OCR A Level Physics A: Dynamics, free-body diagrams, drag and terminal velocity (H556)

This guide covers OCR A Level Physics A (H556), Module 3 Forces and motion, sections 3.2.1 Dynamics and 3.2.2 Motion with non-uniform acceleration. It follows the OCR specification Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. Moments, equilibrium, density and pressure (3.2.3 and 3.2.4) are taught separately.

Related pages: revision notes, practice questions, vectors and resolving, the OCR A Level Physics hub and the printable checklist. Check your level with the free 10-minute diagnostics.

What the specification asks for

Reference	You must be able to
3.2.1(a)	Use net force = mass × acceleration, F = ma (recall it)
3.2.1(b)	Define the newton as the unit of force
3.2.1(c)	Use weight W = mg (recall it)
3.2.1(d)	Use the terms tension, normal contact force, upthrust and friction
3.2.1(e)	Draw and use free-body diagrams
3.2.1(f)	Analyse one- and two-dimensional motion under constant force
3.2.2(a)	Describe drag as the frictional force on an object moving through a fluid
3.2.2(b)	Give factors affecting drag for an object moving through air
3.2.2(c)	Describe motion in a uniform gravitational field with drag
3.2.2(d)	Explain terminal velocity and techniques to determine it in fluids

Net force, F = ma and the newton (3.2.1(a), (b))

The acceleration of an object is proportional to the net force (the vector sum of all forces) and inversely proportional to its mass:

net force = mass × acceleration, F = ma

Acceleration has the same direction as the net force. If the net force is zero, the velocity is constant.

The newton (N) is the net force that produces an acceleration of 1 m s⁻² in a 1 kg mass. So 1 N = 1 kg m s⁻².

Worked example 1. A van of mass 1200 kg has a driving force of 3100 N and resistive forces totalling 900 N.

Net force = 3100 − 900 = 2200 N a = F/m = 2200/1200 = 1.8 m s⁻² (2 significant figures), in the direction of the driving force.

Weight, W = mg (3.2.1(c))

Weight is the gravitational force on an object. Mass (kg) does not change with place. Weight (N) depends on the gravitational field strength g, in N kg⁻¹ (numerically the same as m s⁻²). Near Earth’s surface g = 9.81 N kg⁻¹. Weight acts vertically downwards.

Worked example 2. A rover has mass 180 kg. Take g = 9.81 N kg⁻¹ on Earth and 1.62 N kg⁻¹ on the Moon.

Earth: W = 180 × 9.81 = 1766 N (1.77 kN) Moon: W = 180 × 1.62 = 292 N The mass is 180 kg in both places.

The four contact forces (3.2.1(d))

Force	Meaning	Direction
Tension	Force in a stretched rope, string or cable	Along the rope, pulling on whatever it is attached to
Normal contact force	Force from a surface on an object pressing on it	Perpendicular to the surface
Upthrust	Upward force from a fluid on an object in it	Vertically upward
Friction	Force between surfaces in contact that opposes sliding, or the tendency to slide	Parallel to the surface, against the relative motion

Upthrust is covered in 3.2.4; here treat it as one force in the balance. “Normal” means at right angles.

Free-body diagrams (3.2.1(e))

A free-body diagram isolates one object, shown as a dot or box, with an arrow for every force acting on it.

1. Isolate the object. Do not draw the objects exerting the forces.
2. Draw each force from the object, as an arrow in the direction of the force, labelled with its name (weight, tension, normal contact force and so on).
3. Make arrow lengths roughly proportional to magnitudes. Equal and opposite forces get equal lengths.
4. Include only real forces. A “force of motion” is not a force on the object.
5. Show the net force separately, or state it, once you have resolved.

Situation	Forces on the object
Book at rest on a table	Weight down, normal contact force up (equal)
Lift accelerating upward	Weight down, tension up (tension larger)
Box sliding down a rough slope	Weight (vertically down), normal contact force (perpendicular to slope), friction (up the slope)
Ball bearing falling in oil	Weight down, upthrust up, drag up

Worked example 3. A lift of mass 640 kg hangs from a cable. Find the tension when it (i) accelerates upward at 1.2 m s⁻², (ii) moves at constant speed, (iii) decelerates at 1.2 m s⁻² while still moving upward.

Take upward as positive: T − mg = ma, so T = m(g + a). (i) T = 640 × (9.81 + 1.2) = 7046 N, about 7.0 kN (ii) a = 0: T = 640 × 9.81 = 6278 N, about 6.3 kN (iii) the acceleration is downward, a = −1.2: T = 640 × (9.81 − 1.2) = 5510 N, about 5.5 kN

Motion under constant force in two dimensions (3.2.1(f))

A constant net force gives a constant acceleration. Use F = ma with the equations of constant acceleration. In two dimensions, resolve into perpendicular components using F cos θ and F sin θ, and treat each direction separately. The components are independent: a force in one direction does not change the motion in the perpendicular direction.

Motion on a slope

For a slope at angle θ to the horizontal, resolve the weight:

• Component down the slope: mg sin θ
• Component perpendicular to the slope: mg cos θ, balanced by the normal contact force if the object stays on the surface

Worked example 4. A block of mass 12 kg slides down a slope at 25° to the horizontal. Friction of 18 N acts up the slope.

Down-slope component = 12 × 9.81 × sin 25° = 49.8 N Net force down the slope = 49.8 − 18 = 31.8 N a = 31.8/12 = 2.6 m s⁻² Normal contact force = 12 × 9.81 × cos 25° = 107 N

Projectiles

For a projectile with only weight acting (no drag), the net force is vertical. So the horizontal acceleration is zero (constant horizontal velocity) and the vertical acceleration is g.

Worked example 5. A stone is thrown horizontally at 12.0 m s⁻¹ from a cliff 35.0 m high. Ignore drag.

Vertical: s = ½gt², so t = √(2 × 35.0/9.81) = 2.67 s Horizontal distance = 12.0 × 2.67 = 32.1 m Vertical speed on landing = gt = 9.81 × 2.67 = 26.2 m s⁻¹ Speed = √(12.0² + 26.2²) = 28.8 m s⁻¹, at tan⁻¹(26.2/12.0) = 65.4° below the horizontal.

Drag and the factors affecting it (3.2.2(a), (b))

Drag is the frictional force on an object moving through a fluid (a liquid or a gas). It acts opposite to the object’s velocity relative to the fluid. Air resistance is drag in air.

For an object travelling through air, drag is greater when:

• speed is greater. At low speed drag rises roughly in proportion to speed, and at higher speed it rises faster than that.
• cross-sectional area is larger, because more air must be pushed aside.
• the shape is less streamlined. A pointed or smooth profile gives less drag than a flat face at the same area.
• the air is denser. Drag is lower at altitude than at sea level.

The specification asks for the factors, not a drag equation. Describe each as an increase or decrease, and say what is kept the same.

Falling with drag (3.2.2(c))

Take downward as positive for a falling object. Weight acts down, drag acts up:

net force = mg − D, so a = (mg − D)/m

Worked example 6. A ball of mass 0.25 kg falls and, at one instant, drag is 0.80 N.

Weight = 0.25 × 9.81 = 2.45 N a = (2.45 − 0.80)/0.25 = 6.6 m s⁻² downward (less than g).

The motion in a uniform gravitational field has three stages:

1. Release: speed is zero, so drag is (almost) zero and a = g.
2. Speeding up: as speed rises, drag rises, net force falls, and acceleration falls. The object still speeds up, but more slowly. The velocity–time graph is a curve with a decreasing gradient.
3. Terminal velocity: drag equals weight, net force is zero, a = 0.

Terminal velocity (3.2.2(d)(i))

Terminal velocity is the constant velocity reached by an object falling through a fluid when the resultant force on it is zero. In air, weight = drag. In a liquid, weight = drag + upthrust.

If conditions change, a new terminal velocity follows. If a parachute opens, drag suddenly exceeds weight. The net force is upward, so the parachutist decelerates. Drag then falls with the speed until it equals weight again, at a smaller terminal velocity.

Worked example 7. A steel ball bearing of mass 0.88 g falls through oil. The upthrust is 1.1 mN.

Weight = 0.88 × 10⁻³ × 9.81 = 8.6 mN At release (no drag): net force = 8.6 − 1.1 = 7.5 mN, so a = 7.5 × 10⁻³/0.88 × 10⁻³ = 8.6 m s⁻² At terminal velocity: drag = weight − upthrust = 8.6 − 1.1 = 7.5 mN

Determining terminal velocity (3.2.2(d)(ii))

OCR links this to practical activity group 1 (PAG1), with a ball bearing in a viscous liquid or cones in air as examples.

Ball bearing in a viscous liquid

1. Fill a tall transparent cylinder with glycerol or oil. Mark equal intervals (for example every 0.10 m) well below the surface.
2. Measure the ball’s diameter with a micrometer at several orientations, then drop it just under the surface so it enters centrally.
3. Time the fall through each interval with a stopwatch or, better, a video with a clock in the frame. Repeat and average.
4. Calculate speed = interval/time. Terminal velocity has been reached when the speed stops changing.

Worked example 8. Interval times for successive 0.10 m sections were 0.62, 0.45, 0.38, 0.36, 0.36 and 0.36 s.

Speeds: 0.16, 0.22, 0.26, 0.28, 0.28, 0.28 m s⁻¹ The speed is constant for the last three intervals, so the terminal velocity is 0.28 m s⁻¹. A timing uncertainty of ±0.02 s on 0.36 s is 5.6%, so the result is about 0.28 ± 0.02 m s⁻¹.

Cones in air

Use paper cake cases or paper cones. Drop each from a fixed height of at least 2 m, measure the time over the last stretch with light gates or video, and compute speed. Stacking cases changes the weight while keeping the shape, so you can investigate how terminal velocity depends on weight. Release from rest at the same height each time and avoid draughts.

Reducing error

• Reaction time is large compared with 0.36 s, so use video or light gates.
• A ball near the cylinder wall falls more slowly, so drop it along the axis.
• Temperature changes the viscosity of oil, so keep it steady.

Common errors

• Writing “weight” for mass, or giving weight in kg.
• Drawing the normal contact force vertical on a slope.
• Adding a “force of motion” arrow to a free-body diagram.
• Saying a terminal-velocity object has “no forces”. It has balanced forces.
• Forgetting upthrust in a liquid.
• Saying drag stops the motion at terminal velocity: it ends the acceleration, not the movement.

Next: use the revision notes to compress this, then the practice questions.

Official syllabus

OCR A Level Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 3.2.1 and 3.2.2.