Corrected exercises: units, measurement and vectors (PHYS 101)

a) The litre is defined as the volume of a cube of edge $10$ cm, so $1$ L $=(0.10\ \text{m})^{3}=1\times 10^{-3}$ m$^{3}$. Divide by $1000$: $1$ mL $=1\times 10^{-6}$ m$^{3}$, which is exactly $(0.01\ \text{m})^{3}$, that is $1$ cm$^{3}$. Keep that pair in your head for the whole term: a millilitre and a cubic centimetre are the same volume, and a litre is a thousandth of a cubic metre.

b) Write the conversion as a product of fractions worth $1$, and let the units cancel on the page: $1.90\ \dfrac{\text{g}}{\text{cm}^{3}}\times\dfrac{1\ \text{kg}}{10^{3}\ \text{g}}\times\dfrac{10^{6}\ \text{cm}^{3}}{1\ \text{m}^{3}}=1.90\times 10^{3}$ kg/m$^{3}$. The grams divide by a thousand and the cubic centimetres multiply by a million, so the net factor is $1000$, not $100$ and not $0.001$.

Check it against something you know: water is $1.00$ g/cm$^{3}$, that is $1.00\times 10^{3}$ kg/m$^{3}$. Bone is denser than water, and $1.90\times 10^{3}$ kg/m$^{3}$ is a little less than twice the value for water. That is the sanity check to run every time a density comes out of an algebraic mill.

c) Two conversions in one, one on the volume and one on the time: $4.5\ \dfrac{\text{mL}}{\text{min}}\times\dfrac{10^{-6}\ \text{m}^{3}}{1\ \text{mL}}\times\dfrac{1\ \text{min}}{60\ \text{s}}=\dfrac{4.5\times 10^{-6}}{60}=7.5\times 10^{-8}$ m$^{3}$/s. Notice that the time factor goes upside down compared with the volume factor, because minutes sit in the denominator of the flow. Writing the fractions out is what stops that inversion from happening by accident.

d) The mistake is applying the factor once when the unit carries an exponent $3$. The correct statement is $1$ m$^{3}=(10^{2}\ \text{cm})^{3}=10^{6}$ cm$^{3}$, so $2.0$ m$^{3}=2.0\times 10^{6}$ cm$^{3}$. The student's answer is off by a factor of $10^{4}$, which is not a rounding problem, it is a wrong answer. In an assessment this is the whole mark, and it also destroys every later line of the question, since the volume feeds a density or a mass.

The gesture that never fails: put the conversion inside the same brackets as the unit, $(10^{2})^{3}$, and only then compute. A surface gets the factor squared, a volume gets it cubed, and a density, which carries a volume in its denominator, gets it cubed too.

a) The millimetre marks let you read $7.4$ cm with certainty; the edge sits about a third of the way to the next mark, so the reading is $7.43$ cm. Three significant figures, and the last one, the $3$, is the estimated digit. Writing $7.4$ cm throws away a digit you genuinely measured; writing $7.430$ cm claims a precision of a hundredth of a millimetre that this ruler cannot give. Both are marked wrong, and for opposite reasons.

b) $0.00340$ kg has three: the zeros on the left only place the decimal point, while the final zero is written on purpose and therefore counts. $1.20\times 10^{3}$ s has three, and this is exactly why scientific notation exists, since the trailing zero is unambiguous once the power of ten is separated out. $9500$ m is ambiguous: nothing tells you whether the zeros were measured or are placeholders, and the only cure is to write $9.5\times 10^{3}$, $9.50\times 10^{3}$ or $9.500\times 10^{3}$.

c) For a product or a quotient, the result carries the number of significant figures of the least precise factor. Here $7.43$ has three and $2.1$ has two, so the answer keeps two. The calculator gives $7.43\times 2.1=15.603$, which rounds to $16$ cm$^{2}$. Quoting $15.6$ cm$^{2}$ claims a precision the width never had.

Careful with the other rule, the one for sums: an addition or a subtraction keeps the number of DECIMAL PLACES of the least precise term, not the number of significant figures. $7.43+2.1=9.5$, one decimal place, and that answer has two significant figures while one of its terms had three.

d) Two significant figures means two digits, so the answer is $0.67$. The digit $6$ that follows is dropped, and since $6$ is greater than $5$ the previous digit goes up from $6$ to $7$. Writing $0.7$ keeps only one significant figure, which loses information you had; writing $0.6667$ invents information you never had. On a laboratory report, the first costs precision marks and the second costs credibility.

One warning about the order of operations: round ONCE, at the end. If you round every intermediate result the errors accumulate, and a three-step calculation can end up wrong in its second digit. Keep everything in the calculator and cut only the final line.

a) Write each length in scientific notation first, then compare the leading factor with $3.16$. The red blood cell is $8\times 10^{-6}$ m, and $8>3.16$, so its order of magnitude is $10^{-5}$ m, not $10^{-6}$. The hair is $7\times 10^{-5}$ m, so $10^{-4}$ m. The fingernail is $5\times 10^{-4}$ m, so $10^{-3}$ m. The adult is $1.7\times 10^{0}$ m, and $1.7<3.16$, so $10^{0}$ m. Only the last one keeps the exponent of its scientific notation, and that is the whole trap of this question.

b) A Fermi estimate rounds every input to one digit and never apologises for it. Take $70$ beats per minute. In an hour, $70\times 60=4200$; in a day, $4200\times 24=1.008\times 10^{5}$, call it $10^{5}$ beats per day; in a year, about $3.7\times 10^{7}$; in $80$ years, $2.9\times 10^{9}$ beats. The order of magnitude is $10^{9}$, a few billion beats in a lifetime.

The chain matters more than the arithmetic: minute, hour, day, year, lifetime. Each step is a multiplication by a factor you know by heart ($60$, $24$, $365$), and writing them in a line lets you check at a glance that no factor was used twice or forgotten.

c) Volume per day $=70$ mL per beat $\times\,1.0\times 10^{5}$ beats per day $=7.0\times 10^{6}$ mL $=7.0\times 10^{3}$ L. Since $1$ m$^{3}=1000$ L, that is about $7$ m$^{3}$, the volume of a small bedroom, pumped every single day. The order of magnitude in litres is $10^{4}$, because $7>3.16$.

d) Because the inputs were rounded to one digit before the multiplication. A heart rate quoted as $70$ beats per minute could be $65$ or $78$, which moves the final number by more than ten per cent, so the digits $2$ and $9$ of $2.9\times 10^{9}$ carry no information. What survives the roughness of the inputs is the exponent, and that is exactly what an order of magnitude claims. Writing $2.94336\times 10^{9}$ beats is not more precise, it is simply false precision.

This is the habit that saves whole questions later in the term: before writing a number down, ask what power of ten it should have. An answer that comes out $10^{4}$ times too large usually means a conversion factor was used once instead of being cubed, as in exercise 1.

a) Mass, temperature and the distance walked are scalars: a number and a unit say everything. The displacement from the camp to the lake is a vector, since it needs a bearing as well as a length, and the position of a landmark relative to the camp is a vector for the same reason. Note the pair that students mix up: the distance walked along a winding path is a scalar and it is longer than the magnitude of the displacement, which only cares about the two endpoints.

b) Read the right triangle: the side along $x$ is adjacent to the angle, so it takes the cosine. $A_{x}=8.0\cos 35^{\circ}=6.553$ and $A_{y}=8.0\sin 35^{\circ}=4.589$. Rounded to the two significant figures of the data, $A_{x}=6.6$ and $A_{y}=4.6$. Both are positive, which is what the first quadrant demands, and the figure confirms it before any arithmetic.

c) With $\theta=240^{\circ}$, the formulas take care of the signs on their own: $B_{x}=5.0\cos 240^{\circ}=-2.5$ and $B_{y}=5.0\sin 240^{\circ}=-4.330$, so $B_{y}=-4.3$ to two significant figures. Both components are negative, and the vector points into the third quadrant, exactly as drawn. If your calculator returned two positive numbers, it is in degree mode but you fed it $60^{\circ}$ instead of $240^{\circ}$, or it is in radian mode.

Here is the check that matters more than the arithmetic: the component along the axis the vector leans towards must be the larger one in absolute value. $\vec{B}$ leans towards the negative $y$ axis, and indeed $|{-4.3}|>|{-2.5}|$. On $\vec{A}$, which leans towards $x$, the $x$ component is the larger. Two seconds of looking at the drawing catches a swapped sine and cosine.

d) $\sqrt{6.553^{2}+4.589^{2}}=\sqrt{42.94+21.06}=\sqrt{64.00}=8.00$, which is the magnitude you started from. The check is worth a great deal and it is not free of holes: it confirms that the pair of numbers has the right size, but it would also pass for $(-6.553\,;\,4.589)$, which is a completely different vector. The magnitude never tells you the quadrant, only the signs of the components do.

Cost of getting this wrong: a swapped sine and cosine changes both components and every line built on them, so the question is lost even though each later step is carried out correctly. This is the single most expensive error of the chapter, and the drawing is what prevents it.

a) Head to tail with $\theta=0^{\circ}$, the two arrows lie along the same line in the same direction, so the lengths simply add: $R=7.0$. At $\theta=180^{\circ}$ the second arrow comes back along the first, so the lengths subtract: $R=|4.0-3.0|=1.0$. At $\theta=90^{\circ}$ the triangle is right angled and Pythagoras applies: $R=\sqrt{3.0^{2}+4.0^{2}}=5.0$. Three angles, three answers, and only the first one is $7.0$.

b) Put $\vec{A}=(3.0\,;\,0)$ along the $x$ axis. For $\theta=60^{\circ}$, $\vec{B}=(4.0\cos 60^{\circ}\,;\,4.0\sin 60^{\circ})=(2.000\,;\,3.464)$, so the sum is $(5.000\,;\,3.464)$ and $R=\sqrt{25.00+12.00}=\sqrt{37.00}=6.083$, that is $6.1$. For $\theta=120^{\circ}$, $\vec{B}=(-2.000\,;\,3.464)$, the sum is $(1.000\,;\,3.464)$ and $R=\sqrt{1.000+12.00}=\sqrt{13.00}=3.606$, that is $3.6$.

Notice that only the $x$ component changed sign between the two cases while the $y$ component stayed the same. That is the geometry of the obtuse case, and it explains why the resultant shrinks as the angle opens.

c) With $\vec{A}$ along $x$, the sum has components $(A+B\cos\theta\,;\,B\sin\theta)$. Then $R^{2}=(A+B\cos\theta)^{2}+(B\sin\theta)^{2}=A^{2}+2AB\cos\theta+B^{2}\cos^{2}\theta+B^{2}\sin^{2}\theta$, and since $\cos^{2}\theta+\sin^{2}\theta=1$ this collapses to $R^{2}=A^{2}+B^{2}+2AB\cos\theta$. As $\theta$ runs from $0^{\circ}$ to $180^{\circ}$, $\cos\theta$ decreases from $1$ to $-1$, so $R$ decreases from $A+B=7.0$ down to $|A-B|=1.0$. The resultant is therefore trapped in $1.0\le R\le 7.0$, and the curve of the solution shows how smoothly it slides between the two bounds.

d) Set $R=5.0$: $25.00=9.000+16.00+24.00\cos\theta$, hence $\cos\theta=0$ and $\theta=90^{\circ}$. The value $5.0$ is not the average of the two extremes, it is the perpendicular case, and that is worth remembering: whenever a problem gives you two perpendicular pieces, Pythagoras is enough and no cosine is needed.

The interval $[\,|A-B|\,;\,A+B\,]$ is the five second check to run on every resultant you compute for the rest of the term. A magnitude outside it means a sign error in the components, and a magnitude equal to $A+B$ when the vectors are visibly not parallel means the magnitudes were added straight, which is the error this exercise exists to kill.

a) First leg: due east is the positive $x$ direction, so $\vec{d_{1}}=(40.0\,;\,0)$ m. Second leg: $d_{2x}=85\cos 145^{\circ}=-69.63$ m and $d_{2y}=85\sin 145^{\circ}=48.75$ m, one negative and one positive, as the second quadrant requires. Third leg: due south is the negative $y$ direction, so $\vec{d_{3}}=(0\,;\,-35.0)$ m. Each leg is written with its own line, which is what keeps the bookkeeping honest.

b) Add the $x$ values together, then the $y$ values together, and never mix the two columns: $x$: $40.0-69.63+0=-29.63$ m. $y$: $0+48.75-35.0=13.75$ m. So the net displacement is $(-29.6\,;\,13.8)$ m, to three significant figures. Negative $x$ and positive $y$ place it in the second quadrant, northwest of the start, which the figure confirms.

c) $R=\sqrt{29.63^{2}+13.75^{2}}=\sqrt{877.9+189.1}=\sqrt{1067}=32.7$ m. Two checks before moving on. First, $32.7$ m is far below the $40+85+35=160$ m actually walked, which is right, since the legs partly cancel. Second, the magnitude must be at least the largest single component, and $32.7>29.6$. Both pass.

d) $\tan\theta=\dfrac{13.75}{-29.63}=-0.4642$, and the calculator returns $\arctan(-0.4642)=-24.9^{\circ}$. That angle points into the FOURTH quadrant, east and slightly south, which is the opposite corner of the plane from where the biologist actually is. The arctangent function only returns values between $-90^{\circ}$ and $+90^{\circ}$, so it cannot distinguish $(-29.63\,;\,13.75)$ from $(29.63\,;\,-13.75)$: both ratios are the same number.

The repair is mechanical. Compute the acute angle from the absolute values, $\arctan\dfrac{13.75}{29.63}=24.9^{\circ}$, then place it using the signs you already wrote down. Negative $x$ with positive $y$ is the second quadrant, so $\theta=180^{\circ}-24.9^{\circ}=155.1^{\circ}$. In words, $24.9^{\circ}$ north of west. Write the direction in words as well as in degrees: a marker can see at once whether the quadrant is right, and so can you.

Cost of skipping this step: the magnitude is correct, the arithmetic is correct, and the answer is still wrong by $180^{\circ}$. It is the most common lost mark of the whole chapter, and it costs the entire final part of the question.

a) $|\vec{A}|=\sqrt{8.0^{2}+(-6.0)^{2}}=\sqrt{64.0+36.0}=\sqrt{100.0}=10.0$. The unit vector is the vector divided by that magnitude, component by component: $\hat{u}=\left(\dfrac{8.0}{10.0}\,;\,\dfrac{-6.0}{10.0}\right)=(0.80\,;\,-0.60)$. Check it at once: $\sqrt{0.80^{2}+0.60^{2}}=\sqrt{0.64+0.36}=1.00$. A unit vector that does not come out to $1$ is a division that went to only one of the two components.

b) Multiplying by a scalar multiplies EVERY component: $2.5\,\vec{A}=(2.5\times 8.0\,;\,2.5\times(-6.0))=(20.0\,;\,-15.0)$, and $|2.5\,\vec{A}|=\sqrt{400.0+225.0}=\sqrt{625.0}=25.0$, which is $2.5\times 10.0$ as it must be. The direction has not moved: the unit vector of $2.5\,\vec{A}$ is still $(0.80\,;\,-0.60)$. A positive scalar changes the length and nothing else.

c) $\vec{A}-\vec{B}=(8.0-(-3.0)\,;\,-6.0-4.0)=(11.0\,;\,-10.0)$. The sign trap is in the first component: subtracting a negative number adds. Its magnitude is $\sqrt{121.0+100.0}=\sqrt{221.0}=14.87$, that is $14.9$. Reversing the order gives $\vec{B}-\vec{A}=(-11.0\,;\,10.0)$, the same magnitude with every sign flipped. Subtraction of vectors is not commutative, exactly as with numbers, and the two results are opposite arrows of equal length.

The construction in the solution figure says the same thing without algebra: draw $\vec{A}$, then from its tip draw $-\vec{B}$, and the arrow from the original starting point to the final tip is $\vec{A}-\vec{B}$. Being able to sketch it is what tells you whether the numbers you computed are plausible.

d) The claim gives $10.0-5.0=5.0$, since $|\vec{B}|=\sqrt{9.0+16.0}=5.0$. The true value is $14.9$, nearly three times larger. The claim is false, and it fails for the reason this whole chapter keeps repeating: a magnitude is not a component, and it does not follow the algebra of ordinary numbers. Here $\vec{A}$ and $\vec{B}$ point in almost opposite directions, so subtracting $\vec{B}$ makes the arrow LONGER, not shorter.

What is true is an inequality, not an equality: $\bigl||\vec{A}|-|\vec{B}|\bigr|\le|\vec{A}-\vec{B}|\le|\vec{A}|+|\vec{B}|$, that is $5.0\le 14.9\le 15.0$. Equality on the left happens only when the two vectors point the same way, and on the right only when they point opposite ways. Use the inequality as a check, never the equality as a formula.

a) FALSE. The magnitude of the sum depends on the angle between the two vectors, and it can be anything from $|4.0-3.0|=1.0$ to $4.0+3.0=7.0$. The value $7.0$ is reached only when the two point in the same direction. Correct version: the magnitude of the sum lies between $1.0$ and $7.0$, and it equals $7.0$ only if the two vectors are parallel and in the same sense.

b) TRUE, and both halves of the sentence have to be checked before answering. Leading zeros only locate the decimal point, so the $0$, the $0$ and the $0$ before the $3$ are not significant. What remains is $3$, $4$ and the trailing $0$, which was written deliberately and is significant: three significant figures. Writing the number as $3.40\times 10^{-3}$ kg makes the count impossible to get wrong.

c) TRUE. A magnitude is a length, computed as a square root of a sum of squares, so it cannot be negative. It vanishes only if every component vanishes, which is the definition of the zero vector. The sign of a component says which way the vector leans along one axis; the magnitude says how long it is, and the two must never be confused.

d) FALSE, and this is the expensive one. The centimetres sit inside a cube, so the factor applies three times: $1$ m$^{3}=10^{6}$ cm$^{3}$. Correct version: divide by $10^{3}$ for the grams and multiply by $10^{6}$ for the cubic centimetres, which gives $1.90\times 10^{3}$ kg/m$^{3}$. The student's value, $0.190$ kg/m$^{3}$, describes a substance ten thousand times lighter than air, which the plausibility check catches immediately.

e) FALSE. The arctangent function returns only values between $-90^{\circ}$ and $+90^{\circ}$. For a direction of $240^{\circ}$ the two components are $V\cos 240^{\circ}$ and $V\sin 240^{\circ}$, both negative, and their ratio is $\tan 240^{\circ}=1.732$, the same value as $\tan 60^{\circ}$. The calculator therefore returns $60^{\circ}$. Correct version: the arctangent gives $60^{\circ}$, and since both components are negative the vector is in the third quadrant, so the direction is $60^{\circ}+180^{\circ}=240^{\circ}$.

Two of these five are true, and that is deliberate: correcting a statement that was already correct costs as much as leaving a false one standing. Read the sentence in two halves, check each half, and only then commit.

a) The volume of the fragment is the difference of the two readings: $V=38.5-25.0=13.5$ mL, that is $13.5$ cm$^{3}$. Each reading carries about $\pm 0.5$ mL, and a difference adds the two uncertainties rather than cancelling them, so $\Delta V=0.5+0.5=1.0$ mL. Writing $\pm 0.5$ mL on the difference is the classic loss of half a mark: the subtraction removed the water, not the uncertainty.

b) $\rho=\dfrac{24.63\ \text{g}}{13.5\ \text{cm}^{3}}=1.824$ g/cm$^{3}$. The mass carries four significant figures, the volume only three, and a quotient keeps the smaller count, so $\rho=1.82$ g/cm$^{3}$. In SI units, multiply by $1000$ as in exercise 1: $\rho=1.82\times 10^{3}$ kg/m$^{3}$. The calculator display, $1.824444\ldots$, is not an answer, it is a machine artefact.

c) Percent uncertainty on the volume: $\dfrac{1.0}{13.5}=0.074$, that is $7.4$ %. On the mass: $\dfrac{0.01}{24.63}=0.0004$, that is $0.04$ %. For a product or a quotient, the percent uncertainties add, so the density carries $7.4+0.04\approx 7.4$ %. In absolute terms, $0.074\times 1.82=0.135$, so $\rho=1.82\pm 0.14$ g/cm$^{3}$.

Read that pair of percentages carefully, because it is the point of the exercise: one measurement contributes almost two hundred times more uncertainty than the other. Weighing the fragment more carefully would change nothing at all.

d) The interval $1.82\pm 0.14$ g/cm$^{3}$ runs from $1.68$ to $1.96$ g/cm$^{3}$, which overlaps the expected range of $1.7$ to $2.0$ g/cm$^{3}$ almost entirely, so the result is acceptable. The measurement to improve is the volume, and the cheapest fix is not a better cylinder but a bigger sample: doubling the volume of bone halves the percent uncertainty, since the absolute $\pm 1.0$ mL stays the same while the denominator grows. A narrower cylinder, graduated every $1$ mL, would help for the same reason.

The habit to build here: before quoting a result, work out which single measurement dominates the uncertainty. It tells you where the next hour in the laboratory should be spent, and it is the sentence a marker looks for in a discussion section.

a) The ramp is the hypotenuse of a right triangle whose legs are the rise and the run, so $\tan\theta=\dfrac{0.80}{5.70}=0.1404$ and $\theta=\arctan(0.1404)=7.99^{\circ}$, that is $8.0^{\circ}$. Sanity check: a rise of $0.80$ m over $5.70$ m is a slope of about one in seven, and a small angle is exactly what you expect from a ramp built for a wheelchair.

b) The displacement of $3.50$ m lies along the ramp, so the angle it makes with the horizontal is $\theta$ itself. Horizontal component: $3.50\cos 7.99^{\circ}=3.466$ m, that is $3.47$ m. Vertical component: $3.50\sin 7.99^{\circ}=0.4865$ m, that is $0.486$ m. The horizontal part is almost the whole displacement and the vertical part is small, which is what a shallow ramp should give. If your two numbers come out nearly equal, the calculator is in radian mode.

c) Now the vector is known in horizontal and vertical terms, $(0\,;\,-0.25)$ m, and the axes are tilted. Project it on each new direction. Along the ramp: the component is $-0.25\sin 7.99^{\circ}=-0.0347$ m, negative because lowering the seat moves it slightly DOWN the slope. Perpendicular to the ramp: $-0.25\cos 7.99^{\circ}=-0.2476$ m, that is $-0.248$ m, negative because it moves towards the surface.

The sine and the cosine have swapped roles compared with b), and that is not a trick, it is geometry: the angle between the vertical direction and the perpendicular to the ramp is the same $\theta$ as the angle between the horizontal and the ramp. Draw the two right triangles side by side once, and you will never have to rederive it.

d) $\sqrt{0.0347^{2}+0.2476^{2}}=\sqrt{0.001205+0.061306}=\sqrt{0.06251}=0.2500$ m, exactly the length we started with. That is the check to run every time you change axes: rotating the axes rotates the description of the vector, never the vector itself, so the magnitude is untouched. If it changes, a sine was used where a cosine belonged.

What the tilted axes bought: the motion along the surface of the ramp now has a single component instead of two, so any later question about distance travelled along the ramp becomes a one line calculation. The price is one sentence of bookkeeping, naming which way $x$ and $y$ point, and that sentence is worth marks on its own in an assessment.