July 15, 20269 min read

How to Prepare for Physics 12 in British Columbia

Physics 12 has a reputation for being the course where hard-working students suddenly stop getting the marks they are used to. That reputation is earned, but the reason is usually misunderstood: it is almost never the mathematics.

The algebra in Physics 12 is easier than the algebra in Pre-Calculus 12. What changes is that nobody tells you which equation to use. A question describes a situation — a car on a curve, a ball thrown at an angle, a charge in a field — and the entire difficulty is deciding what is going on before you write anything down.

This guide covers what the course actually demands, the one method that reliably works, and what to review before you start. If you would rather work through it with someone, our Physics 12 tutoring is built around exactly this problem.

What Physics 12 covers in BC

The course builds on Physics 11 and assumes it. The units are broadly:

Vector kinematics — motion in two dimensions, including projectiles
Dynamics — forces, Newton's laws, friction, inclines
Circular motion and gravitation
Momentum and energy conservation
Electrostatics, electric fields and circuits
Magnetism and electromagnetic induction

Notice what is missing from that list: no new mathematics. Everything is Pre-Calculus 11 algebra and basic trigonometry. The difficulty lives entirely in the modelling.

The method that actually works

There is one procedure worth internalising, and it applies to nearly every problem in the course:

Draw the situation. Not a sketch of the object — a diagram of the forces or vectors acting on it.
Choose and label your axes and sign conventions, and commit to them.
Name the physical principle. Newton's second law? Conservation of energy? Not "which formula" — which principle.
Only now write an equation, and solve it.

Students who skip to step 4 are the ones who lose marks, and they lose them consistently. The algebra then executes flawlessly on a wrong model, producing a confident wrong answer that is very hard to spot afterwards.

The picture that fixes circular motion

Before the worked example, one diagram — because circular motion is where Physics 12 loses the most students, and it is almost always the same misconception.

Looking down on the curve. Only one horizontal force acts on the car, and it points inward. Nothing pushes it outward — the crossed-out arrow is the force students draw and the marker deletes. The car does not need a force to keep going forward, only one to stop going straight.

Ask a class what forces act on a car going round a bend and most will draw an arrow pointing outward. It feels undeniable: you are thrown against the door, so something must be pushing you out.

Nothing is. There is no outward force, and drawing one is the fastest way to lose the question.

What you feel is the door pushing you inward. Your body was travelling in a straight line, as bodies do, and the car turned into you. The push you feel on your shoulder is the car making you turn — it points toward the centre, not away from it. The sensation of being flung outward is your straight-line motion continuing while the car deviates from it.

So the only horizontal force on that car is friction, and it points inward. That is worth stating plainly: $F_{c} = m v^{2} / r$ is not a new force to add to the diagram. It is what the net inward force has to equal. Students who write centripetal force as a separate arrow alongside friction are counting the same force twice.

It also explains why the car needs any grip at all. It is not moving forward that requires force — Newton's first law gives you that for nothing. It is $not going straight$ that costs. On this curve the car needs $a = v^{2} / r = 4.5$ m/s² of inward acceleration, and friction is the only thing available to supply it. Take the grip away — ice, gravel, standing water — and the car does not fly outward. It goes straight on, which from inside feels the same and is not.

Worked example: a car on a flat curve

A car rounds a flat (unbanked) curve of radius $r = 50 m$ at $v = 15 m/s$ . What is the minimum coefficient of static friction needed to keep it on the road?

Step 1–3: model it

Draw the car from behind. Vertically: gravity down, normal force up — these balance, so there is no vertical acceleration. Horizontally: the only force pointing toward the centre of the curve is friction. That is the whole insight.

The principle is Newton's second law applied to circular motion: the net centre-pointing force provides the centripetal force.

Step 4: now the algebra

F_{c} = \frac{m v ^{2}}{r} f_{s, m a x} = μ N = μ m g

Setting friction equal to the required centripetal force:

μ m g = \frac{m v ^{2}}{r} ⟹ μ = \frac{v ^{2}}{r g}

μ = \frac{( 15 m/s ) ^{2}}{( 50 m ) ( 9.8 m/ s ^{2} )} = \frac{225}{490} = 0.46

The part worth noticing

The mass cancelled. A loaded truck and a small hatchback need the same coefficient of friction to take that curve at that speed.

That result is genuinely surprising, and it is the kind of thing the exam likes to ask about conceptually. If you had been given $m = 1200 kg$ you would have computed $F_{c} = 5400 N$ and still divided it away. A student who models the problem sees the cancellation coming; a student hunting for a formula never notices.

Worked example: projectile range

A ball is launched at $v_{0} = 20 m/s$ at $3 0^{\circ}$ above the horizontal, over level ground. How far does it land?

The modelling step is the same insight every time: horizontal and vertical motion are independent. Horizontally there is no acceleration; vertically there is gravity. They share only the time.

v_{x} = v_{0} cos θ v_{y} = v_{0} sin θ

Time in the air is set by the vertical motion; range is set by the horizontal. Combining them gives the standard result:

R = \frac{v _{0}^{2} sin ( 2 θ )}{g} = \frac{( 20 ) ^{2} sin 6 0 ^{\circ}}{9.8} = 35.4 m

Worth checking against the long route: the ball is in the air about 2.04 s, travelling horizontally at 17.3 m/s, which gives 35.4 m. Same answer — a good habit, because a formula you cannot derive is a formula you will misapply.

Collisions: when to stop using forces

Kinematics and dynamics can solve almost anything in the course, and for collisions they are the wrong tool entirely. The forces during a crash are enormous, vary wildly, and last a few milliseconds. Nobody can write them down.

Conservation laws are the way out. They let you compare before and after while ignoring the middle completely — which is the whole point, because the middle is unknowable.

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Worked example

A 1200 kg car doing 20 m/s runs into a stationary 800 kg car. They lock together. How fast does the wreckage move?

(1200) (20) + (800) (0) = (2000) v

v = \frac{24 , 000}{2000} = 12 m/s

No forces, no time interval, no details of the crumpling. Two lines.

The part that is examined

Now check the kinetic energy, because this is the trap.

E_{k} = \frac{1}{2} (1200) (20)^{2} = 240, 000 J

E_{k}^{'} = \frac{1}{2} (2000) (12)^{2} = 144, 000 J

96,000 joules — 40% of it — have vanished. Momentum was conserved exactly; kinetic energy was not, and it is not close.

That is not an error. Momentum is always conserved in a collision. Kinetic energy is only conserved in an elastic one, and a crash where the cars stick together is as inelastic as it gets. The missing 96,000 J went into deforming metal, heat, and noise — and that is precisely what a crumple zone is designed to do. A car that bounced off elastically would conserve kinetic energy and kill you.

Momentum: conserved in every collision. Always. It is a vector, so direction and sign matter.
Kinetic energy: conserved only if the question says elastic. If they stick together, expect to lose a large fraction of it.

The exam question here is nearly always conceptual: "is kinetic energy conserved?" A student who assumes both are conserved will get a plausible number and no marks.

Where marks are actually lost

Treating vectors as numbers — adding magnitudes instead of components
Assuming $N = m g$ on an incline. It is not — on a slope, $N = m g cos θ$ , and this single error propagates through the whole question.
Changing sign convention halfway through, usually without noticing
Using constant-acceleration kinematics when the acceleration is not constant
Skipping the free-body diagram to save two minutes, then losing ten

Practice problems

1. A car takes a flat curve of radius $80 m$ at $20 m/s$ . What minimum coefficient of static friction is required?

2. A projectile is launched at $25 m/s$ at $4 0^{\circ}$ over level ground. Find its range.

3. A block slides down a frictionless incline at $3 0^{\circ}$ . Find its acceleration.

Solutions

1. Flat curve

μ = \frac{v ^{2}}{r g} = \frac{( 20 ) ^{2}}{( 80 ) ( 9.8 )} = \frac{400}{784} = 0.51

Again the mass never appears. If you found yourself needing it, the model was wrong rather than the arithmetic.

2. Projectile range

R = \frac{v _{0}^{2} sin ( 2 θ )}{g} = \frac{( 25 ) ^{2} sin 8 0 ^{\circ}}{9.8} = 62.8 m

3. Frictionless incline

Resolve gravity along the slope. The component down the incline is the only unbalanced force:

ma = m g sin θ ⟹ a = g sin θ = 9.8 sin 3 0^{\circ} = 4.9 m/ s^{2}

The mass cancels once more — and note that $N = m g cos θ$ , not $m g$ . That is the error from the list above, met in the wild.

What to review before the course starts

Physics 12 does not revisit these, so a fortnight spent here is worth more than a fortnight spent reading ahead:

Resolving vectors into components, and recombining them
Right-angle trigonometry until it is automatic — sin, cos and tan of a triangle you drew yourself
Rearranging equations before substituting numbers, not after
Unit conversion and significant figures, which quietly cost marks all year

Why Physics 12 matters beyond the grade

Physics 12 is a prerequisite for engineering and most physical-science programs at UBC and SFU, and it is assumed knowledge in first-year physics — those courses do not reteach it. The modelling habit you build here is the thing first year tests, only faster and with calculus attached. The same is true of university-level physics, which starts where this course stops.

The encouraging part: none of the failure modes above are about ability. They are about a step being skipped under time pressure. Skipped steps are fixable.

Getting help with Physics 12

If the pattern here is familiar — understanding the lesson but losing marks on tests — that is almost always a modelling problem rather than a physics problem, and it is one of the fastest things to fix with one-on-one Physics 12 help, because a tutor watches the step where the model goes wrong.

Sessions run in person in Burnaby or online across Metro Vancouver, aligned to the BC curriculum. Book a free 30-minute consultation and we will look at where the marks are going.

Need one-on-one help with this? Our tutors can guide you step by step.

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