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

Quick recall for OCR A Level Physics A (H556), sections 3.2.1 Dynamics and 3.2.2 Motion with non-uniform acceleration (specification Version 3.0, March 2026). For full explanations and worked examples, read the study guide. Then test yourself with the practice questions. Hub: OCR A Level Physics. Tick off topics on the printable checklist, and find weak spots with the free 10-minute diagnostics. Vector resolving is revised in Foundations of Physics.

Key equations and definitions

Item	Statement	Spec
Newton’s second law as used here	net force = mass × acceleration, F = ma	3.2.1(a)
The newton	Net force producing 1 m s⁻² in a 1 kg mass; 1 N = 1 kg m s⁻²	3.2.1(b)
Weight	W = mg, with g = 9.81 N kg⁻¹ near Earth’s surface	3.2.1(c)
Resolving a force	Fx = F cos θ, Fy = F sin θ (θ measured from x)	3.2.1(f)
Slope, weight components	Along slope mg sin θ; perpendicular mg cos θ	3.2.1(f)
Drag	Frictional force on an object moving through a fluid	3.2.2(a)
Terminal velocity	Constant velocity when the resultant force is zero	3.2.2(d)

Recall F = ma and W = mg from memory. The specification expects it.

The four named forces

Force	Direction	Typical mistake
Tension	Along the rope or cable, away from the object	Drawing it for a string that is slack
Normal contact force	Perpendicular to the surface	Drawing it vertical on a slope
Upthrust	Up, from the fluid	Leaving it out for a ball in a liquid
Friction	Parallel to the surface, against the relative motion	Drawing it in the direction of motion

Free-body diagram checklist

• One object only, as a dot or box.
• Arrows start on the object and show forces on it.
• Every force is named: weight, tension, normal contact force, upthrust, friction, drag, applied force.
• Equal and opposite forces are drawn equal in length.
• No arrows for velocity, acceleration or a “force of motion”.
• Resolve on a slope along and perpendicular to the surface.

Method: any constant-force problem

1. Draw the free-body diagram.
2. Choose a positive direction (or two perpendicular axes).
3. Resolve any force at an angle.
4. Write net force = ma in each direction.
5. Use constant-acceleration equations for speed, time or distance.
6. Check units and significant figures.

Dynamics reminders

• Tension in a lift cable moving up with acceleration a: T = m(g + a). With constant speed, T = mg. For deceleration while moving up, T = m(g − a).
• A block on a slope with friction F: net force down the slope = mg sin θ − F.
• Projectile (no drag): horizontal velocity constant, vertical acceleration g. Treat each direction separately.
• A force perpendicular to the motion changes the direction of travel, not the component of speed along the original direction.

Worked reminders

Horizontal pull with friction. A 5.0 kg box is pulled by a horizontal 20 N force against friction of 8.0 N. Net force = 20 − 8.0 = 12 N, so a = 12/5.0 = 2.4 m s⁻².

Smooth slope. A 3.0 kg trolley is on a frictionless slope at 30°. Only the down-slope component of weight gives a net force. a = g sin 30° = 9.81 × 0.50 = 4.9 m s⁻², and the mass cancels.

Projectile. A ball rolls off a table 0.80 m high at 8.0 m s⁻¹ (ignore drag). t = √(2 × 0.80/9.81) = 0.40 s, so the horizontal distance is 8.0 × 0.404 = 3.2 m.

Falling with drag. A 0.10 kg object falls with a drag of 0.40 N at that instant. a = (0.10 × 9.81 − 0.40)/0.10 = 5.8 m s⁻² downward.

Terminal velocity in liquid. A 2.0 g marble falls through oil with upthrust 3.0 mN. Weight = 2.0 × 10⁻³ × 9.81 = 19.6 mN, so drag at terminal velocity = 19.6 − 3.0 = 16.6 mN, about 17 mN.

A 4.5 kg trolley with net force 12 N has a = 12/4.5 = 2.7 m s⁻².

Drag

• Acts opposite to the velocity relative to the fluid.
• In air, drag increases with speed, cross-sectional area, less streamlined shape and air density.
• Always say which factor and in which direction. “More area, so more air pushed aside, so more drag” earns more than “air resistance”.

Falling with drag: three stages

Stage	Drag	Net force	Acceleration
Release	About zero	Weight	g
Speeding up	Rising	Falling	Falling (graph gets shallower)
Terminal velocity	Equals weight (air)	Zero	Zero (graph is horizontal)

In a liquid: weight = drag + upthrust at terminal velocity. If drag changes (parachute opens): drag > weight, so the object decelerates; drag falls with speed until it again equals weight at a lower terminal velocity.

Graphs for a fall with drag

• Velocity–time: starts at zero with gradient g, curves over with a decreasing gradient, then becomes horizontal at the terminal velocity.
• Acceleration–time: starts at g and decreases to zero.
• Distance–time: starts flat, steepens, then becomes a straight line whose gradient is the terminal velocity.

Template for “explain terminal velocity”

1. Weight acts down; drag (and upthrust, in a liquid) acts up.
2. Speed increases, so drag increases.
3. The net force decreases, so the acceleration decreases.
4. Drag (plus upthrust) equals weight, so the net force is zero and the velocity is constant.

Determining terminal velocity (PAG1 examples)

Technique	Key points
Ball bearing in viscous liquid	Tall cylinder; equal marked intervals well below the surface; micrometer for diameter; time each interval by video or light gates; speed constant means terminal velocity reached; drop along the axis; repeat and average
Cones in air	Fixed release height of at least 2 m; time the last stretch; stack cases to change weight; no draughts; repeat

Graph check: a distance–time graph that becomes a straight line shows constant speed. The gradient of that straight section is the terminal velocity.

Must-know distinctions

• Mass (kg, scalar) versus weight (N, vector).
• Net force versus any single force.
• Balanced forces (constant velocity, possibly moving) versus no forces.
• Terminal velocity in air (weight = drag) versus in liquid (weight = drag + upthrust).

Quick self-test

1. A 0.40 kg trolley accelerates at 3.5 m s⁻². What net force acts on it?
2. Find the weight of a 65 kg student, using g = 9.81 N kg⁻¹.
3. Find the mass of an object that weighs 49 N.
4. A force of 50 N acts at 40° above the horizontal. Find its horizontal and vertical components.
5. A lift of mass 640 kg moves upward at a steady speed. State the tension in the cable.
6. State the direction of the normal contact force on a block resting on a slope.
7. A parachutist of weight 710 N falls at terminal velocity. State the drag force.
8. A paper cone of mass 1.2 g falls at terminal velocity in air. Find the drag on it.
9. Explain why the acceleration of a falling ball falls as its speed rises.
10. State two precautions when timing a ball bearing falling in oil.

Answers

1. F = ma = 0.40 × 3.5 = 1.4 N.
2. W = 65 × 9.81 = 638 N.
3. m = W/g = 49/9.81 = 5.0 kg.
4. Horizontal 50 cos 40° = 38 N; vertical 50 sin 40° = 32 N.
5. Constant speed means zero net force, so T = mg = 640 × 9.81 = 6.3 kN.
6. Perpendicular to the slope, away from the surface.
7. 710 N, upward, because the net force is zero.
8. W = 1.2 × 10⁻³ × 9.81 = 0.012 N, so drag is 0.012 N.
9. Drag increases with speed, so net force (weight minus drag) falls, so acceleration falls.
10. Any two: drop along the cylinder axis; wait until the ball is moving at constant speed before timing; use video or light gates instead of a stopwatch; keep the temperature steady; repeat and average.

Where marks are usually lost

• Quoting weight in kg, or mass in N.
• Writing “gravity” for weight, or “friction” for drag in a liquid, when the question wants the named force.
• Forgetting upthrust in a liquid.
• Putting the normal contact force vertical on a slope.
• Using g for the acceleration of a falling object when drag is acting.
• Saying “the forces cancel” for an accelerating object.
• Not stating the direction when a quantity is a vector.
• Giving a terminal velocity from a single timed run with no check that the speed had stopped changing.
• Giving terminal velocity as the point where drag is zero.
• Rounding in the middle of a calculation, then losing a significant figure.
• Describing an experiment without saying how to know that terminal velocity has been reached.

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.