1994-1995 Physics Olympiad Preparation Program

{ University of Toronto {

Problem Set 1: General PhysicsDue October 24, 1994

1. Your physics professor (you're in university now) has babbled all year long in lectures and you couldn'tunderstand a word he said. It's now exam time and, since you didn't bring a cheat sheet (good foryou), you decide to derive the formulas for various physical quantities using dimensional analysis inthe hopes of partial marks. Your babbling professor may give you partial marks but we'll give you fullones if you get all these (as long as you show your assumptions and your work!).

(a) You remember a 2� but can't remember the rest of the expression for the period of a simplependulum. What is the full expression?

(b) A particle of mass m rotates in a circle of radius r with speed v. The particle has an accelerationac called centripetal (centre seeking) acceleration. What is the form of ac?

(c) A gas bubble from a deep explosion under water oscillates with a period T . Known variables arep, the static pressure, �, the water density, and e, the total energy of the explosion. Find thedependence of the period on the other variables.

2. Being the hip physics-dude/dudette you are, you often go to parties with your non-physics friends.You like to be the centre of attention, telling them about the physics of quarks and gluons, but thatusually ends up killing the party. You decide, then to pull a di�erent beast out of your physics hat:rapid estimation. You know that sometimes we don't need exact answers but only order-of-magnitudeestimations. At a recent party, your friends asked you the following:

(a) Jake asked: if I beat everyone on the planet into a pulpy liquid, roughly how deep would the liquidbe on the Earth's surface? Would I need Doc's, rubber boots, hip waders, or a boat to a avoidgetting my new socks dirty? What if I wanted to �ll up containers the size of the SkydomeTM indowntown Toronto with this vile mess; how many containers would I need?

(b) Jane asked: if the entire debt of the Government of Canada were paid o� in loonies, how muchwould it weigh? If I stacked them, how high would it reach? If each loonie was turned into apound of esh, how would that a�ect the answer to Jakes question?

Remember to justify any non-obvious estimations you make; your physics teacher's daughter is at theparty and she'll call your blu� on bad estimations, making you look like a fool in front of your friends.

3. Your Great-Aunt Edna is in town visiting. She's really rich and is about to kick the bucket, so she willbe drawing up her will soon. You want to be in it, you materialistic !@?$%#!, you. Now, Auntie Edna,as you like to call her, made her fortune as a high-school physics and english teacher, so she reallygets P-O'd when people use poor grammar and physically incorrect phrases. You decide to impressher with the following:

(a) Metro Toronto Police as well as local newscasters on TV often describe the latest tra�c fatalityusing the phrase \the car was traveling at a high rate of speed." What is wrong with this?

(b) People describe their watch as being \�ve minutes fast" or \two minutes slow". What is wrongwith this? What do they really mean (or what should they really say)? Is there a correct way ofdescribing a watch's inaccuracy using the words \fast" or \slow"?

(c) Weather announcers and even meteorologists use the phrase \normal high" and \normal low".What do they mean by \normal"? Is that normal? What might be a better word? Discuss.

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X Y

Figure 1: Rectangular wire mesh of in�nite extent.

4. Your Uncle Sal from Italy was in North America for the World Cup. He was terribly disappointed withtheir loss, and is in a blue funk (the blue that is found on the Brazilian ag, as a matter of fact). Youwant to make him a pesto pizza to cheer him up. While pouring the olive oil, you accidentally spillsome into a glass of water, and it spreads out on the surface. Sal, who is watching, is a physics teacherin Italy and he is reminded of an experiment done by Lord Rayleigh. He tells you that he would behappy if only you would solve the following.

Rayleigh1 found that 0.81 mg of olive oil on a water surface produced a mono-molecular layer 84 cmin diameter. What value of Avogadro's Number results?

Note: The approximate composition of olive oil is H(CH2)18COOH, in a linear chain with one end(which?) hydrophilic and the other hydrophobic. Its density is 0:8 g=cm3

5. Ever the consummate student, you are at Woodbine doing some betting||| studying. Of course, youbrought your $500 camera with a telephoto lens. Watching the thoroughbred racers Eric's Idle andPaul's Bunyon thunder down the home straight, you decide to take a picture. You are looking head onthrough the telephoto lens and you notice that the horses seem strangely foreshortened (not as longfrom front to back) as they gallop towards the camera. Explain this observation.

6. Your little brother has been playing with the wire screens and batteries again. Will he ever learn? Heasks you the following.

(a) A rectangular wire mesh of in�nite extent in a plane has 1 A of current fed into it at point X, asin the diagram (Figure 1), and 1 A of current taken from it at point Y. Find the current in thewire XY.

(b) Suppose the wire mesh was made of equilateral triangles joined in the obvious way. What is thecurrent where X and Y are separated by one wire?

1Rayleigh, Proc. Roy. Soc., 47, 364 (1890); Holy smokes, that's 104 years ago! And you're only learning about it now?

Bonus points to anyone who can get me a photocopy of the article.

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. . .

Figure 2: In�nite resistor ladder.

(c) The screen has been cut up and now resembles the circuit in the diagram. What is the resistanceof the circuit shown, an in�nite ladder.

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1994-1995 Physics Olympiad Preparation Program

{ University of Toronto {

Solution Set 1: General Physics

1. Dimensional analysis can be used to check to see if an expression is dimensionally correct or to get the

form of an expression if we don't know it. We are going to do the latter.

(a) Hmm... What could the period of a simple pendulum possibly depend on? There's the mass of

the (ideal point) bob (or is that Bob, founder of the Church of the Sub-Genius?), m, the length of

the (essentially massless ideal) string, l, the angle of the swing, �, and the acceleration of gravity,

g. We might think to include things like frictional forces, but they're small compared to gravity

and besides, that would be too complicated. Let's assume the period T is a function of these four

variables, each raised to some power:

T = Cmwlx�ygz:

C is a dimensionless constant, and w, x, y, and z are exponents for which we wish to solve (no

dangling prepositions here). The dimensional equation for this relationship is

[T ] = [M ]w[L]x[L=T 2]z;

the angle � (measured in radians) has no dimensions. Simplifying,

[T ] = [M ]w[L]x+z[T ]�2z:

To have dimensional consistency, we must have

0 = w

0 = x+ z

1 = �2z

which gives us w = 0, z = �1=2, and x = 1=2. Thus the desired equation is

T = Cpl=gf(�)

It turns out that C = 2� and f(�) ' 1 for small angles, but we can't get that from this analysis.

(b) The general equation for this problem is ac = Cmxryvz , which gives us the dimensional equation

[L=T 2] = [M ]x[L]y[L=T ]z. Simplifying, [L][T ]�2 = [M ]x[L]y+z[T ]�z, which gives us the three

equations 0 = x, 1 = y + z, and �2 = �z, which are solved by x = 0, z = 2, and y = �1. Thusac = Cv2=r. It turns out that C = 1.

(c) The general equation for this problem is T = Cpx�yez , which gives us the dimensional equation

[T ] = [M=LT 2]x[M=L3]y[ML2=T 2]z. Simplifying, [T ] = [M ]x+y+z[L]�x�3y+2z[T ]�2x�2z, which

gives us the three equations 0 = x+ y+ z, 0 = �x� 3y+ 2z, and 1 = �2x� 2z, which are solved

by x = �5=6, y = 1=2, and z = 1=3. Thus T = C��3e2

p5

�1=6.

2. For these ones, we quote numerical values of things from memory or make good guesses, multiply

numbers in our head, rounding them o�, and in general not worry about being too precise. It may also

help to do the problem algebraically �rst, then put in the numbers, since this will allow us to plug in

more precise numbers if we wish.

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(a) In the spirit of rapid estimation, we will give the result as a series of approximate calculations.

This way we only have to remember a few numbers at a time.

Presumably the depth of liquid, d, will not be large compared to the radius of the Earth, RE , so

we can use the formula for the volume of a thin shell1; V = 4�R2Ed. Thus d = V=4�R2

E. What

is V? Well, there are nearly 6 billion people on the planet. Their average mass is, say, about 50

kilograms thus the total mass of people is about 300 billion kilograms. The density of a human

body is close to that of water2, about a tonne per cubic metre, so the total volume is about 300

million cubic metres.

The radius of the Earth is about 4 thousand miles (a good number to remember) or over six

thousand kilometres. Squared, we have about 40 trillion square metres. Multiplying by 4� ' 12:5

we have the surface of the Earth being about 500 trillion square metres. Dividing the volume by

this we get a depth of about a half a micron (�m). I don't even know if this would get your socks

dirty, but you could easily get by without the rubber boots and just wearing shoes.

The politically-correct restatement of the problem is: if every person on the planet decided to go

skinny-dipping in the ocean, then by how much would the ocean level rise? The answer is about

4/3 the answer to the non-politically correct question since the oceans cover about 3/4 of the

planet.

What about the number of SkydomeTM's? We just divide the volume found above by the volume

of a building. The rough area of the football �eld is 60 m by 130 m or about 8 thousand square

metres. The total area is maybe four times that, and the total height is, perhaps, 30 m. Thus the

total volume is on the order of one million cubic metres. Dividing the volume of liquid by this we

get something on the order of several hundred SkydomeTM's.

(b) You may have heard that the federal debt per capita is about $20,000. If you hadn't heard that,

then you should read more newspapers! With 27 million people in Canada, this works out to

about $550 billion. Wow! A loonie has a mass of about 10 grams, I believe. On Earth, 454 grams

weighs a pound, so lets say that 50 loonies weigh one pound. Thus the federal debt weighs about

11 billion pounds in loonies.

Stacking them, they are about 2 mm think so that makes about 1.1 trillion metres or a million

kilometres. That's to the moon and back easily!

Converting them to a pound of esh, we have that a pound is about half a kilogram and the

density is a tonne per cubic metre. This works out to 270 billion kilograms or 270 million cubic

metres. This nearly doubles the volume of liquid for the previous problem, thus it nearly doubles

the depth.

3. (a) What is wrong with the phrase \the car was traveling at a high rate of speed"? Well, the word

rate usually refers to a change in a quantity measured with respect to an interval of time. The

de�nition of speed is the distance traveled per unit of time, i.e. the rate of change of position

with respect to time. Thus the word rate in the above phrase is super uous. One could say \at

a high speed" or one could say \at a high rate" where it is implicit that the rate is position with

respect to time, i.e. speed. Note also that we can't really interpret \at a high rate of speed" as

acceleration since in that case it would be \at a high rate of change of speed".

(b) As in the previous question, the fast and slow refer to rates as well. What most people mean is

\�ve minutes ahead" or \two minutes behind". One can use rate-implying words when talking

about clocks, but in the following context. Suppose we set a watch to coincide with an atomic

1You can derive this in the following way. Let r be the radius of a sphere, and r + �r be the radius of a slightly largersphere. The volume of the shell between the two spheres is the di�erence in volumes which is 4�((r + �r)3 � r3)=3 =4�(r3 + 3r2�r + 3r�r2 + �r3 � r3)=3. This is then equal to 4�(3r2�r + 3r�r2 + �r3)=3 = 4�r2�r(1 + �r=r +�r2=3r2).We then use the fact that �r=r << 1 to get the volume V ' 4�r2�r.

2People oat with the aid of air in their lungs; without that they would sink. Thus people are denser than water, but notby much.

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clock. Thirty days later we check and the watch is ahead one minute. We have gained 1 minute

over a period of 30 days. Thus the watch is about one minute per month fast. Strictly speaking

the units can be canceled to get a unitless number, but it is more meaningful to people to use the

above form. Also note that I have seen people with degrees in physics make the mistake of saying

fast when they mean ahead!

(c) The normal high and normal low are, in more precise language, the average high and low. That

is, they are the averages of the daily highs and lows for a particular day of the year, averaged

over the past 100 years or so for which there are recorded temperatures. What is left out is the

variance of the highs and lows or rather the distribution of the temperatures.

It is possible that the average temperature on August 28 is 25 �C but there may be just as many

August 28's which were 23, 24, 26, and 27. Does it then make sense to use the word normal to

mean average? In such a case is 25 normal? Probably not.

It is really a case of experts not wanting to use precise language because it may end up confusing

people, or so they think.

4. We know that N=M = NA=MA where N is the number of molecules in the sample and M is the

mass of the sample and the subscript refers to a mole of the substance. Thus Avogadro's Number

is NA = NMA=M . To get MA, we take the formula and the molecular masses and �nd the total;

(38� 1 + 19� 12 + 2� 16) = 298 g. To get N , we take the volume of the sample, V , and divide it by

the volume of a single molecule, v. Getting v is the tricky part.

The COOH end of the molecule is hydrophilic while the rest of the molecule is hydrophobic. Thus

the molecules form a layer standing side-by-side vertically on the surface of the water. The diagram

shows approximately what the molecule looks like if we assume that the carbons are at the centres of

tetrahedrons with the neighbouring atoms on the corners. There is an end-on view and a side view.

The distance between two atoms joined to a common carbon is x|this is the length of an edge of

the tetrahedron. It turns out that the end-on area is nearly square|the zig-zag in the carbon chain

compensates for the di�erence between the edge length in one direction and the distance from edge-

centre to opposite edge-centre in the other direction. In all three directions a distance of x is added

to account for the fact that neighbouring molecules will not be butted up against one-another. The x

puts the hydrogen atoms on one molecule a distance x from those on the neighbouring molecules. Thus

the volume of each molecule is about v ' 4t3=121 where t is the layer thickness. There is a measure of

uncertainty in this last equation since we don't know how much space there is between molecules.

The total volume is V = At, thus 1=v ' 121A3=4V 3 and N ' 121A3=4V 2. Now A = �(d=2)2 and

V = M=�; with a little algebra we get

NA '121�3�2d6MA

256M3:

Putting in the numbers,

NA '121�3(0:8 g=cm3)2(84 cm)6(298 g)

256(0:81� 10�3 g)3

:= 2� 1024:

So Avogadro's number is about 1024 molecules per mole. We are o� by a factor of 3 from the accepted

value, which is ok considering how many uncertain numbers there were.

5. A telephoto lens, like a telescope, makes far objects appear near by giving rise to an angular magni�-

cation. Let us approximate a horse (call him Boxy) by a rectangular body with a vertical leg at each

corner. Suppose the separation of the forelegs and of the back legs is 0.5 m, and the distance between

the front and back legs is 2.0 m. If we observe the horse head-on when it is 10 m away, the gap between

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its forelegs will subtend an angle of 1/20 rad at the eye, while the gap between the hind legs subtends

1/24 rad. This di�erence gives most of the impression of the length of the horse.

If the same horse is 100 m away the same two angles become 1/200 rad and 1/204 rad, which are not

very di�erent. However, if we use a telescope with an angular magni�cation of 10�, the angles become

1/20 rad and 1/20.4 rad. The apparent size of the front is the same as that of a horse standing 10 m

away, but now its hind legs appear to be only 0.2 m further back. A very short horse!

6. (a) Break the problem into two parts. Consider the mesh with 1 A going in at X and 1 A coming

out at in�nity. By symmetry, the wire XY must carry 1=4 A of current. Alternately, consider

the mesh with 1 A going in at in�nity and 1 A coming out at Y. Again, by symmetry, 1=4 A

runs through the wire XY. Add the two cases together and you get the stated problem, with the

solution that 1=2 A ows along wire XY.

(b) Similarly, the two cases each have 1=6 A owing along XY, giving a solution of 1=3 A.

(c) Referring to the diagram, you can see that because the circuit is in�nite, it can be written as a

circuit with itself being one of the components. If R0 is the resistance of the entire circuit, then,

using the rules for combining resistors in series and parallel, we have

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R0=

1

R+

1

2R+R0:

Multiplying by all three denominators and rearranging we end up with a quadratic equation for

R0, R02+ 2RR0� 2R2 = 0. We use the quadratic formula to solve, with the negative root thrown

out. The result is R0 = (p3� 1)R.

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