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

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.

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 $0.001$.

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.

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.

Dimensions first, with brackets for the unit of a quantity: $[v]=$ m/s, $[g]=$ m/s$^{2}$, $[H]=$ m, and $h$ must come out in metres.

(1): $\left[\dfrac{v}{g}\right]=\dfrac{\text{m/s}}{\text{m/s}^{2}}=\text{s}$. A time added to a length: rejected on sight, before any number.

(2): $\left[\dfrac{v^{2}}{g}\right]=\dfrac{\text{m}^{2}/\text{s}^{2}}{\text{m/s}^{2}}=\text{m}$. Dimensionally correct.

(3): a length plus a length, so metres. Dimensionally correct too.

Dimensions cannot choose between (2) and (3), and a pure number such as the $2$ is invisible to them. The limiting case can.

Case $v\to 0$: the water just wells out of the nozzle and stays at height $H$. Formula (2) gives $h\to 0$, the floor of the basin: rejected. Formula (3) gives $h\to H$: it survives.

Third check, the direction of variation: in (3) a faster jet goes higher, and a stronger gravity, with $g$ in the denominator, gives a lower jet. Both are what you expect.

$h=H+\dfrac{v^{2}}{2g}=1.20+\dfrac{(6.0)^{2}}{2(9.81)}=1.20+1.83=3.03$ m

Formula (2) is not absurd, it answers another question: it is the rise above the nozzle, $1.83$ m. The case $v\to 0$ is what showed that the classmate measured from the wrong level. You will derive (3) yourself in the next chapter, from $v^{2}=v_{0}^{2}+2a(x-x_{0})$.

Pitfall: trusting a formula because the number it gives looks reasonable. Formula (2) gives $1.83$ m, a perfectly plausible fountain, and it is still wrong for the question asked. The two checks cost ten seconds.

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.

$0.00340$ 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 $9.500\times 10^{3}$.

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.

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.

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.

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.

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 $7>3.16$.

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.

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.

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.

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.

$\sqrt{6.553^{2}+4.589^{2}}=\sqrt{42.94+21.06}=\sqrt{64.00}=8.00$, 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.

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 $7.0$.

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 $3.6$.

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.

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 dot product pairs each component with the matching one, $x$ with $x$ and $y$ with $y$:

$\vec{d}_{1}\cdot\vec{d}_{2}=(300)(600)+(400)(-250)=180\,000-100\,000=8.0\times 10^{4}$ m$^{2}$

The sign answers first: positive, so $\cos\theta>0$ and the angle between the legs is acute. Zero would mean perpendicular, negative obtuse.

Magnitudes: $d_{1}=\sqrt{300^{2}+400^{2}}=500$ m and $d_{2}=\sqrt{600^{2}+250^{2}}=650$ m.

Units check: the dot product is in m$^{2}$, the product $d_{1}d_{2}$ too, so the cosine comes out without unit, as a cosine must.

Unlike the arctangent of exercise 6, the arccosine returns an angle between $0^{\circ}$ and $180^{\circ}$, which is exactly the range of an angle between two vectors: there is no quadrant to repair.

Pitfall: multiplying the magnitudes, $500\times 650$, and calling that the dot product. That is the dot product only for parallel vectors; in general the cosine, or the pairing of the components, carries the geometry.

Which angle goes into the formula? The sheet calls $\theta$ the interior angle of the two VECTORS: the angle between them placed tail to tail. At the waypoint, that is the change of heading, between the dashed extension of the first leg and the second leg: $\theta=75.7^{\circ}$.

$R^{2}=d_{1}^{2}+d_{2}^{2}+2d_{1}d_{2}\cos\theta=250\,000+422\,500+2(500)(650)(0.2462)=832\,500$ m$^{2}$

$R=912$ m

The interior angle of the TRIANGLE at the waypoint is the supplement: $180^{\circ}-75.7^{\circ}=104.3^{\circ}$.

Pitfall: putting $104.3^{\circ}$ into the formula of the sheet gives $R^{2}=672\,500-160\,000$ and $R=716$ m, about $200$ m short. The geometry course writes the law with a minus sign, $c^{2}=a^{2}+b^{2}-2ab\cos C$, precisely because $C$ is the angle of the triangle: since $\cos(180^{\circ}-\theta)=-\cos\theta$, both forms agree as long as each one gets its own angle.

Check by components: $\vec{d}_{1}+\vec{d}_{2}=(900\,;\,150)$ m, so $R=\sqrt{900^{2}+150^{2}}=\sqrt{832\,500}=912$ m. Same number, and it had to be: the term $2d_{1}d_{2}\cos\theta$ is exactly twice the dot product of e).

Bounds of c): $650-500=150\le 912\le 1150$. And since the dot product is positive, $R$ must exceed the perpendicular value $\sqrt{500^{2}+650^{2}}=820$ m: it does.

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.

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.

$R=\sqrt{29.63^{2}+13.75^{2}}=\sqrt{877.9+189.1}=\sqrt{1067}=32.7$ 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.

$\tan\theta=\dfrac{13.75}{-29.63}=-0.4642$, 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.