Key Idea: This topic is all about what forces do to motion. When forces balance (net force zero) an object is in equilibrium; when they don't, the net force gives an acceleration through F = ma. The same idea, rewritten with momentum (p = mv), governs impulse and collisions. It is examined everywhere in Theme A: Paper 1A tests quick free-body diagrams, ratio and graph reasoning; Paper 2 tests multi-step calculations — connected bodies, terminal velocity, vertical circles and conservation-of-momentum problems.
⚖️ Balanced forces — equilibrium & resolving
Draw the object as a dot with one arrow per force (a free-body diagram). In equilibrium the net force is zero in every direction — so resolve every slanted force first, then balance each direction on its own.
- size (magnitude) of the force (N)
- angle the force makes with the horizontal (°)
- horizontal component of the force (N)
- vertical component of the force (N)
In equilibrium…: net force = **0**. left pull = **right pull**. up pull = **down pull**. at rest **or** steady velocity.
NOT in equilibrium…: net force ≠ 0. the forces don't cancel. the object **accelerates**. speeding up, slowing, or turning.
🚀 Newton's laws & F = ma
A net force changes the motion. 1st law: no net force → constant velocity. 2nd law: net force → acceleration. 3rd law: forces come in equal, opposite pairs on different objects. Always use the net force, and remember connected bodies share one acceleration.
- net (resultant) force (N)
- mass (kg)
- acceleration (m s⁻²)
- change in momentum (kg m s⁻¹)
- time interval (s)
🧱 Friction, buoyancy & drag
Three resistive/support forces share one habit: each is fixed by something other than the object alone. Friction depends on the normal force, buoyancy on the fluid and submerged volume, and drag on the speed.
- friction force, parallel to the surface (N)
- coefficient of STATIC friction (no unit — a pure number)
- coefficient of DYNAMIC (sliding) friction (no unit)
- normal force — the support push (N). Booklet writes it F_N.
- buoyancy (upthrust) force (N)
- density of the FLUID (kg m⁻³)
- volume of fluid pushed aside — the submerged volume (m³)
- gravitational field strength (9.8 N kg⁻¹)
- drag (resistive) force (N)
- viscosity of the fluid (Pa s)
- radius of the sphere (m)
- speed through the fluid (m s⁻¹)
🔄 Circular motion
Going round a circle means always changing direction, so there is always an acceleration pointing to the centre. The net force that supplies it is the centripetal force — it is not an extra arrow on the diagram, it IS the resultant (friction, tension, gravity, etc.).
- centripetal acceleration — points to the centre (m s⁻²)
- speed around the circle (m s⁻¹)
- radius of the circle (m)
- angular speed (rad s⁻¹)
- period — time for one full lap (s)
💥 Momentum, impulse & collisions
Momentum (p = mv) measures how hard something is to stop. An impulse (force × time) changes it: J = FΔt = Δp, also read off the area under a force–time graph. With no outside push, total momentum is conserved in every collision and explosion.
- momentum (kg m s⁻¹)
- mass (kg)
- velocity — carries a sign for direction (m s⁻¹)
- impulse — equals the change in momentum (N s = kg m s⁻¹)
- average force acting (N)
- time the force acts for (s)
📝 IB-style worked examples
IB-style question — connected blocks (F = ma)
Two blocks on a smooth floor are joined by a light string: a 3.0 kg block in front, a 5.0 kg block behind. A 32 N force pulls the front block. Find (a) the acceleration of the pair and (b) the tension in the string.
Solution:
(a) Treat the pair as one mass — they share an acceleration. Start with the given law:
Use the net force and the total mass (3.0 + 5.0 = 8.0 kg):
(b) The string is the only horizontal force on the back block — apply F = ma to it alone:
So the tension is:
a = 4.0 m s⁻² (shared by both blocks); tension = 20 N — the net force needed to accelerate the 5.0 kg back block.
IB-style question — tension in a sagging line
A lantern of weight 48 N hangs from the midpoint of a wire stretched between two posts. Each half of the wire makes 6.0° with the horizontal. By balancing the vertical forces, find the tension T in the wire. (sin 6.0° = 0.105)
Solution:
Both halves pull up at 6.0°, so each tension has vertical part T sin 6.0°. Resolve with the given component formula:
Two halves share the weight, so vertical balance gives:
Rearrange for T:
Work it out — keep the unit:
T ≈ 2.3 × 10² N (about 229 N) — far bigger than the 48 N weight, because a nearly-horizontal wire has only a tiny vertical part to do the lifting.
IB-style question — average force from impulse
A 0.20 kg ball travelling at 8.0 m s⁻¹ hits a wall head-on and bounces straight back at 6.0 m s⁻¹. The contact lasts 0.040 s. Find the average force the wall exerts on the ball.
Solution:
Momentum is a vector — take the rebound direction as positive. The ball arrives at −8.0 and leaves at +6.0, so:
Add the speeds because the direction flips:
Impulse equals the change in momentum, and impulse is FΔt — start from the given relation:
Solve for the average force — keep the unit:
average force ≈ 70 N. The classic trap is using v − u = 2.0; because the ball reverses, you ADD the speeds (6.0 + 8.0 = 14).
IB-style question — collision that sticks (perfectly inelastic)
A 1.5 kg trolley moving right at 4.0 m s⁻¹ collides with a stationary 0.50 kg trolley and they stick together. Find (a) their common velocity afterwards and (b) the fraction of kinetic energy lost.
Solution:
(a) Conserve momentum — total before = total after (they move off together at v):
Put in the numbers (the second trolley starts at rest):
(b) Kinetic energy before, using the given Eₖ = ½mv²:
Kinetic energy after (the combined 2.0 kg at 3.0 m s⁻¹):
Fraction lost = (before − after) ÷ before:
common velocity = 3.0 m s⁻¹; 0.25 (25%) of the kinetic energy is lost as heat and sound. Momentum is conserved even though KE is not — that's the signature of an inelastic collision.
✅ Quick self-check
Tap each card to check yourself across the whole topic.
Exam Tips
- Equilibrium ⇒ resolve every slanted force, then set the total to zero in EACH direction separately (left = right, up = down).
- Always find the NET force first, then use a = F ÷ m. Never plug a single force into F = ma.
- A nearly-horizontal rope holding a weight needs a HUGE tension — its vertical part is tiny, so the full pull must be large (T ∝ 1/sinθ).
- Buoyancy uses the FLUID's density and the SUBMERGED volume — not the object's density or its whole volume.
- At terminal velocity, weight = drag, so the acceleration is zero. Just after release there's no drag yet, so a ≈ g.
- Centripetal force is the NET inward force (friction, tension, gravity) — don't add it as an extra arrow.
- Momentum is a VECTOR: a ball that bounces back changes momentum by m(v + u), so you ADD the speeds.
- In any collision momentum is conserved; check KE separately to decide if it is elastic. Sticking together ⇒ KE lost.