi r}{T}$
Centripetal acceleration:$a_c = \dfrac{v^2}{r} = \omega^2 r = \dfrac{4\pi^2 r}{T^2}$
Centripetal force:$F_c = \dfrac{mv^2}{r} = m\omega^2 r = \dfrac{4\pi^2 mr}{T^2}$
Key scenarios
- Conical pendulum (string at angle $\theta$ from vertical): $T\cos\theta = mg$, $T\sin\theta = m\omega^2 r$, so $\tan\theta = \omega^2 r / g$.
- Vertical circle — bottom: $N - mg = mv^2/r$ ($N$ is large here).
- Vertical circle — top: $N + mg = mv^2/r$ ($N$ is small here).
- Minimum speed at top of loop: $v_{\min} = \sqrt{gr}$ (when $N = 0$).
- Inside a rotating cylinder wall: $N = mv^2/r$ (centripetal); $\mu_d N = mg$ (friction supports weight).
TrapCentripetal force is not a new, separate force. It is the net inward force, supplied by tension, gravity, normal force, friction, etc. Never draw it as an extra arrow on a free-body diagram.
Trick$F_c \propto \omega^2 r$. Doubling $\omega$ quadruples $F_c$; doubling $r$ at the same $\omega$ doubles $F_c$.
NoteCentripetal force is always perpendicular to velocity, so it changes direction but not speed. The work done by the centripetal force is zero.
§7 — Exam Attack Plan A.2 — All sections
| Question trigger | Reach for |
|---|
| "Block on an incline" with friction | Resolve along + perpendicular to slope; $F_N = mg\cos\theta$ |
| "Connected blocks / pulley" | Free-body each block, $F = ma$ for each, common $a$ |
| "Spring stretched / compressed by $x$" | Hooke's law $F = kx$; PE later $= \tfrac{1}{2}k x^2$ |
| "Falls through fluid, find terminal speed" | Balance weight, buoyancy, Stokes drag |
| "Charged particle between plates" | $E = V/d$; $qE = mg$ for balance |
| "Particle deflected by magnetic field" | $F = qvB$, perpendicular to $v$, circular motion |
| "Two objects collide and stick" | Perfectly inelastic; conserve momentum, lose KE |
| "Object explodes / recoils / fires bullet" | Conserve momentum from rest; lighter fragment has more KE |
| "Force vs time graph, find $\Delta p$" | Area under graph $= J = \Delta p$ |
| "Loop, swing, satellite, conical pendulum" | Identify the real forces; set net inward $= mv^2/r$ |
| "Minimum speed to maintain contact at top" | $N = 0 \Rightarrow v_{\min} = \sqrt{gr}$ |
Worked Example — IB-Style 2-D Collision
Question (HL Paper 2 style — 7 marks)
A trolley A of mass 2.0 kg moves east at 6.0 m s⁻¹ on a smooth horizontal track. It collides head-on with a stationary trolley B of mass 4.0 kg. After the collision, A rebounds (moves west) at 1.0 m s⁻¹. Calculate (a) the velocity of B after the collision, and (b) the kinetic energy lost. State whether the collision is elastic.
Solution
- Take east as positive. Before: $p_i = (2.0)(+6.0) + (4.0)(0) = +12.0$ kg m s⁻¹. (M1)
- After: $p_f = (2.0)(-1.0) + (4.0)\,v_B = -2.0 + 4.0\,v_B$. (M1)
- Conservation of momentum: $-2.0 + 4.0\,v_B = +12.0$ ⇒ $v_B = +3.5$ m s⁻¹ (east). (A1) [part (a)]
- KE before $= \tfrac{1}{2}(2.0)(6.0)^2 = 36.0$ J. (B1)
- KE after $= \tfrac{1}{2}(2.0)(1.0)^2 + \tfrac{1}{2}(4.0)(3.5)^2 = 1.0 + 24.5 = 25.5$ J. (B1)
- Energy lost $= 36.0 - 25.5 = 10.5$ J. The collision is inelastic (KE not conserved). (A1)(R1) [part (b)]
Examiner's note: The single most common error here is forgetting the sign on A's rebound velocity — writing $+1.0$ instead of $-1.0$ gives $v_B = +1.5$ and a "negative energy lost". Always set up axes first and write all velocities as signed numbers. The conclusion mark (R1) requires you to compare KE before and after explicitly and state "inelastic".
Common Student Questions
Are weight and normal force a Newton's third law pair?
No. Weight and normal force act on the same body (the block) — N3L pairs must act on different bodies. The N3L partner of the block's weight (Earth pulling block down) is the block pulling Earth up. The N3L partner of the normal force (table pushing block up) is the block pushing the table down. Same magnitude, opposite direction, same force type, different bodies — all four conditions must hold.
Why do I lose marks on bouncing-ball impulse questions?
Because the velocities have opposite signs. If a ball hits the wall at $+v$ and rebounds at $-v$, then $\Delta p = m(-v) - m(+v) = -2mv$. The magnitude of impulse is $2mv$, NOT $mv$. Always set up axes, take inward as positive, and write velocities as signed quantities. The impulse magnitude is $m(v_f + v_i)$ when the ball reverses, not $m(v_f - v_i)$.
Is centripetal force a separate new force?
No — and putting it on a free-body diagram is the most common circular-motion error. Centripetal force is the net inward force, supplied by existing forces — tension in a string, gravity in orbit, friction on a road, normal force inside a loop. Always identify the real forces first, write Newton's second law in the radial direction, and set the net $= mv^2/r$.
What is the minimum speed at the top of a vertical loop?
At the top of a vertical loop, both gravity and the normal force point inward (toward the centre). Newton's second law gives $N + mg = mv^2/r$. The minimum speed is when $N = 0$ (the object is just barely in contact), so $v_{\min} = \sqrt{gr}$. Below this speed the object falls away from the track. Many students forget that gravity contributes to centripetal force at the top.
In an explosion, which fragment has more kinetic energy?
The lighter fragment. Momentum conservation gives $m_1 v_1 = m_2 v_2$ in magnitude, so the lighter fragment moves faster. Using $E_k = p^2/(2m)$ with equal momenta, the lighter mass has more kinetic energy. This is a frequent Paper 1 MCQ trap — students intuitively pick the heavier fragment because it is "bigger".
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy A.2 Forces and Momentum library (only available to enrolled students or via the resource library subscription) adds:
- Notes PDF — every concept worked through in full, with proofs and free-body-diagram walkthroughs.
- Tutorial booklet — IB-style questions sequenced from Newton's laws through 2-D collisions and vertical loops.
- Tutorial Solutions — full mark-scheme-style worked solutions with M1/A1/B1 annotations.
- Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
- Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
Get the printable PDF version
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