Module 1: Mathematical Foundations · Practice · complements Foundations 2, 3, 4 and 7
Seven examples on the one question this module keeps returning to: what belongs to the physics, and what belongs to the description.
What this set is for
The first worked-examples post took one problem from each Foundation and covered the standard techniques. This one is narrower and goes deeper into a single idea: that a physical quantity and its numerical description are different things, and that most confusion in mechanics comes from treating them as the same.
Same three-part structure. The Physics before any symbol. The Mathematics in full. The Contemplation on what the working revealed. These examples are more conceptual than the first set, but none of them is answerable without doing the calculation.
On this page
1. One force, three sets of components Foundation 2
Problem. A force of magnitude 10 N acts at 30° above the positive x-axis. Write its components in the given axes; in axes rotated 30° anticlockwise; and in axes rotated 90° anticlockwise. In each case compute the magnitude from the components. What, if anything, has happened to the force?
The physics
Nothing is being done to the force. No one is turning it, adding to it, or moving the object it acts on — the only thing changing between the three parts is where we have chosen to draw our axes, which is a decision made on paper. So whatever the numbers do, the force must come through unchanged, and there should be some quantity computed from the components that refuses to move.
The mathematics
Original axes. Resolving at 30°:
Fx = 10 cos 30° = 5√3 ≈ 8.66 N Fy = 10 sin 30° = 5 N
|F| = √(75 + 25) = √100 = 10 N
Axes rotated by 30°. Use the rotation formulas from Foundation 3, with θ = 30°:
F′x = 8.66 cos 30° + 5 sin 30° = 7.5 + 2.5 = 10 N
F′y = −8.66 sin 30° + 5 cos 30° = −4.33 + 4.33 = 0
|F| = √(100 + 0) = 10 N
The force now lies entirely along one axis, and one component has vanished. That is not a simplification of the physics — it is a convenient choice of paper.
Axes rotated by 90°.
F′x = 8.66(0) + 5(1) = 5 N F′y = −8.66(1) + 5(0) = −8.66 N
|F| = √(25 + 75) = 10 N
Three descriptions — (8.66, 5), (10, 0) and (5, −8.66) — and one magnitude throughout.
The contemplation
In the second set of axes one component was exactly zero. A student meeting only that line might conclude the force has no vertical part — but “vertical” was never a property of the force, only of the page it was drawn on. This is why a vector is defined as an arrow rather than as a pair of numbers: the arrow was the same object in all three calculations, and the pairs were three ways of naming it. The magnitude survived because it is a property of the arrow. Anything that survives every rotation is a candidate for being physical; anything that does not is a feature of your description.
2. A constant vector with changing components Foundation 3
Problem. A constant vector A = Aî points along the positive x-axis and never changes. Express it in the polar basis (r̂, θ̂) at an arbitrary angle θ, and verify that its magnitude is independent of θ.
The physics
Example 1 rotated the axes once and got new numbers. Here something stranger happens: the vector is fixed and the axes are fixed too, but the polar basis is defined locally, so it points differently at every position in the plane. The same unchanging vector will therefore have different polar components depending on where in the plane we ask. Nothing about A varies with θ. Only the ruler does.
The mathematics
We know the polar unit vectors in Cartesian terms:
r̂ = cos θ î + sin θ ĵ
θ̂ = −sin θ î + cos θ ĵ
We need î the other way round. Multiply the first by cos θ, the second by −sin θ, and add — the ĵ terms cancel:
cos θ r̂ − sin θ θ̂ = (cos²θ + sin²θ) î = î
So the constant vector, written in the local polar basis, is
A = A cos θ r̂ − A sin θ θ̂
so Ar = A cos θ and Aθ = −A sin θ
Both components depend on θ. Yet:
|A| = √(A²cos²θ + A²sin²θ) = A√(cos²θ + sin²θ) = A
The contemplation
Here is a component that varies with position while the quantity it describes is constant. If you were handed Aθ = −A sin θ with no context, you would reasonably assume something was changing. Nothing was. This is worth carrying into Foundation 8, where differentiating in polar coordinates produces terms that look like new physics and are partly the basis turning underneath. The identity sin² + cos² = 1 is what rescued the magnitude — and it is doing that job everywhere in physics, quietly, whenever something has to survive a change of description.
3. The force that has only one component Foundation 3
Problem. A particle at position (x, y) is attracted toward the origin by a central force of magnitude k/r², where r is its distance from the origin. Write the force in Cartesian components, then in polar components, and compare the two descriptions.
The physics
The force points at the origin from wherever the particle happens to be, and its size depends only on how far away that is. The situation has a centre and no preferred direction around it. Cartesian axes impose two preferred directions that the physics does not have, so expect them to fit badly. Polar coordinates share the symmetry, so expect them to fit well — and specifically, expect the transverse component to vanish, because there is nothing in the problem that could make a force point sideways.
The mathematics
Cartesian. The force has magnitude k/r² directed along −r̂, and r̂ has components (x/r, y/r). So:
Fx = −kx / (x² + y²)3/2
Fy = −ky / (x² + y²)3/2
Two components, each a function of both variables, each with a three-halves power in the denominator.
Polar. The same force, described in a basis that knows where the centre is:
Fr = −k/r² Fθ = 0
F = −(k/r²) r̂
One component, one variable, no fractional powers. The two descriptions are of the same force and are equally correct.
The contemplation
The Cartesian version is not harder because the physics is harder there. It is harder because the description carries information the situation does not — two special directions, invented by us, which then have to be tracked through every subsequent line of algebra. The polar version has one component because the problem genuinely has one degree of freedom in the force. When a calculation feels laborious, it is worth asking whether the difficulty belongs to the physics or to the coordinates, because the second kind can be removed and the first cannot.
4. Choosing the frame before solving Foundation 3
Problem. A block of mass m slides down a frictionless plane inclined at φ. Set the problem up twice — once with horizontal and vertical axes, once with axes along and perpendicular to the slope — and compare the work each requires.
The physics
Two forces act: weight straight down, and the normal force perpendicular to the surface. The block can only move along the slope — that is the constraint, and it is the most important fact in the problem. A set of axes that knows about the constraint should do less work than one that does not.
The mathematics
Horizontal and vertical axes. The weight is (0, −mg), which is easy. The normal force must be resolved, since it points perpendicular to the slope: (N sin φ, N cos φ). Newton’s second law then gives two coupled equations:
N sin φ = max
N cos φ − mg = may
Both ax and ay are unknown, and so is N. Three unknowns, two equations — the third relation must come from the constraint that the block stays on the slope, which has to be imposed separately as ay/ax = −tan φ.
Axes along and across the slope. Now the weight must be resolved instead: mg sin φ down the slope and mg cos φ into it. The normal force needs no resolving at all — it lies entirely along one axis.
Along the slope: mg sin φ = ma, so a = g sin φ
Across the slope: N − mg cos φ = 0, so N = mg cos φ
Two independent equations, each with one unknown, and the answer in a single line. The constraint did not need imposing separately because the axes were built to respect it — motion across the slope is zero by construction.
The contemplation
Both routes reach a = g sin φ. The block does not know which one you took, and would slide identically if you had drawn no axes at all. What differed was how much of your effort went into the physics and how much into undoing a choice you made at the start. The tilted axes were not a clever trick; they were the ordinary consequence of noticing the constraint before setting up equations rather than after. Most of what looks like cleverness in mechanics is this — a decision made early, in the reading of the problem rather than the solving of it.
5. Two motions that never speak to each other Foundation 4
Problem. A ball is launched at 20 m/s at 30° above the horizontal from ground level. A second ball is simply dropped from the highest point the first one reaches, at the instant the first one arrives there. Take g = 10 m/s². Which lands first?
The physics
At the top of its flight the projectile is still moving horizontally, but its vertical velocity is zero — the same vertical state as a ball released from rest. If the horizontal and vertical motions genuinely have nothing to say to each other, the two balls should fall in exactly the same way, and the horizontal motion of the first should be irrelevant to when it lands. That prediction is worth making before calculating, because it is the thing being tested.
The mathematics
Resolve the launch.
ux = 20 cos 30° = 10√3 ≈ 17.3 m/s uy = 20 sin 30° = 10 m/s
Time to the top, from the vertical motion alone:
0 = 10 − 10t, so t = 1 s
Height reached: h = 10(1) − ½(10)(1)² = 5 m
Ball 1, falling from the top. Vertical velocity zero, height 5 m:
5 = ½(10)t², so t = 1 s
Ball 2, dropped from 5 m. Vertical velocity zero, height 5 m — the identical equation, and therefore the identical answer, t = 1 s.
They land together. Ball 1 arrives 17.3 m away and ball 2 arrives directly below the release point, but the vertical calculation never referred to that horizontal motion, and the horizontal motion never referred to gravity.
The contemplation
Two balls in visibly different states of motion — one travelling at 17 m/s, one hanging still — fall in perfect step. The result feels like it should be a coincidence and is not: the vertical equation contains no x and the horizontal equation contains no y, so neither can influence the other. Notice what this rests on. Gravity supplies a constant vertical acceleration regardless of horizontal speed, and Newton’s law is linear, so the two components separate cleanly into two one-dimensional problems. It is easy to hear this as a fact about forces acting independently in nature. It is a fact about the equations.
6. When the independence breaks Foundation 4
Problem. Repeat the previous comparison, but now include air resistance, modelled as a drag force of magnitude bv directed opposite to the velocity, where v is the total speed. Write the two component equations and decide whether the two balls still land together.
The physics
Drag opposes the direction of travel, and the direction of travel depends on both components of the velocity at once. So the drag on a ball moving fast horizontally is not the same as the drag on a ball hanging still — even if their vertical speeds match. The two motions now have something to say to each other. We do not need to solve the resulting equations; we only need to see whether they can be separated.
The mathematics
The drag force is −bv, so its components are −bvx and −bvy. Newton’s second law gives:
m dvx/dt = −bvx
m dvy/dt = −mg − bvy
For this drag law the two equations still contain only their own variable, so they do separate — the linear model is the exception, which is exactly why textbooks use it. But now take the more realistic quadratic drag, magnitude bv² opposing the motion. Its components are −bv vx and −bv vy, where v = √(vx² + vy²):
m dvx/dt = −b√(vx² + vy²) vx
m dvy/dt = −mg − b√(vx² + vy²) vy
The vertical equation now contains vx. The horizontal motion is affecting the vertical one, the two cannot be solved apart, and the ball still moving horizontally experiences more drag than the one hanging still. They do not land together.
The contemplation
The previous example’s tidy result did not survive contact with a slightly more honest force law. That is worth sitting with, because the independence of perpendicular components is taught as though it were a principle, and every JEE projectile question relies on it. It is not a principle — it is a consequence of linearity, and it holds because the syllabus works in a regime where drag is neglected. The formula did not become wrong; the situation moved outside the range where it applied. This is Foundation 10’s claim, met early: every result carries conditions, and the ones you are never told about are the dangerous ones.
7. Two parabolas that are not the same curve Foundation 7
Problem. For the projectile of example 5 — launched at 20 m/s at 30°, with g = 10 m/s² — derive both y(t), the height against time, and y(x), the height against horizontal distance. Both are parabolas. Show that they are different curves, and say what each one is a picture of.
The physics
One flight, two things you might plot. The first is what a camera on the ground would record if it traced the ball through the air: a shape in space, which you could walk along. The second is a record of where the ball was at each moment, which exists nowhere — nothing ever travelled along it. Under uniform gravity both happen to come out parabolic, which is precisely why students merge them.
The mathematics
With ux = 10√3 and uy = 10 from example 5:
y(t) = 10t − 5t²
For the path, eliminate t using x = 10√3 t, so t = x/(10√3):
y = 10[x/(10√3)] − 5[x/(10√3)]²
y(x) = x/√3 − x²/60
Now compare them. Both reach the same maximum height of 5 m, as they must — it is the same flight. But y(t) peaks at t = 1 and returns to zero at t = 2, so its width is 2 seconds. And y(x) peaks at x = 10√3 ≈ 17.3 and returns to zero at x = 20√3 ≈ 34.6, so its width is 34.6 metres. Different coefficients, different widths, different units on the horizontal axis — two distinct curves.
And a check that settles it: the gradient of y(t) at launch is 10, with units of m/s — that is the initial vertical velocity. The gradient of y(x) at launch is 1/√3 ≈ 0.577, which is dimensionless — that is tan 30°, the launch angle. Two gradients, two entirely different physical meanings.
The contemplation
The two curves look alike and mean entirely different things, and the only thing distinguishing them is the label on the horizontal axis. A student who reads the shape and skips the axes has, in effect, read a sentence without noticing which language it was in. Note where the check came from: the units of the gradients, m/s in one case and nothing at all in the other. Units settled a question about meaning, just as they settled the arithmetic in the first example of the previous post. They are the most under-used diagnostic tool a physics student has.
What these seven had in common
In every one of them, something changed and something did not. Components changed while the force did not. Polar components varied with position while the vector was constant. Two coordinate choices gave the same acceleration down the slope. Two balls with different horizontal motion fell in step — until a more honest force law removed the reason they had to.
Learning to tell those two apart is most of what this module was for. The quantities that survive a change of description are the physics. The rest is bookkeeping — necessary, chosen by you, and worth choosing well.
Complements: Foundations 2, 3, 4 and 7 · Type: Worked examples · Reading: 25 min · Best used: after the first worked-examples set