PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: units, measurement and vectors (PHYS 101)

This sheet is not a summary of the first chapter of PHYS 101: you already have the course. It answers one question, what loses marks on units, measurements and vectors in an assessment, and which precise gesture prevents each loss.

Everything here is done with algebra and right triangle trigonometry, which is the level at which PHYS 101 is taught at McGill University. A solution found online that begins by differentiating is not a PHYS 101 solution, even when its final number is correct.

It also covers the two vector lines of the PHYS 101 formula sheet, the dot product and the law of cosines, with the trap hidden in the words interior angle, and the check PHYS 101 expects on every formula: its dimensions first, then a limiting case.

Updated on October 24, 2026

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A physical quantity is never a bare number: it carries a unit, a precision and, when it is a vector, a direction. Every mark lost on this chapter comes from dropping one of the three, and the direction is the expensive one, because magnitudes of $3.0$ and $4.0$ add to anything between $1.0$ and $7.0$ .

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

SI base units
Dimensional analysis
Unit conversion factor
Significant figures
Order of magnitude
Scalar and vector quantities
Components of a vector
Vector addition by components
Resultant magnitude and direction
Unit vector
Dot product
Law of cosines for vectors

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• Dot product: $\vec{a}\cdot\vec{b}=ab\cos\theta$ , and by components $\vec{a}\cdot\vec{b}=a_{x}b_{x}+a_{y}b_{y}$ . The result is a SCALAR, in the product of the two units: m $^{2}$ for two displacements.
• Its sign reads the angle before any cosine: positive, acute; zero, perpendicular; negative, obtuse.
• Angle between two vectors: $\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{ab}$ . The arccosine returns $0^{\circ}$ to $180^{\circ}$ , exactly the range of an angle between two vectors, so no quadrant needs repairing.
• Resultant without components: $|\vec{a}+\vec{b}|^{2}=a^{2}+b^{2}+2ab\cos\theta$ . The term $2ab\cos\theta$ is twice the dot product.
• The sheet's $\theta$ , the interior angle of the two VECTORS, is the angle between them placed TAIL TO TAIL. In the head to tail triangle the angle at the joint is $180^{\circ}-\theta$ , and the geometry form $c^{2}=a^{2}+b^{2}-2ab\cos C$ takes that one, hence its minus sign.

Same

\vec{a}

, three directions for

\vec{b}

: the sign of

\vec{a}\cdot\vec{b}=ab\cos\theta

reads the angle at once, positive when acute, zero at the right angle, negative when obtuse.

Work, later in the course, is the same product: $W=\vec{F}\cdot\vec{d}=Fd\cos\theta$ . A force at right angles to the motion does no work because its dot product with the displacement is zero.

Units and digits: what a measurement is allowed to claim

• SI base units of mechanics: the metre, the kilogram and the second. Every other unit is a product of their powers, and reading it out loud tells you which quantities were divided.
• A conversion multiplies by a fraction worth exactly $1$ , raised to the SAME power as the unit: $1$ m $^{3}=(10^{2}\ \text{cm})^{3}=10^{6}$ cm $^{3}$ , and $1$ mL $=1$ cm $^{3}$ .
• Significant figures: a product or a quotient keeps the SMALLEST count of significant figures, a sum or a difference keeps the smallest number of DECIMAL PLACES. Round once, at the end.
• Order of magnitude: the nearest power of ten, with the threshold at $\sqrt{10}\approx 3.16$ and not at $5$ . So $8\times 10^{-6}$ m is of order $10^{-5}$ m, while $1.7$ m is of order $10^{0}$ m.
• A Fermi estimate rounds every input to one digit and chains the conversion factors. Only the power of ten in the answer is claimed, never the digits in front of it.

Dimensional analysis is the free check: if the two sides of an equation do not carry the same units, no amount of algebra will save the line, and a marker sees it in one second.

A vector, and the only legal way to add two of them

• Components, with $\theta$ measured counterclockwise from the positive $x$ axis: $V_{x}=V\cos\theta$ and $V_{y}=V\sin\theta$ . Written that way, the formulas carry the correct signs in all four quadrants on their own.
• Back the other way: $V=\sqrt{V_{x}^{2}+V_{y}^{2}}$ , which is never negative, and the direction comes from the acute angle $\arctan\dfrac{|V_{y}|}{|V_{x}|}$ PLACED by the signs of the components.
• Addition: $\vec{R}=\vec{A}+\vec{B}$ means $R_{x}=A_{x}+B_{x}$ and $R_{y}=A_{y}+B_{y}$ , two separate columns that never mix. Subtraction is $\vec{A}+(-\vec{B})$ .
• Unit vector: $\hat{u}=\dfrac{\vec{V}}{V}$ , length exactly $1$ , direction only. Multiplying a vector by a positive scalar changes the length and nothing else.
• Bounds on any resultant: $|A-B|\le R\le A+B$ , with equality only when the two vectors are parallel. When only the angle $\theta$ BETWEEN them is known, $R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}$ .

The side ALONG the axis from which

\theta

is measured is the adjacent one, so it takes the cosine. Reading the triangle beats memorising which function goes with which letter.

No calculus anywhere in PHYS 101: everything above is algebra and right triangle trigonometry, which is exactly why a solution found online that starts by differentiating is useless to you.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Adding two magnitudes as if they were two numbers

the whole question, plus every line built on the resultant afterwards

What not to write

“The two displacements are $3.0$ m and $4.0$ m, so the resultant is $3.0+4.0=7.0$ m.”

What to write

“The two displacements are perpendicular, so $R=\sqrt{3.0^{2}+4.0^{2}}=5.0$ m.”

Same two arrows,

3.0

and

4.0

, three different angles: the resultant slides from

7.0

down to

2.1

. Only the left picture gives the sum of the magnitudes.

Why: The magnitudes only add when the two vectors point the same way. At $90^{\circ}$ the resultant is $5.0$ , at $150^{\circ}$ it is $2.1$ , and at $180^{\circ}$ it is $1.0$ : the same two numbers give three different answers because the angle decides how much of each vector points along the other.

2. Using a conversion factor once when the unit carries an exponent

the whole question: the answer is out by a factor of

10^{4}

, which no rounding can excuse

What not to write

“ $1$ m $=10^{2}$ cm, so $2.0$ m $^{3}=2.0\times 10^{2}$ cm $^{3}$ .”

What to write

“ $1$ m $^{3}=(10^{2}\ \text{cm})^{3}=10^{6}$ cm $^{3}$ , so $2.0$ m $^{3}=2.0\times 10^{6}$ cm $^{3}$ .”

Why: The exponent belongs to the unit, so it belongs to the factor as well. Put the conversion inside the brackets, $(10^{2})^{3}$ , and only then compute. An area takes the factor squared, a volume takes it cubed, and a density, which hides a volume in its denominator, takes it cubed too.

3. Trusting the angle the calculator returns for an arctangent

the final part of the question, and it happens on almost every paper

What not to write

“ $\tan\theta=\dfrac{13.8}{-29.6}=-0.464$ , so $\theta=-24.9^{\circ}$ .”

What to write

“The acute angle is $\arctan\dfrac{13.8}{29.6}=24.9^{\circ}$ ; $R_{x}<0$ and $R_{y}>0$ place the vector in the second quadrant, so $\theta=180^{\circ}-24.9^{\circ}=155.1^{\circ}$ .”

Two opposite vectors on one line: the ratio of the components is the same for both, so the calculator answers

-53^{\circ}

twice. Only the signs separate them.

Why: The arctangent only returns values between $-90^{\circ}$ and $+90^{\circ}$ , so it cannot tell the second quadrant from the fourth: $(-3\,;\,4)$ and $(3\,;\,-4)$ give the same ratio and therefore the same display. The quadrant is information that lives in the SIGNS of the components, and you have already written them down.

4. Copying the calculator display instead of the digits the data earned

half a mark per line, and the credibility of a whole laboratory report

What not to write

“ $\rho=\dfrac{24.63\ \text{g}}{13.5\ \text{cm}^{3}}=1.824444$ g/cm $^{3}$ .”

What to write

“ $\rho=1.82$ g/cm $^{3}$ by the counting rule, then $\rho=1.8\pm 0.1$ g/cm $^{3}$ once the $7.4$ % uncertainty is computed.”

Why: A quotient keeps the smallest count of significant figures among its terms. Writing six digits claims a precision of one part in a million from a cylinder readable to half a millilitre, which is a factor of ten thousand of invented information. And once an uncertainty has been computed it outranks the counting rule: $\pm 0.14$ g/cm $^{3}$ makes the tenths digit the estimated one, so the result stops there.

5. Taking the cosine for the horizontal component whatever the angle is measured from

both components, so the whole question, even though every later step is carried out correctly

What not to write

“The vector makes $20^{\circ}$ with the VERTICAL, so $V_{x}=V\cos 20^{\circ}$ .”

What to write

“The angle with the vertical is $20^{\circ}$ , so the angle with the $x$ axis is $70^{\circ}$ and $V_{x}=V\cos 70^{\circ}=V\sin 20^{\circ}$ .”

Here the angle

\varphi

is measured from the VERTICAL: the side along the vertical is the adjacent one, so the vertical component takes the cosine and the horizontal one the sine. Read the triangle, not a habit.

Why: Cosine goes with the side ADJACENT to the angle, not with the horizontal. When the statement measures its angle from the vertical, or from an inclined surface, the two functions swap. The cure is not a rule to memorise but a sketch: draw the right triangle and see which side touches the angle.

6. Leaving the calculator in radian mode

every trigonometric line on the paper, so typically two or three questions at once

What not to write

“ $A_{x}=8.0\cos 35=-7.2$ , so the vector points to the left.”

What to write

“In degree mode, $A_{x}=8.0\cos 35^{\circ}=6.6$ , positive, as the drawing requires.”

Why: In radian mode $\cos 35$ means $35$ radians, about five and a half turns, and the sign comes out wherever that lands. The symptom is unmistakable: components whose signs contradict your own sketch. Check the mode once at the start of every assessment, and sketch before computing so the contradiction is visible.

7. Reading the order of magnitude off the exponent of the scientific notation

the estimate question, usually two marks, and it is entirely avoidable

What not to write

“A red blood cell is $8$ µm $=8\times 10^{-6}$ m, so its order of magnitude is $10^{-6}$ m.”

What to write

“ $8>\sqrt{10}\approx 3.16$ , so the nearest power of ten is $10^{-5}$ m.”

Why: The order of magnitude is the CLOSEST power of ten, and closeness on a logarithmic scale is decided at $\sqrt{10}$ , not at $5$ . Above $3.16$ you round up. Note that $1.7$ m does stay at $10^{0}$ m, so the rule is not one of those that always pushes upwards.

8. Giving a magnitude a minus sign to say which way the vector points

one mark on the spot, and it signals the confusion that loses the rest

What not to write

“ $\vec{B}$ points along the negative $x$ axis, so $|\vec{B}|=-5.0$ .”

What to write

“ $|\vec{B}|=5.0$ , and the direction is carried by the components $(-5.0\,;\,0)$ .”

Why: A magnitude is a length, obtained as a square root of a sum of squares, so it cannot be negative and it vanishes only for the zero vector. Direction lives in the signs of the components, or in an angle, never in the size. A negative magnitude on a paper tells the marker that vector and component have been merged into one idea.

9. Putting the triangle's angle into the law of cosines of the formula sheet

the whole question, about 200 m off, when a component check would have caught it in one line

What not to write

“The legs are $500$ m and $650$ m and the triangle has $104.3^{\circ}$ at the waypoint, so $R^{2}=500^{2}+650^{2}+2(500)(650)\cos 104.3^{\circ}$ and $R=716$ m.”

What to write

“The angle between the legs placed tail to tail is $180^{\circ}-104.3^{\circ}=75.7^{\circ}$ , so $R^{2}=500^{2}+650^{2}+2(500)(650)\cos 75.7^{\circ}$ and $R=912$ m.”

The sheet's

\theta

is the green angle, between the legs placed tail to tail; the red one is the triangle's angle at the waypoint. With the plus sign, only the green one gives

R=912

Why: The plus sign of the sheet goes with the angle between the vectors; the minus sign of the geometry course goes with the angle of the triangle. Mixing the sign of one with the angle of the other flips the cosine term. Check by components: $(300\,;\,400)+(600\,;\,-250)=(900\,;\,150)$ , of length $912$ m.

10. Accepting a formula because its number looks reasonable

the whole question, and a plausible number hides the error from a quick glance

What not to write

“ $h=\dfrac{v^{2}}{2g}=\dfrac{(6.0)^{2}}{2(9.81)}=1.83$ m, a sensible height for a fountain.”

What to write

“At $v=0$ the water stays at the nozzle, so $h$ must reduce to $H$ : $h=H+\dfrac{v^{2}}{2g}=1.20+1.83=3.03$ m.”

Three heights on one jet:

H

to the nozzle,

v^{2}/2g

of rise above it,

h

in all. At

v=0

the rise vanishes and

h

must fall back to

H

, not to the floor.

Why: Dimensions rule out a formula such as $H+\dfrac{v}{2g}$ , a length plus a time, but they cannot see a missing term that has the right units. The limiting case can: set to zero the variable that makes the situation trivial and ask what must happen. Then the direction of variation: a faster jet must go higher, a stronger $g$ lower.

Which method to choose

Which gesture, according to what the statement hands you

Read what the statement gives, not what the chapter is called

The last branch, an angle measured from an inclined surface: with the axes along and across the slope, the vertical weight splits into

W\sin 25^{\circ}

along the slope and

W\cos 25^{\circ}

across it.

If a magnitude and an angle from the $x$ axis, and something to add → project first with $V\cos\theta$ and $V\sin\theta$ , add the two columns, recombine only at the very end

Example: $8.0$ at $35^{\circ}$ gives $(6.6\,;\,4.6)$
If two vectors and the angle BETWEEN them, no axes in sight → $R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}$ with $\theta$ between the vectors placed tail to tail, never the triangle's angle; one line, no components needed

Example: $3.0$ and $4.0$ at $60^{\circ}$ give $R=6.1$
If two vectors given by components, and the angle between them asked → dot product: $\cos\theta=\dfrac{a_{x}b_{x}+a_{y}b_{y}}{ab}$ ; the arccosine already returns $0^{\circ}$ to $180^{\circ}$ , no quadrant to repair

Example: $(300\,;\,400)$ m and $(600\,;\,-250)$ m give $75.7^{\circ}$
If components in hand and a direction asked → acute angle from the ABSOLUTE values, then the quadrant from the signs

Example: $(-29.6\,;\,13.8)$ gives $155.1^{\circ}$
If an angle measured from the vertical, or from an inclined surface → tilt the axes to follow the surface, or convert the angle to one measured from $x$ ; never mix the two conventions in the same line

Example: $20^{\circ}$ from the vertical is $70^{\circ}$ from $x$
If a number with a unit to convert → a fraction worth $1$ , raised to the power that the unit carries, and let the units cancel on the page

Example: $1.90$ g/cm $^{3}=1.90\times 10^{3}$ kg/m $^{3}$
If a quantity to estimate with no data at all → one digit per input, chain the conversion factors, keep only the power of ten

Example: $70$ beats per minute gives $10^{9}$ beats in a lifetime

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing a vector addition that collects every mark

When to use it: The statement gives two or more vectors by magnitude and direction and asks for the resultant, its size and its direction.

1 Sketch the vectors roughly to scale and state in ONE sentence which way $x$ and $y$ point. That sentence is worth a mark on its own and it is what makes every sign checkable.
2 Build a table of components, one line per vector: $V_{x}=V\cos\theta$ , $V_{y}=V\sin\theta$ , each with its sign. Keep full calculator precision here.
3 Add the two columns SEPARATELY and write the result as a pair, for instance $(-29.6\,;\,13.8)$ m.
4 Compute $R=\sqrt{R_{x}^{2}+R_{y}^{2}}$ , then check the bounds on the spot: for two vectors $|A-B|\le R\le A+B$ ; for more, $R$ is at most the sum of the lengths and at least the longest length minus all the others.
5 Give the direction as an acute angle placed by the signs, in degrees AND in words, then round everything to the precision of the data.

Concluding sentence

“Taking $x$ due east and $y$ due north, the components add to $(-29.6\,;\,13.8)$ m, so $R=32.7$ m at $155.1^{\circ}$ from east, that is $24.9^{\circ}$ north of west.”

Marking: Typically 1 mark for the axes, 2 for the components with their signs, 1 for the magnitude, 2 for the direction with the correct quadrant.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The resultant is trapped between two bounds

Before writing the final line, compare $R$ with $|A-B|$ and $A+B$ . Outside that interval, a component has the wrong sign; equal to $A+B$ on vectors that are visibly not parallel, the magnitudes were added straight.

$3.0$ and $4.0$ must give $R$ between $1.0$ and $7.0$ , so $R=8.0$ is impossible.
The signs of the components match your own sketch

Look at the drawing, name the quadrant, then look at the two signs. A vector drawn down and to the left must have two negative components. Ten seconds here catch a swapped sine and cosine and a calculator left in radian mode.

At $240^{\circ}$ , both components are negative: $(-2.5\,;\,-4.3)$ .
The bigger component is on the axis the vector leans towards

A vector closer to the $y$ axis must have the larger $y$ component in absolute value. This is the fastest way to spot a sine written where a cosine belonged, and it needs no arithmetic at all.

$8.0$ at $35^{\circ}$ leans towards $x$ , and indeed $6.6>4.6$ .
The digits match the least precise datum

Count the significant figures of every input before writing the answer. A quotient keeps the smallest count, a difference keeps the smallest number of decimal places, and the rounding happens once, on the last line. Once an uncertainty has been computed, it decides instead: the last digit written sits at the place of the uncertainty.

$\dfrac{24.63}{13.5}$ gives $1.82$ , not $1.824444$ , and with $\pm 0.14$ of uncertainty the result is written $1.8\pm 0.1$ .
The units cancel on the page

Write the conversion factors as fractions and strike out the units that cancel. If the surviving unit is not the one the question asked for, the algebra is wrong and no amount of recomputing will fix it.

$4.5$ mL/min becomes $7.5\times 10^{-8}$ m $^{3}$ /s, and the minutes cancel upside down.
The formula survives its units and its trivial case

Before using a formula you derived or were handed, check that both sides carry the same units, then set to zero the variable that makes the situation trivial and see whether the formula says what must happen. Last, ask whether it moves the right way when a quantity grows.

$h=H+\dfrac{v^{2}}{2g}$ is in metres, gives $h=H$ at $v=0$ , and grows with $v$ .

The typical problem, taken apart

Two displacements, and the quadrant the calculator will get wrong

A displacement $\vec{A}$ of magnitude $5.0$ m points at $30^{\circ}$ from the positive $x$ axis, and a displacement $\vec{B}$ of magnitude $6.0$ m points at $150^{\circ}$ .

Find the magnitude and the direction of $\vec{A}+\vec{B}$ .

The two arrows straddle the

y

axis, so the resultant will point almost straight up and slightly to the left: a direction just past

90^{\circ}

, which is what the signs will have to confirm.

Step 1

Axes: $x$ to the right, $y$ upwards, angles counterclockwise from $x$ .

Why

One sentence, one mark, and every sign in the rest of the solution becomes checkable by the marker and by you.

Step 2

$A_{x}=5.0\cos 30^{\circ}=4.330$ and $A_{y}=5.0\sin 30^{\circ}=2.500$ ; $B_{x}=6.0\cos 150^{\circ}=-5.196$ and $B_{y}=6.0\sin 150^{\circ}=3.000$ .

Why

The formulas produce the signs on their own, so there is nothing to decide by hand. Note $B_{x}<0$ : the second quadrant is already visible in the numbers.

Step 3

$R_{x}=4.330-5.196=-0.866$ and $R_{y}=2.500+3.000=5.500$ .

Why

Two separate columns. The whole chapter is here: the $x$ values never meet the $y$ values, and the resultant is a PAIR, not a single number.

Step 4

$R=\sqrt{0.866^{2}+5.500^{2}}=\sqrt{0.750+30.25}=\sqrt{31.00}=5.568$ m, so $5.6$ m.

Why

Check the bounds immediately: $|6.0-5.0|=1.0\le 5.6\le 11.0$ . The resultant is much smaller than $11$ m because the two vectors partly oppose each other along $x$ .

Step 5

$\arctan\dfrac{5.500}{0.866}=81.0^{\circ}$ , and since $R_{x}<0$ with $R_{y}>0$ the vector is in the second quadrant: $\theta=180^{\circ}-81.0^{\circ}=99.0^{\circ}$ .

Why

The calculator, fed the signed ratio, would have answered $-81.0^{\circ}$ , pointing down and to the right. The acute angle plus the signs is the only reliable route, and the sketch confirms the answer sits just past the $y$ axis.

The conclusion, written out

“The resultant has magnitude $5.6$ m and points at $99^{\circ}$ from the positive $x$ axis, that is $9^{\circ}$ past the vertical towards the negative $x$ side.”

Learn by heart

• $V_{x}=V\cos\theta$ and $V_{y}=V\sin\theta$ , with $\theta$ counterclockwise from the positive $x$ axis.
• $V=\sqrt{V_{x}^{2}+V_{y}^{2}}$ , never negative, zero only for the zero vector.
• Direction: acute angle from the absolute values, quadrant from the signs. The arctangent alone is never the answer.
• $|A-B|\le R\le A+B$ , and $R^{2}=A^{2}+B^{2}+2AB\cos\theta$ with $\theta$ the angle between the vectors placed tail to tail.
• $\vec{a}\cdot\vec{b}=ab\cos\theta=a_{x}b_{x}+a_{y}b_{y}$ : positive when the angle is acute, zero when perpendicular, negative when obtuse.
• $1$ mL $=1$ cm $^{3}$ , $1$ L $=1\times 10^{-3}$ m $^{3}$ , $1$ m $^{3}=10^{6}$ cm $^{3}$ , $1$ g/cm $^{3}=10^{3}$ kg/m $^{3}$ .
• Products and quotients keep the fewest significant figures; sums and differences keep the fewest decimal places.
• Order of magnitude: the nearest power of ten, threshold $\sqrt{10}\approx 3.16$ .
• Unit vector $\hat{u}=\dfrac{\vec{V}}{V}$ , of length exactly $1$ ; a positive scalar changes the length and nothing else.
• A formula is checked three ways: the same units on both sides, a limiting case ( $v\to 0$ ), the direction of variation.

Frequently asked questions

How do I know whether a quantity is a vector or a scalar?

Ask whether the quantity would be fully described without a direction. A mass, a temperature, a volume and the distance walked along a path are complete as a number with a unit, so they are scalars. A displacement or the position of one point relative to another needs a direction as well, so it is a vector. The test is practical: if reversing the direction would change what you are describing, you are holding a vector.

Why can I not simply add the magnitudes of two vectors?

Because the angle between them decides how much of each one points along the other. Two displacements of three metres and four metres give seven metres only if they point the same way. At right angles they give five, at one hundred and fifty degrees about two, and head to head they give one. The safe route is always the same: project both on the axes, add the columns separately, and rebuild a magnitude at the end.

How many significant figures should my final answer have?

Count the significant figures of every number the question gave you. If your answer came from multiplications and divisions, keep the smallest of those counts. If it came from additions and subtractions, keep instead the smallest number of decimal places among the terms. Round once, on the last line, never at every step, because rounding early makes the errors pile up.

Why does my calculator give the wrong angle for a vector?

The arctangent function only returns angles between minus ninety and plus ninety degrees, so it cannot tell the second quadrant from the fourth, nor the third from the first. Two opposite vectors give the same ratio of components and therefore the same display. Compute the acute angle from the absolute values, then use the signs of the two components to decide which quadrant the vector really lives in.

What is an order of magnitude estimate actually for?

It tells you what size the answer should be before you compute it, which is how you catch a conversion factor used once instead of cubed or a decimal point in the wrong place. Round every input to one digit, chain the factors you know by heart, and quote only the power of ten. The digits in front of it carry no information, because the inputs were rough on purpose.

How do I find the angle between two vectors in PHYS 101?

Use the dot product. Multiply the x components together, multiply the y components together and add the two products, then divide by the product of the two magnitudes. That ratio is the cosine of the angle between the vectors, and the inverse cosine gives an angle between zero and one hundred and eighty degrees, so there is no quadrant to fix. A positive dot product already tells you the angle is acute, a negative one that it is obtuse.

Practise it

Corrected exercises: Units, measurement and vectors, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Next sheet Motion in a straight line

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-units-vectors. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a PHYS 101 tutor in Montreal?

Get in touch for a first session. We go back over the points of method that lose marks on units and vectors, then put them to work on problems set at the real level of the McGill midterm.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: units, measurement and vectors (PHYS 101)

The essentials

On your formula sheet: this chapter's lines, and when each one holds

Units and digits: what a measurement is allowed to claim

A vector, and the only legal way to add two of them

The mistakes that cost marks

1. Adding two magnitudes as if they were two numbers

2. Using a conversion factor once when the unit carries an exponent

3. Trusting the angle the calculator returns for an arctangent

4. Copying the calculator display instead of the digits the data earned

5. Taking the cosine for the horizontal component whatever the angle is measured from

6. Leaving the calculator in radian mode

7. Reading the order of magnitude off the exponent of the scientific notation

8. Giving a magnitude a minus sign to say which way the vector points

9. Putting the triangle's angle into the law of cosines of the formula sheet

10. Accepting a formula because its number looks reasonable

Which method to choose

Which gesture, according to what the statement hands you

How the answer is expected to be written

Writing a vector addition that collects every mark

Check before you hand in

The typical problem, taken apart

Two displacements, and the quadrant the calculator will get wrong

Learn by heart

Frequently asked questions

How do I know whether a quantity is a vector or a scalar?

Why can I not simply add the magnitudes of two vectors?

How many significant figures should my final answer have?

Why does my calculator give the wrong angle for a vector?

What is an order of magnitude estimate actually for?

How do I find the angle between two vectors in PHYS 101?

Corrected exercises: Units, measurement and vectors, PHYS 101 at McGill

See also

Looking for a PHYS 101 tutor in Montreal?