OXFORD New Century Senior Physics  by Richard Walding, Greg Rapkins and Glenn Rossiter 

NOVEL CHALLENGE ANSWERS
 from Richard Walding
This page has the answers to all the Novel Challenge questions in the text boxes in the
margins.
Text box page 2
NOVEL CHALLENGE
Here are a few “Fermi” Questions (named after US physicist Enrico Fermi who used to drive his students nuts with them).
(a) How fast does hair grow?
Answer: 0.44 mm/day (http://science.howstuffworks.com/question251.htm). Hair growth is fastest from the age of sixteen to the late twenties. New hairs grow faster and the growth rate slows down with the increasing length (almost half the rate when the hair is over three feet long). http://www.stophairlossnow.co.uk/Hair%20Structure.htm
(b) How many piano tuners in your capital city? Answer: 41 in the Brisbane Yellow Pages (www.yellowpages.com.au)
(c) How many pingpong balls can you fit in a suitcase?
Answer: The new international standard size for a table tennis (ping pong) ball is 40mm diameter. The volume of a cub with a side of 40 mm is 64000 mm^{3}. A typical large Chinesemade suitcase is 24” x 17” x 9” (600 mm x 400 mm x 230 mm) with a volume of 5.52 x 10^{7} mm^{3}. Hence 862 balls will fit into this suitcase.
(d) How fast does grass grow? Answer: Blue couch: Between 2050mm every 714 days in Spring/Summer; Green couch: between 1525mm every 7 days in Spring/Summer (http://www.centenarylandscaping.com/static/YourTurfChoice.htm)
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NOVEL CHALLENGE
The four compass directions North, East, South, West are derived from old foreign words.
Can you match up the original meanings with the compass directions:
A – Indoeuropean wes = sun goes ‘down’; B  Italian nerto =
‘to the left’ as one faces the sun; C – German suntha = region in
which the ‘sun’ appears in the northern hemisphere; D – Indoeuropean aus
= sun ‘rises’.
ANSWER: A = west; B = north; C = south; D = east.
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NOVEL CHALLENGE
If you were transported in a time machine to an unknown date in Australian history, how
could you work out the date?
ANSWER: Several ways: (a) you could look at the fauna (eg dinosaurs) and relate this to
their extinction history; (b) you could look at the position of the Sun with respect to
the constellations. This would give you some idea but because the "precession"
cycle repeats itself, you’d have to know which billion years you’re in
beforehand; (c) if you knew the changes in sea level you could work out a rough time, but
only if you knew which global warming cycle you were in; (d) you could check which version
of Windows your school was using and add 10 years on to that. Any other suggestions? Check
if any sausage rolls at the tuckshop had gold watches for long service?
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NOVEL CHALLENGE
You have two 100page volumes of a dictionary on your shelf.
A worm eats its way from Volume 1 page 1 through to Volume 2 page 100. How many pages does
it eat through?
ANSWER: None (or two at the most).
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NOVEL CHALLENGE
Consider the Earth to have a circumference of 40 000 km and a ribbon to be put tightly
around it. If you cut the ribbon and inserted a 30 cm piece, how far will the ribbon be
from the Earth if it was evenly spaced?
ANSWER: 4.8 cm (the 40 000 km circumference makes no difference). Use C = 2p r and for a circumference of 30 cm, r works out to 4.8 cm. If you
try it the long way you won’t get a significant difference.
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NOVEL CHALLENGE
Quick now – is a physics lesson longer or shorter than a microcentury?
ANSWER: Shorter (well, presuming your lesson is less than 52 minutes). Forty minutes is
long enough.
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NOVEL CHALLENGE
In the first paragraph of Charles Dickens’ The Pickwick Papers, he states that
he was at the bottom of a deep well and could see the stars in the daytime. Aristotle made
the same claim in On the Generation of Animals in 350 BC. Is this possible? Propose
points for and against this idea.
ANSWER: No, you can see the stars. All you see is a brilliant white light. We asked a
Cornish tin miner what he recalled seeing and he said "bright sunlight". So
there!
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NOVEL CHALLENGE
What time is it at the South Pole.
ANSWER: They use Eastern Standard Time (Brisbane).
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NOVEL CHALLENGE
This book is printed on paper classified as 80 gsm (80 grams per square metre). The cover is made of 249 gsm paper. What should be the mass of this book? Check it and see. What went wrong?
Answer: Size 276mm x 209mm x 43mm; 800 pages = 400 sheets with an area of 0.0577 m^{2} at 0.080 kg per sq. metre = 1.8459 kg.
Covers (2) at 0.249 kg/m^{2} = 0.0287 kg; spine 0.043m x 0.276m at 0.249 kg/m^{2} = 0.0029 kg. Total mass = 1.8775 kg. Actual mass = 1.91 kg. Pretty close. Maybe it’s the glue and stitching (?).
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NOVEL CHALLENGE
Under the system of measurement adopted during the reign of Queen Elizabeth I:
1 mouthful = 1 cubic inch
1 handful = 2 mouthful
1 jack = 2 handful
1 gill = 2 jacks
1 cup = 2 gills
1 pint = 2 cups
1 quart = 2 pints
If 1 cubic inch = 14.7 mL, how many cups to 1 litre?
ANSWER: 1 L = 68 cubic inches = 68 mouthfull = 34 handfull = 17 jacks = 8.5 jills = 4.25
cups.
Let me tell you a story about jacks and jills. Remember the nursery rhyme about:
Jack and Jill went up the hill,
to fetch a pail of water.
Jack fell down and broke his crown,
and Jill came tumbling after.
In the mid1600s, English king Charles I placed a tax on beer and spirits to raise money
for his own pleasure. At that time drinks were sold by the jackpot which had the
volume of one jack, and the jill (or gill). His subjects resented this new tax. They resented
it even more when he reduced the size of the jack and jill to increase his revenue even
further. Under his tough rule protests had to be disguised so the people made up the rhyme
about jack and jill. The words about "jack and jill went up the hill" refers to
the increase in price for a jack and jill of drink as the volume was decreased; "jack
came down" is about the measure returning to its original size; "broke his
crown" refers to Charles I being almost toppled from the throne as the people
revolted; and "jill came tumbling after" means that the jill returned to the
original size.
If you know what being patched with vinegar and brown paper means, please tell us.
Update: Physics student Amy of Brisbane writes: My Mum says that "Vinegar and
Brown Paper" was an old remedy by the English for healing that was used as frequently
then as bandaids are used now. I also found out from my own research that the four line
ending to the poem was actually added later on to make the rhyme more suitable for
children to sing, and hence the lyrics above were never in the original version.
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NOVEL CHALLENGE
Question about Schneider vs Taylor.
ANSWER: Back in the 13^{th} century, surnames were first used and these were
related to occupations. People who made clothes (tailors) adopted the surname Tailor or
Taylor (Schneider in German); people who were blacksmiths took the name Smith (or Schmidt
in German). Smiths were big strong men whereas Tailors were generally smaller and not so
strong. The average mass of Smiths was higher than for Tailor (by about 10 kg back then).
This difference is now down to about 3 kg. People with the surname King – don’t
think your ancestors were kings. During the May (spring) Festival, people often dressed up
as kings, queens, soldiers etc for fun and sometimes took their surname from this.
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10
NOVEL
CHALLENGE
A testable hypothesis: that students in the top half of my maths class will
.....
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10
NOVEL
CHALLENGE
People not only shrink in height as they get older, but also during the day.
Some of our students shrink by 1½ cm between first and last lesson. What is the
reason for this? Can you find factual support for your suggestion? Do you think
taller people shrink more than shorter ones? Does everyone shrink by a certain
percentage? Do younger and older people shrink by the same percentage?
Answer: Using a stadiometer, stature was measured with an accuracy of I mm in
eight young adults. The mean circadian variation was 19.3 mm (1.1% of stature).
Fiftyfour percent of the diurnal loss in stature occurred in the first hour
after rising. Approximately 70% was regained during the first half of the night.
http://humanicses.com/stadiometer.htm
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NOVEL CHALLENGE
Humans have 10^{14} cells at a diameter of 0.01 mm each. If they were placed in a line, how many times around the Earth would they go? The radius of the Earth is 6.38 x 10^{6} m.
Answer: 10^{14} x 0.01 x 10^{6} m = 1 x 10^{6}m
Circumference = 2 x 3.14 x 6.38 x 10^{6} = 4.0 x 10^{7} m; No. of times = 40.
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NOVEL CHALLENGE
A brochure for the 1963 Ford Falcon said it averaged 26 miles per gallon of petrol. A 2002 Falcon is reported to use 12 L of petrol per 100 km. (a) which is the more economical; (b) develop a conversion formula from mpg to L/100km; (c) in 1963, the standard Falcon engine had a capacity of 170 cubic inches whereas the 2002 Falcon has a 4.5 litre engine. Which is the bigger?
Answer: (a) Old Falcon: 26 mpg = 26 x 1.61 km/gallon = 41.86 km/gallon = 41.86/4.546 L = 10.87 L/100km; New Falcon is 12L/100km so old Falcon is more economical.
(b) Conversion: L/100km = 282.5/mpg;
(c) 1 cu. in. = (2.54)^{3} = 16.445 cm^{3}; 170 cu. in. = 2.786 L
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NOVEL CHALLENGE
Number sense question.
ANSWER: The hypothesis being tested was that the further a number is away from a given
number the easier it is to say whether it is greater or less than the given number. What's
more  it's true that it is harder when the numbers are close together. Even scientists
take longer to figure it out.
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NOVEL CHALLENGE
Famous biologist Charles Darwin described the size of a canary finch…etc. Convert the
original measurement to centimetres using the correct number of significant figures.
ANSWER: A few possibilities: (a) his ruler was in 64ths and this was the usual way to
express measurements; (b) 3 ½ inches has only 2 significant figures and Darwin wanted to
be more precise so he used 3 ^{32}/_{64} which has 3 sf. In centimetres,
the measurement is 3 ^{32}/_{64} x 2.54 = 8.89 cm (to 3 sf).
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NOVEL CHALLENGE
In the English translation of a manual on violin playing by the great HungarianGerman
teacher Carl Flesch, it told budding violinists to "lift your fingers 0.3937 inches
from the fingerboard". Why is this funny? What do you suppose the original
measurement was? Rewrite the inches measurement with the correct number of significant
figures.
ANSWER: It is funny (to us anyway – maybe not to you) because it has a ridiculous
number of significant figures. A distance of 0.3937 inches = 0.3937 x 2.54 cm (= 1 cm).
Carl Flesch would have written "lift your fingers 1 cm from the fingerboard"
which has one significant figure. In inches this would be 0.4 inches.
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NOVEL CHALLENGE
In a shop in North Walsham, Norfolk, the height restriction to its carpark is written as
2300 mm. Is there anything wrong with this? Explain!
ANSWER: They obviously meant the measurement to be 2.3 m and so strictly speaking it is
correct. But to most people it would look like a car 2299 mm high would fit but not one
2301 mm high. This is plainly absurd. They should have written 2.3 m.
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NOVEL CHALLENGE
Muttabuttasaurus question.
The age estimate 30 million years would be 30 million give or take a few million (probably
2535 million) seeing it has only one significant figure. A change of 20 years would not
make this number any different. To one significant figure it would still be 30 million
years. (We think the real age of the fossil in the Queensland Museum is 100 million years
old – they’d died out about 60 million years ago).
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NOVEL CHALLENGE
The rate at which hydrogen is consumed on the Sun is proportional to the Kelvin temperature raised to the power of 20 (Rate µ T^{20}). How much faster is the rate at 6000 Kelvin than it would be at 5000 Kelvin?
Answer: Rate = 6000^{2}/5000^{20} = 1.2^{20} = 38 times.
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NOVEL CHALLENGE
19 is about 20.
ANSWER: 20 has one significant figure so really could mean anything from 15 to 25. The
number 19 has 2 sf so could only mean something from 18.5 to 19.5.
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NOVEL CHALLENGE
Tesco lemonade question….
ANSWER: The 8 kJ per 100 mL may have been 8.4 kJ rounded off to 1 sf so that two lots of
8.4 = 16.8 kJ which is 17 kJ to 1 sf. Just drink the 100 mL size and you won't get as fat.
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NOVEL CHALLENGE
(a) How many golf
balls will fit in a suitcase?
(b) How many hairs are there on a human head?
Answer: The average number of hairs on
the scalp of an adult is 100,000. Blondes have more hairs with around 140,000
while redheads have the least with about 90,000. (http://www.stophairlossnow.co.uk/Hair%20Structure.htm)
(c). How fast does human hair grow (in km/hr)? Answer: ½ inch/month or
0.44mm/day.
(d) If all the people of the world were crowded together, how much area would we
cover?
Answer: Population in January 2005
was 6.446 x 10^{9} people. Say the average person approximates a
rectangle 0.45 m x 0.35m (= 0.1575 m^{2}). Total area = 1.015 x 10^{9}
m^{2} = 1015 km^{2}.
(e) What is the relative cost of fuel (per kilometer) of rickshaws and cars?
Answer: A car typically used 10L per 100km so that's 0.1 L/km. At say 90c/L that
costs 9c. A rickshaw would typically travel at 10 m/s which means 100 s per km.
A rickshaw driver might eat $10 worth of food per day to sustain himself, so 100
s would cost about 1.1 cents.
(f) How far does a car travel before a one molecule layer of rubber is worn off
the tires?
Answer: The tires on a passenger car are meant to last 60,000 to 100,000 km
http://people.howstuffworks.com/champcar5.htm. A typical tyre for a Holden
Commodore Executive is the Bridgestone RE92 which has a diameter of 726 mm and
is 205 mm wide and a tread depth of 10 mm. So it takes say 60000 km to wear off
8 mm down to the legal minimum of 2 mm. An average molecule of rubber
(isoprene) is about 6 nm long when coiled and 180 nm long extended. Its width is
in the order of 1 nm. So if it takes 60000 km to wear off 8 x 10^{3} m,
how long to wear off 1 x 10^{9} m? This works out at 7.5 m per molecule
thickness.
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NOVEL CHALLENGE
A lizard runs 30 m West, rests and heads 40 m North where it meets the base of a tree. It scampers 5 m straight up the tree. What is the magnitude of its displacement? How are you going to indicate the angle?
Answer: Displacement = 51 m; heading W53°N, elevation 11.3° to horizontal.
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NOVEL CHALLENGE
Confirm or refute the following statement: “when an object is moved, its displacement can be smaller than the distance traveled, but the distance travelled can never be smaller than the displacement”. Statement is true.
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33
NOVEL CHALLENGE
The 08:00
express from Cleveland to Brisbane arrives at 9:00, while the 08:30 from
Brisbane to Cleveland arrives at 9:30. Assuming both trains travel at constant
speed, at what time should they pass each other? Answer: 8.45am
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NOVEL CHALLENGE
Here’s an interesting theory that could be investigated experimentally. R. McNeill Alexander from Leeds University, England, measured the speed at which animals switched from walking to running. For humans, the speed is about 8 km/h. He developed a rule which stated mathematically is: v^{2} = ½ g d_{H} where v is the speed at which an animal switches, d_{H} is the distance from the hip to the ground, and g is the acceleration due to gravity. His rule applies to animals from insects to humans. Can you confirm this rule by experiment?
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NOVEL CHALLENGE
You have learnt that the rate of change of position with respect to time is velocity, and the rate of change of velocity is acceleration. Did you know that the rate of change of acceleration is known as jerk (symbol j). Jerk is important when evaluating the destructive effect of motion on a mechanism or the discomfort caused to passengers in a vehicle. The movement of delicate instruments needs to be kept within specified limits of jerk as well as acceleration to avoid damage. When designing a train the engineers will typically be required to keep the jerk less than 2 m s^{3} for passenger comfort. In the aerospace industry they even have such a thing as a jerkmeter; an instrument for measuring jerk. In the case of the Hubble space telescope, the engineers specified limits on the magnitude of the rate of change of jerk. There is no universally accepted name for this fourth derivative. Question: Is the slope of, or area under, an a/t graph related to jerk. Does the slope of, or area under, a jerk/time graph mean anything?
Answer: No!
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NOVEL CHALLENGE
A car travels from A to B at an average speed of 100 km/h and returns at 60 km/h. What is
the average speed for the journey?
ANSWER: 75 km/h. This is an old trap. Assume the distance between A and B was 600 km. It
would take 6 hours to go there and 10 hours to come back. That’s a total of 16 h for
a 1200 km trip. Hence 1200/16 = 75 km/h. Simple huh?
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NOVEL CHALLENGE
A column of troops 3 km long is marching along a road. An officer rides from the rear to
the head of the column and back once, and he reaches the rear of the column just as an
advance of 4 km has been made from where he first left. How far did he ride?
ANSWER: 8 km (see working below):
1. Let the officer on the horse be "O"
2. Let the rider at the rear of the column be “R"
3. The velocity of the officer to the ground (v_{OG}) = velocity of officer to rear (v_{OR}) plus velocity of rear with respect to ground (v_{RG}). That is, v_{OG} = v_{OR} + v_{RG}
4. Let time for O to reach head of column be t_{1}. Hence t_{1} = s_{1}/v_{OG}
5. In this time R has moved a distance s_{2}. Hence t_{1} = s_{2}/v_{RG}
6. Both are equal to t_{1}, hence: s_{1}/v_{OG} = s_{2}/v_{RG}
7. But s_{1}= 3 + s_{2} thus (3 + s_{2})/v_{OG} = s_{2}/v_{RG}
8. Rearranging:(3 + s_{2})/s_{2} = v_{OG}/v_{RG}
9 Let the distance O travels from the head of the column to the rear be s_{3}
10. Let the time for O to travel from head of column back to rear = t_{2} (= s_{3}/v_{OG})
11. In this time R has moved a distance s_{4} in a time t_{2} (= s_{4}/v_{RG})
12. But s_{4} = 4  s_{2}
13. Substituting into equation in line 11: t_{2} = (4  s_{2})/v_{RG}
14. But s_{4} = 3  s_{3} = 4  s_{2}. Hence s_{3} = s_{2} 1.
15. Substituting into line 10: t_{2} = s_{3}/v_{OG} = (s_{2}  1)/v_{OG}
16. Hence t_{2} = (s_{2}  1)/v_{OG} = (4  s_{2})/v_{RG}
17. Rearranging: v_{OG}/v_{RG }= (s_{2 } 1)/(4  s_{2}), but this equals (3 + s_{2})/s_{2} from line 8.
18. Hence (s_{2}  1)/(4  s_{2}) = (3 + s_{2})/s_{2}
19. Rearranging: (s_{2}  1)s_{2} = (3 + s_{2})(4  s_{2})
20. s_{2}^{2}  s_{2} = 12  3s_{2} + 4s_{2}  s_{2}^{2}
21. 2s_{2}^{2}  2s_{2} 12 = 0
22. Simplify: s_{2}^{2}  s_{2}  6 = 0
23. Factorise: (s_{2}  3)(s_{2} + 2) = 0. Hence s_{2} = 3 (ignore 2)
24. If s_{2} = 3, then s_{1} = 6, s_{3} = 2, s_{4} = 1
THUS  the officer travels s_{1} + s_{3} = 6 + 2 = 8 km
By the way, the velocity of the officer and column can be found:
25. Substituting into equation 7: v_{RG} = v_{OG}/2
26. But v_{OG} = v_{OR} + v_{RG}, hence 2v_{RG} = v_{OR} + v_{RG}, thus v_{RG} = v_{OR}
27. In the total time of t_{1} + t_{2}, R has traveled 4 km, hence: v_{RG} = 4/(t_{1}+t_{2})
28. Substituting for t_{1} (equation 4) and t_{2} (equation 15): (using actual values of s_{1} and s_{3}):
v_{RG} = 4 /((6/v_{OG}) + 2/v_{OG})) = 8/v_{OG}
29. Hence v_{RG} = 8/v_{OG} = v_{OG}/2 (from equation 25)
30. v_{OG}^{2} = 16, hence v_{OG} = 4, and v_{RG} = 2.
The officer travels at 4 km/h and the column travels at 2 km/h.
The times are t_{1} = 1.5 hr, t_{2} = 0.5 hr. Total time 2 hours.
Text box page 40
NOVEL CHALLENGE
A man goes from A to B at 30 km/h. How fast must he return to average 60 km/h for the
whole trip?
ANSWER: Impossible. He’d have to return at an infinite speed (that is – in zero
time). If you said 90 km/h then all we can say is "sucked in".
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NOVEL CHALLENGE
A boy is carried up an escalator in 1 minute. He can walk up a stationary escalator in 3
minutes. How long will it take him to walk up a moving escalator?
ANSWER: Let the distance up the escalator be x metres.
The velocity of the boy with respect to the escalator v_{BE} = x/3 metres/minute;
The velocity of the escalator with respect to the ground v_{EG }= x/1
metres/minute;
Let the velocity of the boy with respect to the ground be v_{BG} (to be found).
Hence: v_{BG} = v_{BE} + v_{EG} = x/3 + x/1 = 4x/3 metres/minute;
But v_{BG} = x/t, hence x/t = 4x/3; thus t = ¾ minute.
The question is hypothetical  the boy would probably be armwrestled to the ground by the
security men and chucked out of the shopping centre.
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NOVEL CHALLENGE
If you put a row of coins on a 1 metre ruler which has one end on the ground and let the
other end fall – which coins will stay on the ruler and which ones will be left
behind?
ANSWER: All the ones up to the centre will stay put. When the ruler falls, the centre of
gravity will fall with an acceleration of 9.8 m s^{2} whereas the end touching
the table will stay at rest and the free end will accelerate faster than 9.8 m s^{2}.
Try it.
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NOVEL CHALLENGE
If a flea was as big as a man, how high could it jump (approximately)?
ANSWER: A flea can jump maybe 30 cm but humans are about 1800 mm high and a flea is about
1/10 mm high (we guess). So we’re 18000 times as high as a flea so if a flea can
normally jump 30 cm, a humansized flea could jump 18000 x 30 cm = 5.4 km. Now that’s
high.
See other answer below.
NOVEL CHALLENGE
A
flea (Pulex irritans) can jump about 4 m high. If the flea was a big as a
person, how high would it be able to jump (proportionally)?
Answer: a flea
is about 2.5 mm long and just as high. I don’t really think it can jump 4 m high
– more like 90 cm. However, if it could it would be jumping 4000/2.5 times its
own height (= 1600 times). A human of 1.8 m height could jump 2.88 km high.
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NOVEL CHALLENGE
A ranger at Mt Mungo National Park published an booklet entitled "Twenty Family Walks". In the introduction he wrote, "The walks are short, ranging from a kilometre and a half to five kilometres; the average is two and a half kilometres."
1. What is the total length of all twenty walks?
2. What is the greatest possible number of walks over four km long?
3. If there are three walks of 5 km each, what is the greatest possible number of walks less than 5 km but over 2½ km?
Answer: (1) 50 km; (2) 6 walks over 4 km; (3) 9
If you get Q2 wrong, like I did first time around, the logic is thus: The 50 km = 1 walk of 5km + N walks of 4.001km + (19N) walks of 1.5km. Hence: 5 + N*4 + (19N)*1.5 = 50. Therefore N = 6.6, which means 6 full walks of at least 4km. Testing:
8 walks of 4.000000000001 + 11 at 5km + 1 at 5km = 53.5km (too long)
7 walks of 4.000000000001 + 12 at 5km + 1 at 5km = 51.0km (too long)
6 walks of 4.000000000001 + 11 at 5km + 1 at 5km = 48.5km (okay)
Thank you to Chris Murphy for pointing out the mistake.
Text box page 60
NOVEL CHALLENGE
Run or walk in the rain?
ANSWER: The number of raindrops that strike your body is made up of two components: (i)
the number of drops that fall on you head and shoulders (D_{HS}) and (ii) the
number that you collect on the front of your body as you sweep through the volume of air
containing the falling drops (D_{SV}). The total number of drops on your body (D_{T})
= D_{HS} + D_{SV}. The D_{HS }is independent of your speed –
you always sweep out the same volume of air whether you run or walk. The D_{HS}
however, depends on how long you are in the rain. The faster you run the fewer drops
strike your head. So – in vertical rain – run as fast as you can. Rain at an
angle is too complex. Wear a raincoat and run.
Text box page 63
NOVEL CHALLENGE
A passenger on a train travelling at 60 km/h observes that it requires 4 s for another
train 100 m long to pass her by. What is the speed of the second train?
ANSWER: 30 km/h
Let train A be travelling at 60 km/h with respect to the ground (= v_{AG});
The velocity of train B with respect to train A is 100m/4s = 25 m/s or 90 km/h;
Because train B is travelling in the opposite direction to train A: v_{BA }= 90
km/h or v_{AB} = 90 km/h;
v_{AB} = v_{AG} + v_{GB }
v_{AB} = v_{AG} + v_{BG }
90 = 60 + v_{BG }
v_{BG} = 30
hence v_{BG} = 30 km/h
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74
NOVEL
CHALLENGE
The
average mass of a Sumo wrestler in 1974 was 126 kg. In 2003 their average mass
had risen to 157 kg. If this trend continues, when will they no longer be able
to stand up (the maximum mass two legs can stand is 180 kg)? Answer: at 24.6
kg/year he would reach this mass in 2027.
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NOVEL CHALLENGE
A fan blows a cart with a sail attached.
If the fan and the sail are on the same cart what happens. Explain why.
ANSWER: The air pushes the sail one way but the fan pushes against the air (Newton’s
third law) the other way. So they sit there doing nothing. Zilch!
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NOVEL
CHALLENGE
Isaac
Newton’s mother said that he would fit into a “quart pot” at birth. If the
density of a baby is 1020 kg m^{3}, calculate his mass. Answer: 1 quart
= 1.136 L = 1.1365 x 10^{3} m^{3}; mass = DV = 1020 x 1.1365 x
10^{3} = 1.159 kg.
Text box page 80
NOVEL CHALLENGE
Issac Newton’s early childhood was marked by rejection and hatred. He has been ranked
as the second most influential person in the world (influential not important). Develop an
argument for who might be 1^{st} and 3^{rd} ?
ANSWER: In Michael Hart’s book "The 100", subtitled "The most
influential persons in history", he lists Mohammad first, and Jesus third. Check it
out in the bookstore (published by Simon and Schuster, 1999). Don’t email us with
your complaints.
Text box page 80
NOVEL CHALLENGE
(a) If you took a beaker of wet sand from the beach, would it weigh more or less than the
same volume of dry sand?
(b) Stand on some wet sand at the beach and it goes dry around your feet. Why is this?
ANSWER: (a) More (but we’re not game to check).
(b) Because it is disturbed from its closest packing arrangement by the weight of your
body. The grains are forced into a less compact arrangement and puffs outwards and looks
dry.
Text box page 81
NOVEL CHALLENGE
Newton and science.
ANSWER: We think Newton invented them. Laws only exist because they have been
created by someone. Radioactivity gets discovered, the laws of radioactive decay get
invented. What about laws that make you wear a helmet when riding a bicycle? Were they
discovered by Mr Stackhat?
Text box page
81
NOVEL
CHALLENGE
For m_{1}
and m_{2} to remain in the same positions relative to the cart, what
force (F) has to be applied.
Answer: Assume no friction. F on m_{1} = m_{2}g; a of m_{1}
= F/m = m_{2}g/(m_{1}+m_{2}). F applied has to
accelerate cart and m_{1} and m_{2} at m_{2}g/(m_{1}+m_{2}).
Hence: F = ma = (m_{1}+ m_{2} + m_{cart} x m_{2}g/(m_{1}+m_{2})).
Text box page 82
NOVEL CHALLENGE
Quick now, could you lift a ball of cork 1.5 m in diameter? Now work out its mass. Answer: D_{cork} = 240 kg m^{3}, V = 4/3pr^{3} = 1.767 m^{3}; m = 1.767 x 240 = 424 kg.
Text box page 83
NOVEL CHALLENGE
For most cars, the rear tyres support more weight than the front tyres. For example a Toyota Corolla has 43% of its weight supported by the front tyres and 57% by the rear. When a Corolla brakes, the weight on the front increases to about 69% and reduces to 31% on the rear. Why is this and why do cars dip when the brakes are applied? Answer: Hmmmm!
Text box page 83
NOVEL CHALLENGE
If a 1 kg mass hanging on a spring balance shows a weight of 10 N as in the figure on the left, will the figure to the right be correct? It shows a 1 kg mass suspended over a pulley by a string tied to the table. Explain. Answer: Floor applies a force of 10 N to string and mass applies a force of 10 N to string. Total downward force = 20N.
Text box page 84
NOVEL CHALLENGE
In the summer of 1693, John Bernoulli posed the following problem which still hadn’t
been solved 6 months later. The day Newton heard it he solved it. The problem: you have
three paths for a ball to roll down – which is the fastest?
ANSWER: The cycloid (middle) path is similar to the path traced out by a point on the rim
of a wheel as it rolls along a horizontal surface. It is the path of least time.
Text box page 84
NOVEL CHALLENGE
Armstrong’s weight question.
ANSWER: On Earth, he would weigh 3 x 1000 ounces = 3000/16 pounds = 187.5 lb. This is
equivalent to a mass of 187.5/2.2 = 85.2 kg.
Text box page 86
NOVEL CHALLENGE
A man comes up to a bridge which can just support his weight and 1 of the 4 balls he is
carrying. He decides to juggle as he crosses so that only 1 ball will be in his hands at
any one time. What do you think of his solution?
ANSWER: Terrible. Although three are in the air, he has imparted a force to the 4^{th}
ball greater than its weight (to make it accelerate upwards). This is greater than the
bridge can stand.
Text box page 89
NOVEL CHALLENGE
Throwing a ball while parachuting.
ANSWER: the drag on a tennis ball is more than on a skydiver in freefall (see page 87 for
some terminal velocities) so the ball would appear to fly upwards as they fell. Once they
opened their parachutes the drag on the ball would be less than on them (see table, page
87) so the ball would start to fall downwards towards them.
Text box page 89
NOVEL CHALLENGE
Imagine you were to drop a book which has a paper napkin resting on the top. How will they
both fall?
ANSWER: The napkin stays on the top of the book. There is a decrease in pressure on the
top of the book which keeps the paper napkin in place. It like cyclists being
‘sucked’ along behind a truck. Don’t try it.
Text box page 89
NOVEL CHALLENGE
If you dropped a marble, a big styrofoam ball and a small one (from a bean bag) together
from chest height, two hit the ground at the same time. Which two and would they be faster
or slower than the other one? Why? Try it and see!
ANSWER: The marble and the small ball hit together. There is proportionally more air
resistance for the big styrofoam ball than for the small one (air resistance is
proportional to surface area which varies with the square of the radius). The big ball is
slower.
Text box page 90
NOVEL CHALLENGE
If the objective of a parachute is to slow the descent of a falling object in air, why do
parachutes have a hole (the apex vent) in the top allowing air to escape. In World
War II they didn’t have an apex vent and they swung like pendulums as they descended
(watch an old war movie – you’ll see). What is the physics behind this?
ANSWER: The air has to escape from the parachute somehow so if there is no apex vent it
tilts to the side to let it out. Then it tilts the other way to let it out the other side.
Text box page
90
NOVEL
CHALLENGE
If an
elephant, a man and a mouse fell from the 20^{th} story of a highrise
building, the elephant would splatter on impact and die, the human would be
crushed and die, and the mouse would walk away. Now why is this? Answer: mouse
has lowest terminal velocity as it’s surface area is high compared to it’s
volume.
Text box page 93
NOVEL CHALLENGE
(a) Which box requires the smaller force to lift?
(b) Which box requires less work to raise 1 metre?
ANSWER: The top box requires less force.
They both need the same amount of work to lift them 1 metre.
Why? Because in the top figure when the rope is pulled 1 metre the box only lifts up 0.5
m, This is called "double purchase". In the bottom figure (single purchase) the
box rises 1 metre when the rope is pulled 1 metre. Hence in the top figure, half the force
is required over twice the rope distance to move it compared to the bottom figure. The joy
of pulleys.
Text box page 90
NOVEL CHALLENGE
A monkey has the same mass as a box. He climbs a rope. Who will reach the pulley first?
ANSWER: They both reach it together. As the monkey pulls down on the rope and accelerates
himself upwards a bit, the tension in the rope increases. This will pull the box up with
the same acceleration and will move the same distance up the rope in the same time. As the
monkey pulls again, the same thing happens so they both arrive at the top together. If it
was a spider monkey, he'd probably steal your sunglasses.
Text box page 92
NOVEL CHALLENGE
Motorcycle tyres come in two main types – sport and touring. Sport tyres are
vshaped, touring are rounded. The figure below shows the profiles of Dunlop’s
megasuccessful D207GP race tyre which has been cleaningup in supersport competitions for
some time now.
(a) Plot a graph of the lean angle (from 5° to 50° in 5° increments) against the cosec of
the lean angle. Note when q = 30° ,
cosec 30° = 1/sin 30° = 2. A
computer spreadsheet may speed things up. How does the curve compare to the groove shape
in the D207GP tyre?
(b) As riders using the vshaped tyre go into a corner they flip the bike on to its side
rather than doing a uniform lean. Why is this? The upper limit of contact between the tyre
and the road is called the ‘hero line’. What do you suspect this means?
(c) Cruising bikes like the Harley Davidson have tyres with a flat profile. This must
impair their cornering at speed so why do they have it?
ANSWER: (a) it has the same shape (see figure below).
(b) to get maximum contact area straight away.
(c) a measure of what sort of hero you are is how far over you can lean. There’s a
fine line between being a hero and an idiot.
Text box page 96
NOVEL CHALLENGE
A piece of pine dowel is placed in an electric drill and rotated against a piece of
hardwood. Which will catch on fire first? Try it.
ANSWER: The pine. We tried it. The fibres in hardwood are more tightly packed and act as
an insulator against the heat spreading.
Text box page 96
NOVEL CHALLENGE
A block of wood is placed on top of a smooth cylinder.
When the block gets to 45° it slides off. If a ball was used
and allowed to roll down the surface, would it fall off at a bigger or smaller angle than
45° ? Justify your answer.
ANSWER: The rolling ball will stay on longer. If you can do the mathematics (we
can’t), the sliding body leaves at 48° and the rolling
body leaves at 54° . We are told that the sliding block leaves
when the cosine of the angle is 2/3 and the rolling ball leaves when the cosine is 10/17.
Simple huh?
Text box page 96
NOVEL CHALLENGE
A cubical block of mass 10 kg is placed on surface where m =
1.0. Where about on the block would you have to push it so that it was on the verge of
tipping and sliding?
ANSWER: Push it in the middle of the vertical face. This will make it be on the point of
tipping and sliding. If you push it higher it will tip; lower and it will slide.
Text box page
98
NOVEL
CHALLENGE
In an
extreme skiing competition in
Alaska
in 1995, a New Zealand woman tumbled 400 m down a 50°
slope. The slope used in the Olympics is 35°. She ended up with severe head trauma which
tended to improve humanity’s gene pool. How fast would she be going if the
coefficient of friction was (a) zero, (b) 0.10?
Text box page 96
NOVEL CHALLENGE
A wooden block is put on an electronic balance and it reads 100 g (weight = 1 N). A string
is attached by a spring balanced and a force of 0.4 N applied at an angle of 45° . What will the balance read? Try it to check.
ANSWER: The vertical component of the force applied is 0.4 sin 45°
= 0.28 N;
Hence the balance registers the weight less the upward pull (1 – 0.28 = 0.72 N);
The balance would then read 72 grams. We do this demonstration in class and it works like
a charm.
Text box page 100
NOVEL CHALLENGE
A 10 kg block …etc.
ANSWER: Push on the 5 kg block.
Solution: Consider the push on the 5 kg block:
Total friction = F_{f} of 10 kg block + F_{f} of 5 kg block
= m x 100 + m (mg + mg sin 30° ) = 50 + 0.5(50 + 50 x 0.5) = 87.5 N
Consider the push on the 10 kg block:
Total friction = F_{f} on 10 kg block + F_{f} on 5 kg block
= m (mg + mg sin30° ) + m x 50 = 0.5(mg + mg sin30° ) + 0.5 x 50
= 75 + 25 = 100 N
Text box page 100
NOVEL CHALLENGE
Fermi Question: What force is required to break a blade of grass by pulling at each end?
Text box page 101
NOVEL CHALLENGE
With the automatic gearbox in “drive” a Toyota 1200 kg RAV 4 will remain stationery facing uphill on a 5° slope. What would its initial acceleration be on the flat (assuming the driver’s foot was not on the accelerator)? We got 0.87 m s^{2}.
Text box page
104
NOVEL CHALLENGE
In 1999, a 19year old Gold Coast man tried to stop a car starting to roll down a driveway slope. He didn’t succeed and was run over. Why couldn’t he stop the car – after all, it was only on a 10° slope? How much force can you push with? Try this: have someone hold a set of bathroom scales against the wall at about hip height and you should push them with your hands as hard as you can. You’ll probably only push to a scale reading of 40 kg (400 N). Calculate the maximum angle of an incline that you could stop a 1500 kg car from rolling down. Surprising huh?
Answer: F_{W}sinq = 400N; 1500 x 10 x sinq = 400. Angle q = 1.53°
Text box page
105
NOVEL CHALLENGE
I live at the top of a road that has a 5° downhill slope. When I let my car roll down the slope it reaches 25 km h^{1} by the time it gets to the bottom, 400 m away. What frictional force must be acting? Answer: need the mass of the car (try 2000kg).
Text box page 105
NOVEL CHALLENGE
A lawyer emailed me (RW) wanting to know what was meant when someone was “in an accident and suffered highg forces”. I told him in a few paragraphs. How would you explain it? By the way, he never offered to pay but I bet he charged his client. Hmmm!
Text box page
111
NOVEL CHALLENGE
A flea can jump 18.4 cm high when jumping at 45°. How far horizontally will it go?
Text box page 112
NOVEL CHALLENGE
Acapulco cliff divers jump off a cliff 35 m high and just miss rocks 5 m out from the base. What is their minimum pushoff speed? Answer: 1.9 m s^{1}.
Text box page 112
NOVEL CHALLENGE
On the Moon, astronauts hit a golf ball 180 m. If they hit the same ball on Earth with the same speed and angle, how far would it go (neglect air resistance). Note g_{moon} = 1.6 m s^{2}. By the way – there are three golf balls still on the Moon. Learn this off by heart – it could be useful.
Answer:
On Moon: s_{H} = v_{H} t
180 = v cosq x t
s_{v} = ut + ½ at^{2}
0 = v sinq t + 0.93t^{2}
t = vsinq/0.93
180 = vcosq x vsinq/0.93
On Earth: s_{earth} = vcosq t_{earth}
s_{earth} = v cosq vsinq/5 = 0.93 x 180/5 = 33.48 m
Text box page 113
NOVEL CHALLENGE
The following graphs show how the range and altitude of a projectile changes with
elevation angle in the presence of air. Plot a graph of maximum altitude versus elevation
angle and predict maximum altitude for angle of 90° . Should
the graph pass through the origin (0,0). Why?
ANSWER: It should pass through the origin. If you shoot it into the ground it will go zero
metres. The graphs are as follows:
Text box page 114
NOVEL CHALLENGE
The world speed record for an archery shot over 100 m is 1.64 seconds (220 km h^{1}). Calculate the elevation angle of the arrow so that it hits the bulls eye at the same height from which it was fired (shoulder high).
Answer: v_{H} = 60.97 m s^{1} = v cos q
s_{v} = 0 = u t + ½ at^{2} = v sin q x 1.64 + 5 x 1.64^{2};
v sin q = 8.2
v = 8.2/sinq
60.97 = 8.2 cosq/sinq = 8.2/tanq
tanq = 8.2/60.97; q = 7.66°
Textbox page 114
NOVEL CHALLENGE
A really hard one! A cannonball is fired and after travelling 5 m horizontally, it has
reached half its maximum height. After what horizontal distance will it land?
ANSWER: 33.3 metres (or 34.14 to be more exact)
This is a difficult, difficult problem and will defy the best physics tutor around.
I (RW) tried several ways but you might find a quicker way.
ONE METHOD:
You can work out what fraction of the time a projectile spends in the bottom half of its
journey (it doesn’t matter what the launch speed is, the fraction will always be the
same). I worked out that it spends 30% of its time in the bottom half. Hence, it travels 5
m horizontally in 30% of its time to get to maximum height. Therefore, it will travel
16.65 m horizontally when it reaches its top of flight. By the time it lands again it will
have travelled 33.3 m.
Alternative solution provided by Steve Shellshear (you’ve really got to get out more Steve).
Angle of projection = q
Initial vertical velocity component u_{v}= vsinq
Initial horizontal velocity component v_{H}= vcosq
Rearranging v_{H} = s_{H}/t gives s_{H} = v_{H} t = vcosq
Substituting: 5 = vcosq t
Rearranging: t = 5/ vcosq (EQUATION 1)
Using s_{v} = u_{v}t + ½at^{2}
h_{max}/2 = vsinq t + ^{}5t^{2} (EQN 2)
h_{max}/2 = vsinq 5/ vcosq + ^{}5(5/ vcosq)^{2}
Multiply both sides by 2:
h_{max}/2 = 10sinq/cosq + ^{}250/(vcosq)^{2} (EQN 3)
0 = (vsinq)^{2} – 20 h_{max}
Rearranging:
20h_{max} = (vsinq)^{2}
Divide both sides by 20:
h_{max} = (vsinq)^{2}/20 (EQN 4)
(vsinq)^{2}/20 = 10sinq/cosq + ^{}250/(vcosq)^{2}
Multiply both sides by 20 (vcosq)^{2}
(vsinq)^{2}(vcosq)^{2} = 200sinq (vcosq)^{2}/cosq – 250 x 20
v^{4} (sinqcosq)^{2} = 200sinqcosq v^{2}  5000
Prepare for identity conversion of 2 sinqcosq = sin2q
v^{4} (2sinqcosq)^{2}/4 = 100 x 2sinqcosq v^{2}  5000
Convert and multiply both sides by 4:
v^{4} (sin2q)^{2} = 400 v^{2} sin 2q  20000
Subtract 400v^{2}sin2q from, and add 20000 to both sides:
v^{4} (sin2q)^{2} – 400v^{2}sin2q + 20000 = 0 (Equation 5)
Vertically, using s_{v} = u_{v}t + ½at^{2}
0 = vsinq t – 5t^{2}
Add 5t^{2} and subtract vsinqt from both sides:
5t^{2}  vsinqt = 0
Divide both sides by 5:
t^{2}  vsinqt/5 = 0
Factorise:
T(tvsinq/5) = 0
Therefore, t = o t = vsinq/5
We are interested in t = vsinq/5 (EQN 6)
R = vcosqt (EQN 7)
R = vcosq vsinq/5
Convert to sin 2q
R = v^{2} (2sinq cosq)/10
R = v^{2} sin2q/10 (EQN 8)
Substitute EQN 8 into EQN 5
100R^{2} – 4000R + 20000 = 0
Divide both sides by 100
R^{2} – 40R + 200 = 0
Using quadratic formula:
R =
(40 ± Ö(40^{2}
– 4 x 1 x 200)/(2 x 1)
R =
(40 ± 28.284)/2
R =
34.14m or 5.86m
Alternative solution by John Liu.
The Cartesian form of the parabolic equation for projectiles is
y= xtanAg(xsecA)^{2}/2v^{2},
where x is horizontal distance, A angle of projection, g acceleration due to gravity (9.8
ms^{2}), and v speed of projection.
From the question, at x=5 the particle reaches half the maximum height, or
y(max)/2=5tanA(25g(secA)^{2})/2v^{2}. Multiply through by 2:
y(max)=10tanA(25g(secA)^{2})/v^{2}
On the other hand, by what we learn in year 12 mathematics, y(max) is ((vsinA)^{2})/2g.
Equating this to the y(max) given above,
10tanA(25g(secA)^{2})/v^{2}=((vsinA)^{2})/2g. Multiplying through by 2gv^{2},
20gv^{2 }tanA50g^{2}(secA)^{2}=v^{4}(sinA)^{2}. Rearranging terms,
v^{4}(sinA)^{2}20gv^{2}tanA+50g^{2}(secA)^{2}=0.
The range, also from year 12 mathematics, is v^{2}sin2A/g. Since the equation above is a
quadratic in v^{2}, solve it for v^{2} gives
v^{2}=(20gtanA+/Sqrt(400g^{2}(tanA)^{2}200g^{2}(tanA)^{2}))/(2(sinA)^{2}), where +/ means plus or
minus, Sqrt means square root of.
This is
(20gtanA+/10gtanASqrt2)/(2(sinA)^{2}), or
(20gsinA+/10gsinASqrt2)/(2sinAcosA), or
(10+/5Ö2)g/sin2A.
Substitute this value into range, we get
v^{2}sin2A/g=(20+/10Ö2)g/sin2A
´ sin2A/g=(20+/10Ö2).
Because 2010Ö2=5.86, this indicates the projectile fell to the ground less than one
meter after taking 5 meters to reach half the maximum height! Therefore it cannot be the
range the projectile travelled. So we must take 20+10Ö2=34.14 as the range.
Answer: the range travelled by the projectile is 34.14 meters.
Alternative by Year 12 student Chao Quio:
Elegant alternative solution
 in MS Word format. Thanks Chao.
Text box page 117
NOVEL CHALLENGE
When a wheel rolls along, is any point at rest?
ANSWER: Yes – where it makes contact.
Text box page 118
NOVEL CHALLENGE
Some coins were placed on a turntable and then it was turned on. What do you predict will
happen?
ANSWER: The outer ones will fly off because they have a higher velocity and therefore a
larger centripetal force is needed to keep them travelling in a circle. The friction
provides the centripetal force but if the centripetal force needed exceeds friction they
will fly off.
Text box page 118
NOVEL CHALLENGE
How many revolutions will coin A do while rotating around coin B? Try it. You’ll be
surprised.
ANSWER: Two
Text box page 118
NOVEL CHALLENGE
Government steamer Relief and it’s whirling windscreen.
ANSWER: The water was flung off by the spinning disk. Car wipers wouldn’t be quick
enough to keep the screen clear. One problem with a spinning disk is getting a good seal
so water doesn’t leak in.
Text box page 113
NOVEL CHALLENGE
A wheel is rolling along with constant speed and a lump of mud is thrown off its hindmost
point. Will it touch the wheel again?
ANSWER: Yes. It follows a parabola.
Text box page 125
NOVEL CHALLENGE
A candle with a nail through the middle is supported on two glasses and lit at both ends.
How could you check if the resulting motion is SHM or just periodic?
ANSWER: Hmmm! This is a tough one. Perhaps you could videotape it or set a weighted spring
oscillating upanddown beside it and adjust the mass to see if you can get its motion to
match the candle.
Text box page 126
NOVEL CHALLENGE
A pendulum has a bucket for the bob and it is halffilled with water. When it freezes,
predict what happens to the period of the pendulum?
ANSWER: Period decreases because the water expands upwards in the bucket as it freezes and
the centre of gravity of the bucket is higher than before. This results in a faster
pendulum.
Text box page 128
NOVEL CHALLENGE
Two sidebyside pendulums are oscillating. One has a period of 6 s and the other a period
of 7 s. If the bobs are touching at one time, how much longer must you wait until they
come together again.
ANSWER: 42 seconds.
Text box page 128
NOVEL CHALLENGE
How to make a swing go higher.
ANSWER: Your answer should be about why you lean back just as you get to the maximum
amplitude backwards. When you lean back you pull on the rope and this has two components:
one component (vertical) lifts you up out of the seat (no use); the other (horizontal)
pulls you forward and increases your speed.
You could also analyse this movement in terms of the conversion of GPE to KE. The higher
your centre of gravity when you are at maximum amplitude, the great your GPE and hence
there is more energy to convert to KE. To get a high centre of gravity you could stand up
or kick your legs high in the air. You should try to stay low as possible at the bottom of
the swing so you tuck your lrgs under the seat. How do you slow down without skidding your
feet? Do you just do the reverse?
Text box page 142
NOVEL CHALLENGE
ANSWER:
The figure below shows the relative positions of some ink dots on the surface of
a balloon as it is inflated. If you keep dot “d” directly in front of you, all
the other dots appear to be moving away from “d” as the balloon get bigger.
Imagine diagram II shows the positions 1 second after diagram I. (a) Calculate
the speed of dots e, f and g by measuring the distances with a ruler. Plot a
graph of speed (yaxis) versus distance from “d”. Relate this graph to Hubble’s
Law. (b) Imagine we are object “d” in the Milky Way galaxy and object “g” is the
Hydra galaxy. If Hydra is moving at a speed of 60 500 km s1, calculate the time
elapsed between Diagrams I and II.
Answer: (a) Points d, e, f, g go from 0, 6.9, 13.4 and 20.1 mm to 0, 13.6,
26.9, 40.2 mm in the 1 second. Their velocities are 0, 6.7, 13.5, 20.1 m/s
respectively. This gives a linear graph. (b) 32.4 years (Hydra is 1960 Mly away)
Text box page 144
NOVEL CHALLENGE
A marathon runner....
ANSWER: There will be numerous answers depending on the relative positions of
Earth and Jupiter in their orbits. Let us consider the two extreme cases where
the Earth and Jupiter are closest to each other (inferior conjunction) and when
they are furthest apart (superior conjunction). The distances apart can be found
from the table on page 144 showing the orbital radii of the planets. For
inferior conjunction the distance apart is the difference in radii (7.78 x 10^{11}
m  1.49 x 10^{11} m = 6.29 x 10^{11} m). For superior
conjunction the distance apart is the sum of the two radii ( = 9.27 x 10^{11}
m). The time taken for a radio signal to travel between the planets ( t = s/v)
is: t_{inferior} = 6.29 x 10^{11} m / 3.0 x 10^{8} =
2097 s. Thus total time (out & back) = 4193 s. For superior conjunction, t_{superior
}= 9.27 x 10^{11} / 3.0 x 10^{8} = 3090 s; total time =
6180 s.
The speed of a marathon runner can be calculated by knowing that (for men) the
40 km race has a record time of about 2 h 10 m. Thus, speed = s/t = 40000/(130 x
60) = 5.1 m s^{1}. Hence the distances a male marathon runner can
travel in the two times above are: (a) for inferior conjunction: s_{runner}
= v x t = 5.1 x 4193 = 21.5 km; (b) for superior conjunction: s_{runner}
= 5.1 x 6180 = 31.5 km. So the complete answer would be that a champion male
marathon runner could run somewhere between 21.5 km and 31.5 km. An "A+" for
this question would require the student to discuss all assumptions (male/female,
champion/novice, inferior/superior).
Text box page 146
NOVEL CHALLENGE
Another new planet outside our solar system has recently been discovered. It
lies within the Oort Cloud with an orbital radius of 0.4 ly. It has a mass 1.5
to 6 times that of Jupiter and a period of 6 million years. How does its r^{3}/T^{2}
compare to that of our Solar System?
ANSWER: r^{3}/T^{2} = (0.4 x 3 x 10^{8}
x 365 x 24 x 3600)^{3} / (6 x 10^{6} x 365 x 24 x 3600)^{2
}= 5.4 x 10^{46} / 3.6 x 10^{28
} = 1.5 x 10^{18} m^{3}/s^{2
}This is about half the value for our Solar System (3.4 x 10^{18}).
Text box page 148
NOVEL CHALLENGE
Newton said rationem vero harum Gravitatis propietartum ex phenomenis nondrum potui
decucere "But I have not been able to discover the reason for this property of
gravitation from the phenomena". What did he mean?
ANSWER: He could work out the mathematical relationships but could not work out the cause
of the attraction.
Text box page 148
NOVEL CHALLENGE
Two spherical drops of mercury are resting on a frictionless surface. The only force
between them is that of gravitation. What would you need to know to be able to calculate
how much time it would take for them to touch?
ANSWER: You would need to know the radius of both drops, their individual
masses, their distance apart and the equation to calculate the time.
I don't know how to work out the time but a Year 12 student from Brisbane
State High sent me his solution. He said "You need to derive a complex equation
using calculus" and his solution goes like this:
Let the two drops initially be a distance of "r" apart between centres.
Let there be "x" number of distance intervals before the drops touch.
Let "d" = distance of each interval. Hence d = r/x
As the number of intervals → ¥,
d → 0
Let q_{1} be the radius of mass 1, and q_{2}
be the radius of mass 2.
Using g = Gm/r^{2}
Mass 2's gfield accelerates mass 1 at a_{1} = Gm_{2/r}^{2}
Mass 1's gfield accelerates mass 2 at a_{2} = Gm_{1/r}^{2}
The acceleration of both objects towards one another is therefore a_{3} = G(m_{1} + m_{2})/r^{2}
As the distance d between the masses decreases the equation becomes: a_{3} = G(m_{1} + m_{2})/(distance apart)^{2}Let's measure the gravitational attraction at intervals of distance "d" and let d → 0. This allows us to take into account all changes in velocity during the event.
Let c = G(m_{1} + m_{2}) for tidiness, and
t_{1} = time taken to travel d at v_{1}
t_{2} = time taken to travel d at v_{2}
v = u + atv_{1} = 0 + c t_{1}/r^{2}
v_{2} = c t_{1}/(rd)^{2 }+ c t_{2}/(r2d)^{2 }
v_{3} = c t_{1}/(rd)^{2 }+ c t_{2}/(r2d)^{2 } + c t_{3}/(r3d)^{2 }+ c t_{4}/(r4d)^{2 } + ....
Because distance d approaches 0, the time taken for the masses to travel distance d also approaches 0
With an infinite number of time periods, the difference between neighbouring t values becomes so incredibly small that we can pretend that they are equal. Now we can find the time t for each mass to pass distance d at each velocity (v_{1}, v_{2}, v_{3}...). I will only find t_{2} to save space.
let t_{2} = t_{1 }
v_{2} = c t_{1}/(rd)^{2 }+ c t_{2}/(r2d)^{2 }thus
v_{2} = c t_{2}/(rd)^{2 }+ c t_{2}/(r2d)^{2}
But v_{2} = d/t_{2 }thus:
d/t_{2} = c t_{2}/(rd)^{2 }+ c t_{2}/(r2d)^{2}
t_{2}/d = [c t_{2}/(rd)^{2 }+ c t_{2}/(r2d)^{2}] ^{1} (invert)
t_{2}/d = (c t_{2})^{1} ´ [1/(rd)^{2} + 1/(r2d)^{2}]^{ 1 } factorize (c t_{2}) out
(t_{2})^{2 }= d/c ´ [1/(rd)^{2} + 1/(r2d)^{2}]^{ 1 }
^{t}_{2} = Ö(d/c) ´ Ö[1/(rd)^{2} + 1/(r2d)^{2}]^{.} This gives the time travelled over d at v_{2}.
Similarly, other values can be solved in this manner, all following the same pattern and form.
Thus: T_{total} = t_{1} + t_{2} + t_{3} + t_{4} + ..... t_{¥}
BUT!!! This is calculating the time until the two centres of the spheres meet. We want to find when the spheres touch.
let x be the number of d intervals until sphere touch = number of Ö [ ] terms in the sequence.
x = [r (q+1+ + q+2+]/d
T_{total }= t_{1} + t_{2} + t_{3} + t_{4} + ..... t_{x}
T_{total }= Ö(d/c) ´ {Ö[1/(rd)^{2}] + Ö[1/(rd)^{2} + 1/(r2d)^{2}] + Ö[1/(rd)^{2} + 1/(r2d)^{2} + 1/(r3d)^{2}] .... Ö[1/(rd)^{2} + 1/(r2d)^{2} + 1/(r3d)^{2} + ...Ö[1/(rxd)] }
This sequence continues in the same pattern forever. Just punch in the values for r and d for the answer.
More d intervals = more accuracy = more time spent solving problem.
Text box page 150
NOVEL CHALLENGE
Two balls are rolled around the inside surface and outside surface of a pipe whose
diameter is 3 times the diameter of the balls. How many times will the outside one turn to
complete one rotation of the pipe compared to the inside one?
ANSWER: Twice. The outer one rotates 4 times while the inner one rotates twice. The
circumference of the circles traced out by the centre of each ball is found by 2p r. For the outer ball, r = 2 x diameter of the ball; for the inner
one, r = diameter of the ball.
Text box page 152
NOVEL CHALLENGE
How far would you have to travel horizontally out from the Earth for your altitude to be 1
km?
ANSWER: 113 km. The distance AC = 6.38 x 10^{6} m (=EC); AB = 1000 m; hence BC =
6.379 x 10^{6} m.
Using Pythagoras’ Theorem, BE = 113 000 m.
Text box page 152
NOVEL CHALLENGE
Text box page
153
NOVEL CHALLENGE
In 1971 Apollo 15 astronauts David Scott and James Irwin drove the 4WD lunar vehicle around for 30 km on the Moon. Would they have used as much fuel as on Earth? No, less air resistance.
Text box page 153
NOVEL CHALLENGE
During a Lunar Eclipse, the shadow of the Moon on the Earth consists of a black region 100 km wide and travels at 3000 km h^{1}. Scientists and thrill seekers try to stay in the shadow zone as long as possible to make observations. What is the maximum time you stay in the shadow zone if you were in a plane that could travel at 1000 km h^{1}? Answer: 3 minutes.
Text box page 154
NOVEL CHALLENGE
If the Earth stopped rotating, how much would a 60 kg (590 N) person weigh? What would
their bathroom scales read?
ANSWER: About 60.4 kg (592 N)
Apparent weight (F_{N}) = F_{W}  F_{C} = 590 – (4p ^{2} r m / T^{2})
590 = F_{N} – 2
F_{N} = 592 N; m = F/g = 592/9.8 = 60.4 kg
Text box page 158
NOVEL CHALLENGE
What would you hear 1 km from the Big Bang.
ANSWER: (a) the easy one: there is no air so there would be no sound; (b) the one we said
you’d miss: at the time of the Big Bang, there was no ‘outside’ of the Big
Bang. Space was being created by the expansion of the explosion. So you couldn’t have
been 1 km outside. Question: could God have been 1 km outside the Big Bang?
Text box page 165
NOVEL CHALLENGE
Which of the following ball bearings will fall slowest?
ANSWER: The one in water at 10° C. The water will be frozen.
Text box page 165
NOVEL CHALLENGE
You can exert a force of 250 N with your incisor teeth and 1220 N with your molars. Which
do you estimate to produce the higher pressure? Note: your front incisors are about 8 mm x
0.2 mm and your molars are about 8 mm x 8 mm.
ANSWER: Incisors have higher pressure.
P (incisors) = F/A = 250/(0.008 x 0.0002) = 1.56 x 10^{8} Pa
P (molars) = F/A = 1220/(0.008 x 0.008) = 1.91 x 10^{7} Pa
Text box page 173
NOVEL CHALLENGE
The relative density of two types of wood is as follows: ironbark 1.3, balsa
0.24. Balsa is specified as 10 pound per cubic foot. Is this about right?
Answer: D = 159.7 kg m3; Relative Density = 159.7/1000 = 0.16, not 0.24.
Text box page 174
NOVEL CHALLENGE
One of our students can float up to her earlobe in a swimming pool when floating erect. (a) Estimate her body density; (b) how high would she float in seawater (SG 1.030 g cm^{3}); (c) would she float higher or lower in cold water; (d) would a boy of the same height and mass float the same way? Answer: (a) 909 kg m^{3}; (b) 19.3 cm from top of head; (c) Cold water more dense therefore would rise; (d) Yes, even though fat is distributed differently in women, al of it is underwater. Top of head position is no different.
Text box page 174
NOVEL CHALLENGE
A boy is 1.8 m tall and can float upright in pool water with only 5 cm above the
water. When he breathes in, his body rises so that it is 25 cm above the
surface. Can you estimate the change in density of his body?
Text box page 174
NOVEL CHALLENGE
Continents float on the liquid mantle of the Earth. Continents have a density of
2800 kg m3 whereas the mantle has a density of 3300 kg m3. If a continent is
35 km thick, prove that the top of the continent is 5.3 km above the mantle
surface. Answer: 2800/3300 x 35 = 29.6 km, therefore 5.3 km above surface.
Text box page 177
NOVEL CHALLENGE
A plastic soft drink bottle is half full of water. A small piece of cork is held just
under the surface by a piece of string glued to the bottom. The bottle is slid to the
right. Which way does the cork move relative to the bottle – forward, backward,
sideways, no movement? You’ll be shocked if you try.
ANSWER: It moves the same way as the bottle. Look at the sloping surface of the water for
a clue as to why. The cork moves to the part of the surface where it can float and have
zero nett forces on it along the surface.
Text box page 178
NOVEL CHALLENGE
A hollow steel ball….etc
ANSWER: This time we get 0.187 cm.
Let final radius of ball be r; and radius of inside is 4 cm.
V_{steel} = 4/3 p r^{3}  4/3 p 4^{3} = 4/3 p (r^{3}
– 4^{3})
m_{steel} = 4/3 p (r^{3} – 4^{3})
x 7.8 grams;
V_{ball} = 4/3 p r^{3 }
When density = 1 it just floats; D = m/V, hence:
1 = 4/3 p (r^{3} – 4^{3}) x 7.8 / 4/3 p r^{3};
solve for r = 4.19 cm; therefore thickness = 0.19 cm.
I don’t know how I got 0.25 cm last time. Must have been dreaming!
Text box page 179
NOVEL CHALLENGE
A 30 cm wooden ruler was placed on a bench with 10 cm overhanging. A page of the Courier
Mail (58 cm x 40 cm) was placed on top of the ruler on the bench. Calculate the weight of
air on the paper and predict what will happen if the overhang is given a sharp blow with
your fist. You won’t believe the weight. Quick now – is it more than the weight
of three Toyota Corollas?
ANSWER: F = PA = 101.3 x 10^{3} Pa x 0.58 m x 0.40 m = 23502 N. A Corolla has a
mass of about 1200 kg (12000 N) so the weight of air is more than two Corollas but fewer
than three.
Text box page 179
NOVEL CHALLENGE
A cardboard tube is placed halfway into a container of puffed wheat.
What do you think will happen when you blow air across the top of the tube. But why and
what does Bernoulli have to do with it?
ANSWER: the puffed wheat gets sucked up and out the top of the tube. Bernoulli found that
when air passes over a column of air you get a lowering of pressure (venturi effect). This
is how a carburretor in a car works.
Text box page 180
NOVEL CHALLENGE
A
square pond measures 100 m by 100 m. A block of ice with a mass of 1000 kg is
floating freely in the pond. How far will the water level rise when the ice
melts? You won’t like the answer.
(The density of water is 1000 kg m^{3}, and ice 917 kg m^{3}.)
Answer: No change. The 1000 kg of ice has weight of 10000N and therefore
displaces 10000 L of water. Even though the ice has a volume of 1000/917 = 1.09
m^{3}, when it melts it produces 1 m^{3} and replaces was
displaced exactly.
Text box page 185
NOVEL CHALLENGE
Imagine you place a finger under either end of a ruler and a coin is placed on one end.
You pull away both fingers and the ruler and coin fall together staying in contact. But if
you just pull away the finger under the coin something odd happens. What and why? Try it.
ANSWER: The ruler falls faster than the coin. This is because one end of the ruler stays
still so the other end accelerates faster than normal (the centre of gravity of the ruler
– the 50 cm mark – accelerates at the normal rate).
Text box page 186
NOVEL CHALLENGE
There are several types of ‘crooked’ dice used by cheats. For each one
described, deduce why they are crooked.
1. Green’s Load (1880) – two spots drilled out and mercury added.
2. Tapping dice – hollow centre filled with mercury but with small tube to one
corner. Tap to make them crooked.
3. Bevelled – rounded on some edges.
4. Slick – one surface highly polished.
5. Hot iron – ridge along one edge.
6. Capped – one face capped with rubber.
ANSWER: 1. That face is heavier and will tend to stay facedown; 2. That corner will be
heavier when tapped; 3. The bevelled edges make rolling easier so those corners will tend
to roll on the surface and go faceup; 4. Polished surface will tend to roll and go
faceup; 5. The edge with the ridge will not tend to roll and so it will go facedown; 6.
The rubber edge will tend to bounce and go faceup. If you want to find out all the
cheating methods in dice games read "Scarne on Dice" by John Skarne. It's up the
8th edition and may be at your local library.
Text box page 186
NOVEL CHALLENGE
It’s easy to stand a pencil up on its base but impossible to stand up on its point.
But why? What if you could put it in a sealed container free of air currents and arranged
it so that it’s centre of mass was exactly over the point  could you do it then?
Still no! But what’s the physics behind the failure?
ANSWER: One answer is that the Earth is rotating so the bottom is being dragged sideways
by the surface. Another answer is that the random motion of electrons in the wood will
sometimes be lopsided so that more electrons just happen to be on one side for a fraction
of a second. This side will start to fall.
Text box page 188
NOVEL CHALLENGE
Can you jump off a chair on to the floor while holding a cup full of water without
spilling any. Plan how you should land to do this. Hmmm! It sounds good in theory
but….
ANSWER: I would hold it above my head and land with bent knees, gradually bringing my arm
down so that the deceleration could be over as long a time period as possible. Just hope
the Principal doesn't walk in.
Text box page 188
NOVEL CHALLENGE
Imagine you are standing on some bathroom scales and you bend your knees quickly. Predict
what would happen to the scale reading. But why?
ANSWER: The scale reading would decrease because the top half of your body would
remain motionless for a fraction of a second (inertia) as your legs moved upwards. Then
you would start to fall down and the scale reading would increase a bit and then go
back to normal. If this were a complex reasoning question you’d need all three
parts for full marks.
Text box page 190
NOVEL CHALLENGE
A very lightweight boat 24 m long and mass of 30 kg lies still on a quiet pond. A 90 kg
man walks from bow (front) to stern. How far does the boat move relative to the pond? The
answer is not 72 m.
ANSWER: 18 metres.
The best way to do this is by considering centre of gravity. In isolated systems the
centre of gravity remains in the same position. The centre of gravity of the boast is at
the midpoint (12 m from the end); when the man stands on the end, the centre of gravity
for the whole system is 3 m from the man and 9 m from the centre. When the man walks to
the other end, the centre of gravity stays in the same place and so the boat moves until a
point 3 m from the other end is in the same place as the original centre of gravity.
We’d do some diagrams here but they are too difficult to draw with Paintshop Pro. You
do one and email it to us.
Text box page 191
NOVEL CHALLENGE
A superball is tied to a 1.5 m string and suspended vertically from a hook. It is pulled
back and allowed to strike a wooden block standing on the floor. The experiment is
repeated with a lump of plasticine of the same mass as the ball. One knocks the block
over, one doesn’t. Which is which and why?
ANSWER: With the superball, the rebound means there is a high change in velocity (from +
to ) which means a high change in momentum. This is transferred to the wood and knocks it
over. With the plasticine, it just comes to a halt (v goes from + to 0) so the change in
velocity or momentum is not as high.
Text box page 192
NOVEL CHALLENGE
A superball is placed on top of a tennis ball and they are dropped together. Wow –
what a funny rebound. Predict what happens and why.
ANSWER: As the tennis ball starts to rebound upwards, the superball is still coming down
so it gets a huge kick upwards (more than just bouncing off a stationary surface). The
superball should shoot up to the ceiling.
Text box page 193
NOVEL CHALLENGE
A fly crashes into the front windscreen of a train and reverses its direction. Therefore,
at one instant it’s velocity is zero but as it is squashed on to the window, the
window’s velocity must also be zero for a short time. How could a fly stop a speeding
locomotive?
ANSWER: The fly doesn’t stop the entire locomotive – just one part. The window
glass would flex backwards a tiny fraction for a short time and then flex forward to catch
up to the rest of the train. And what's the last thing to go through the fly's mind as it
crashes into the windscreen. That's right  its abdomen.
Text box page 193
NOVEL CHALLENGE
An egg timer is placed on an electronic balance and it’s mass recorded when all the
sand is at the bottom. It is inverted. As the sand falls, will it weigh the same or less?
ANSWER: Weighs the same. Even though some sand is in the air, other sand is crashing into
the bottom and transfers momentum (adds extra weight) to the balance. These two cancel
out.
Text box page 197
NOVEL CHALLENGE
When you blow through a bent drinking straw it recoils away from the jet of air. But when
you suck air in does the reverse happen? I think not! Explain that one in terms of
momentum. Try it and see.
ANSWER: When you blow, the air comes out in a stream in one direction so there is a nett
force in the opposite direction. When you suck, the air comes in from all directions so
there is no nett force.
Text box page 202
NOVEL CHALLENGE
The torque to remove a lid off a jar of food has been set at 2 Nm by food manufacturers. This torque is such that all 20 – 40 yearolds and 97% of 50 – 94 yearolds can remove it. Estimate the force necessary to manage this assuming the radius of a typical lid is 3.5 cm. Why does it help to put a tight lid under a hot water tap?
Answer: T = Fr; F = T/r = 2/0.035 = 57.1N. The hot water expands the metal more than the glass. Could also help melt some of the gunk under the rim.
Text box page 203
NOVEL CHALLENGE
When removing a cork out of a champagne bottle, it is easier if you hold the cork and rotate the base of the bottle rather than holding the base and rotating the cork as most people do. Now why is this easier – it seems to defy logic?
Answer: The radius of the bottom is 44 mm and the radius of the cork is about 15 mm. As torque is Fr, the bottle bottom has a radius 3x that of the cork so only 1/3 as much force is needed on the bottom.
Text box page 204
NOVEL CHALLENGE
The Earth gains 100 thousand tonnes each day as interstellar dust settles on this beautiful planet of ours. Calculate how much longer each day will be because of this. You may need to do this longhand as most calculators won’t show an answer unless you know some tricks.
Text box page 204
NOVEL CHALLENGE
If everyone faced the same way on Earth and took a step at the same time would the
Earth’s rotation change? What data would you need to calculate this mathematically?
ANSWER: Radius of the Earth; force applied; duration of force. Calculate torque (T = Fr).
Do calculation on change in rotational momentum of the spherical Earth. With modern
methods it could be observed. We already know that the length of the day has changed by
0.0047 s in 20 years.
Text box page 205
NOVEL CHALLENGE
A piece of wire has an eye at one end with a small wire hoop through it.
When it is rotated in a drill the hoop lifts up. Now why is this?
ANSWER: In Physics, there is a principle that when energy input continues into an object,
the object tends to go to the state of maximum inertia, so the loop goes into the lifted
position. Rotational inertia is the sum of the product of mass times radius squared for
each atom in the object. Physicists have simplified this into simple formulas for
regularshaped objects. If you could get the formulas for rotational inertia you’d
see that the horizontal loop has a higher rotational inertia. The adage "I is a
maximum" is how physicists explain it. This is different if the object doesn’t
have energy continually put into it. See the next question.
Textbox page 206
NOVEL CHALLENGE
If you take a chocolate covered almond and spin it around while it is lying flat an
amazing thing happens (usually) – it stands on its end. Explain this phenomena in
terms of rotational inertia. Before you try it, predict if the fat end or the pointy end
will be on the top. What would happen with a Smartie or M&M?
ANSWER: When left lying on a table, a chocolate covered almond lies on its side with its
long axis horizontal. This is its "natural" position (with its centre of gravity
low as possible) because the energy of a system tends towards a minimum. Its GPE is low as
it can be. But when it is spun it stands on its end with the long axis vertical. The
centre of gravity is now much higher. This seems to violate the "lowest energy"
principle (see the question above). However, in cases where energy is no longer being put
in (and the object is said to be dissipating energy), the physics principle
"rotational inertia tends towards a minimum". Lower rotational inertia is
in the upright position. Yes it happens to Smarties, M&Ms and footballs. Summary:
Dissipative systems – Inertia goes minimum; conservative systems – Inertia goes
maximum.
Text box page 208
NOVEL CHALLENGE
A large ball bearing is placed…etc
ANSWER: The ball will stay in its same place relative to the desk. Friction causes the
ball to rotate but it should just stay in its same place because of its rotational inertia
(conservation of rotational momentum). Try it and you’ll see.
Text box page 194
NOVEL CHALLENGE
Why do you lean forward when you get up out of a chair?
ANSWER: You need to keep your centre of gravity directly above your feet otherwise there
will be a rotating force causing you to tumble backwards. Leaning forward brings your body
mass over your feet.
Text box page 194
NOVEL CHALLENGE
In 1916, a Dr Taylor observed a man carrying 40 kg ‘pigs’ of iron 11 m up a 2.4
m high incline to a train carriage. He carried 1156 pigs in 10 hours. The man’s mass
was 65 kg and he rested for 15% of the time. What was his average power output for the 8
½ hours? On a later occasion and without a rest he could only carry 305 pigs in the 10
hours. By what factor was his power output increased when he had proper rest? Suggest why
cyclists use a sprintcoastsprint sequence.
ANSWER: (a) 93 watts. P = W/t = E_{P}/t = mgh/t = (65 + 40) x 9.8 x 2.4 x 1156
trips/30600 seconds = 93 W; (b) 1156/305 = 3.79 times
Text box page 200
NOVEL CHALLENGE
Rod Cross, a physicist from University of Sydney, has analysed the power centres of a
tennis racquet.
Dead Spot – 5 cm from top end. No bounce but maximum power for serve. All KE goes
into ball.
Centre of percussion – centre of head. No ringing. Medium bounce.
Power spot – 5 cm in from bottom of head. Good for returning a fast ball but no good
for serve. Maximum bounce.
Power servers like Pete Sampras want to get maximum speed for the ball so they use the
dead spot for the serve for two reasons. One is mentioned above (what is it?). But what is
the other one? Clamp a racquet and drop a tennis ball from 50 cm on to the three points.
How does the bounce compare?
ANSWER: Maximum power; ball comes down over net at steep angle.
Text box page 201
NOVEL CHALLENGE
Blocks A and B are of mass 2m and m respectively. Block A is allowed to slide down the curved incline until it hits block B in an elastic collision. Which block will travel the furthest out? How far away from each other will they strike the ground? We say 1.6 m – how about you? We think you’ll need conservation of energy and momentum to do this ‘un.
E_{K}(A) at position B= mgh = 1/2mv^{2}; v = 3.46 m s^{1}
Conservation of momentum: m_{a}u_{a} + m_{b}u_{b} = m_{a}v_{a} + m_{b}v_{b}
2m x 3.464 + 0 = 2m v_{a} + mv_{b}
6.93m = 2mv_{a} = mv_{b}
v_{a} = (6.93m – mv_{b})/2m = 3.464 – ½ v_{b} (Equation 1)
Conservation of energy:
E_{Ki}(A) + E_{Ki}(B) = E_{Kf}(A) + E_{Kf}(B)
2mg x 0.3 + 0 = ½ 2mv_{a}^{2} + ½ mv_{b}^{2}
6m = mv_{a}^{2} + 1/ mv_{b}^{2}_{ }(Equation 2)
Substitute Equation 1 into Equation 2
6m = m(3.464 – ½ v_{b})^{2} + ½ mv_{b}^{2}
6 = 12 – 3.464v_{b} + ¼ v_{b}^{2} + ½ v_{b}^{2}
¾ v_{b}^{2} – 3.464 v_{b} + ^{}6 = 0
Use quadratic formula: v_{b} = 5.96 m s^{1}
Hence: v_{a} = 3.45 – ½ v_{b} = 0.484 m s^{1}
Time of fall: s_{v} = ½ at^{2}; 0.9 = 5t^{2}; t = 0.424 s.
s_{horiz}(A) = v_{a}t = 0.205 m
s_{horiz}(B) = v_{b}t = 2.52 m
Difference = 2.52 – 0.205 = 2.32 m (not 1.6 m as question said).
Text box page 204
NOVEL CHALLENGE
A plastic measuring cylinder has two holes near the base, one twice the diameter of the
other. The cylinder is filled with water and the small stopper removed. It takes t_{1}
seconds to empty. When repeated with just the big stopper removed it takes t_{2}
seconds. How long will it take with both stoppers removed? Do it algebraically first
before you wreck a good measuring cylinder.
ANSWER: 1/5 of t_{1} (= 4/5 t_{2})
The small hole has an area of p r^{2};
The big hole has an area of p (2r)^{2} = 4p r^{2};
The big hole has 4 times the area so has 4 times the flow rate of the small hole;
So 1/5 of the total volume flows through the small hole in the same time 4/5 of the total
volume flows through the big hole;
Hence 1/5 of t_{1} = 4/5_{ }of t_{2} = total time.
Text box page 206
NOVEL CHALLENGE
Evaluate the conclusion: "The Less you Know, the More you Make" etc.
ANSWER: It’s true. Bill Gates found this out when he was 20.
Text box page 209
NOVEL CHALLENGE
NOTE: Typing error. The length of x should read "3/5 L, not 2.5 L". And point P
is meant to be the pivot point not the suspension point. Sorry folks.
A pendulum bob on a string is allowed to fall but the string strikes a peg.
If the length of x is 3/5 L, the bob has zero velocity when it gets to the pivot
point P. Can you prove this mathematically?
ANSWER: This pendulum is called Galileo’s pendulum after the inventor. The GPE
of the ball at release is mgL. At the bottom of its travel, the bob has converted all this
GPE to kinetic energy ( ½ mv^{2}). To just return to a position directly above
the pivot on a shortened string of radius "h", the bob must have critical
velocity at the bottom of its travel (v_{crit} = Ö gr;
hence:
the initial GPE (mgL) = mg2h + ½ mv^{2}; mgL = mg2h + ½ mgh; hence h = 0.4 L (or
2/5 L).
Text box page 212
NOVEL CHALLENGE
Per kilogram, humans…etc.
ANSWER: The Sun puts out more total heat.
Solution. Humans: Assume the average human has a mass of 50 kg. The energy output per
kilogram in a day is thus 12 MJ/50kg = 2.4 x 10^{5} J/kg. There are about 7
billion people on Earth so the total energy output is 7 x 10^{9} x 50 x 12 x 10^{6}
J/day = 4.2 x 10^{18 }J/day.
The Sun: has 1/10000 the output per kg of humans, hence the Sun puts out 2.4 x 10^{5}
J/10000 = 2.4 x 10^{1} J = 24 J/kg per day. The Sun has a mass of 2 x 10^{30}
kg (see Chapter 6) hence the total output of the Sun is 24 x 2 x 10^{30} = 4.8 x
10^{31} J/day. This is greater than all the humans put out by 10^{13}
times (10 trillion).
Text box page 213
NOVEL CHALLENGE
A spring has an unladen length of 10 cm. With a mass of 1 kg added it stretches to 20 cm.
Three springs identical to the first one are arranged in the pattern shown as in the
diagram. How far down will they stretch when 1 kg is added?
ANSWER: Down to the 35 cm mark.
The top spring will stretch 10 cm; the bottom two will stretch 5 cm.
Text box page 214
NOVEL CHALLENGE
A spring 20 cm long has a spring constant (modulus) of 86 N m^{1}. If the spring
is cut into two 10 cm lengths, what will the modulus of each half be?
ANSWER: the same (86 N m^{1})
Text box page 215
NOVEL CHALLENGE
A spring is 10 cm long. When a 0.6 kg mass is attached and let fall the spring stretches
by 0.76 cm. Do you have enough information to calculate the spring constant?
ANSWER: Yes. The 0.6 kg mass falls through 0.66 m and hence transfers gravitational
potential energy to the spring. The gravitational potential energy GPE = mgh = 0.6 x 10 x
0.66 = 3.96 J. This is transferred to elastic potential energy EPE = ½ kx^{2}.
Hence 3.96 = ½ k (0.66)^{2}; k = 18.2 N m^{1}.
Text box page 224
NOVEL CHALLENGE
During World War II, Nazi scientists threw many prisoners overboard into the freezing
waters of the North Sea to see how fast their body temperatures dropped and how long it
would take for them to die. Today, such data is needed by ocean rescue researchers to help
develop safety devices for ocean users. Should we use this despicable ‘Nazi
science’? Develop an argument for or against its use.
ANSWER: the main lines of argument usually involve: "the data has been collected so
why not use it to save lives and the deaths of the prisoners would not be totally in
vain" or "using it says that what was done was okay and may encourage others to
do the same".
Text box page 224
NOVEL CHALLENGE
The table below shows the effect of changes to body temperature…etc.
Death is defined as a failure to revive on rewarming above 32°
C.
When people freeze to death in cold water it has been reported that they don’t seem
to be in pain as they die. They often seem relaxed. What could be happening here?
ANSWER: hallucinations probably.
Text box page 225
NOVEL CHALLENGE
At what temperature will ° C and °
F readings be the same? Could the kelvin temperature reading ever be the same as ° C or ° F reading?
ANSWER: (a) 40° C = 40° F. Use ° C = (° F –32) x 5/9; or T =
(T32)5/9; T = 40; (b) The kelvin temperature could not be the same as the Celsius as T =
T – 273 has no solution; The kelvin temperature could be the same as the Fahrenheit
temperature because –654K = 654° F, but this may be okay
to your Maths teacher but not to a physicist (you can’t have negative kelvin
temperatures).
Text box page 228
NOVEL CHALLENGE
When you eat ice…etc.
ANSWER: Yes eating ice would help you lose weigh; you could say it has negative joules. I
suppose drinking liquid nitrogen (196° C) would do the same.
You’d have to be a bit of a loser to try any of these. There’s an old story that
celery has negative joules because the energy expended in chewing is more that the energy
content of the food ingested. It must be true…John Laws said so.
Text box page
228
NOVEL CHALLENGE
In 1700, Dr Charles Blagden took some friends, a dog and a raw beefsteak into a
room at 127°C
for ¾ hour. They all came out unharmed except for the steak which was cooked.
Why?
Answer: humans perspire and this removes heat from the body.
Text box page 230
NOVEL CHALLENGE
At the Le Mans race in France there is 14 km of track and cars reach 360 km h^{1} (100 m s^{1}). At the end of the straight drivers approach Mulsanne Corner at 250 km h^{1} and jam on their brakes to go through the corner at 56 km h^{1}. The disc rotors glow red hot and sometimes reach 800°C. If a disc rotor (there are four) is 30 cm diameter steel 0.8 cm thick (density 7.8 g/cm^{3}) and is 400°C before the corner, show that 95% of the kinetic energy is transferred to heat.
Answer: Mass of a rotor = DV = pr^{2 }x thickness = 0.07065 x 0.008 = 5.65 x 10^{4} m^{3} x 7900 kg m^{3} = 4.408 kg. Four rotors = 17.6 kg.
Heat gain by rotors = mcDT = 17.6 x 460 x 400 = 3.245 MJ.
Mass of a typical Le Mans car is 675 kg. For example, the MG LOLA has this mass –
KE = ½ mv^{2} = ½ x 675 x 69.4^{2} = 1.625 MJ. Can't do the rest.
Text box page 236
NOVEL CHALLENGE
Imagine a 1 cm square….etc.
ANSWER: No, you can’t detect it because your body is so used to it. Just like you
didn’t reaslise that the chair you were sitting on exerts pressure on your backside
until we just mentioned it. If the air blows stopped you would notice it. You’d also
feel yourself puffing up like a balloon as the outside pressure decreased. But you
wouldn’t explode like people say as your skin is strong enough. But you’re ears
and eyes would bleed. They used to kill cats and dogs this way but not anymore.
Text box page 237
NOVEL CHALLENGE
A matchstick is placed in a testtube of water. When you place your thumb over the top and
press down, the match sinks. Propose a reason for this and test it to see if we’re
lying or not.
ANSWER: The gas inside the pine match compresses and the match becomes less buoyant.
Text box page 244
NOVEL CHALLENGE
French scientist….etc.
ANSWER: Yes, he was correct. You could have two equal masses of gases in two equal sized
containers and have different pressures. For example, 32 grams of oxygen gas (1 mole) in a
22.4 L container exerts a pressure of 1 atm (at 0° C). If you
put 32 grams of hydrogen gas in there, you would have 16 moles and the pressure would be
16 atm.
Text box page 247
NOVEL CHALLENGE
When a rod made up of four metals as shown in the diagram is heated, which of the
following diagrams represents its final shape. What about if it is cooled?
ANSWER: C; if cooled then B
Text box page 247
NOVEL CHALLENGE
The General Electric building…etc.
ANSWER: (a) a = 3/(6000 x 61) = 8.2 x 10^{6};
(b) use slots rather than holes;
(c) imagine the centre bows out so that each half is like a long skinny triangle: then the
bow out length will be 94.9 mm (whew, that’s a big bow).
Text box page 249
NOVEL CHALLENGE
The coefficient of volume expansion (b ) for iron is 0.36 x 10^{4}.
How many times greater is this than the coefficient of linear expansion (a )? Propose a proof for why this should be.
ANSWER: three times. It is difficult to show the proof of this but we’ll try.
Let the original length of each side be L and the expansion be D
L.
The increase in volume (D V) is made up by the expansion of the
three faces 3(L x L x DL) + the expansion of the three side
corners 3(D L^{2} x L) + the expansion represented by
the small cube in the top front right corner DL^{3}.
Hence: D V = 3(L x L x D L) + 3(D L^{2} x L) + D L^{3}.
Because D L^{2} and D L^{3}
are so incredibly tiny, we can cancel them out of the equation. That leaves us with D V = 3(L x L x D L); and as b = D V/VD T,
this becomes 3(L x L x D L)/L^{3D
}T or 3D L/LD T which equals 3a .
Text box page 248
NOVEL CHALLENGE
In days gone by, warships had cannons mounted on the decks with the iron cannonballs
resting in shallow brass ashtrayshaped containers called ‘brass monkeys’. These
would invariably fill with water as the sea washed over the decks. In the Atlantic ocean
sometimes it would get so cold that sailors would say: "It’s cold enough to
freeze the balls off a brass monkey". What do you suspect they meant?
ANSWER: when the weather was cold the water on deck and in the brass monkeys would freeze
and expand. This would cause the balls to roll off, hence the saying.
Text box page 249
NOVEL CHALLENGE
A steel ruler has a hole in one end. When the ruler is heated does the hole get bigger, smaller or stay the same. Be careful – even some science texts get it wrong.
Answer: hole gets bigger.
Text box page 253
NOVEL CHALLENGE
Cooks sometimes put a metal skewer through potatoes to make them cook quicker. Would you
speed things up by using a skewer of twice the diameter or two of the smaller skewers? If
you used two, where is the optimal place to put them? Why?
ANSWER: We would say use two skewers, each about onethird of the way in from the skin. It
is pretty subjective but the four sources of heat (left and right side and left and right
skewer) are about equally spaced. It would make a good experiment.
Text box page 253
NOVEL CHALLENGE
When you get out of bed in the morning, carpet feels warmer on your feet than tiles. Why is this if they are both at the same temperature?
Answer: heat travels from your feet into carpet slower than it does into tiles (the tiles have onetwelfth the thermal conductivity (0.05 W/m/K) of tiles (0.6 W/m/K).
Text box page 254
NOVEL CHALLENGE
A copper rod is placed through a hole in a piece of pine and heated. Charring occurs more
along the grain that across it. Now why is this? Propose a physics explanation.
ANSWER: The grain of wood is the direction the water and nutrient tubes run. When wood is
heated, heated, the heat can travel more easily along these tubes than it can across the
grain where the wood fibres are packed tightly together and act as an insulator. If you
want more information ring the Timber Research Advisory Committee in Brisbane on 3358
1400.
Text box page 254
NOVEL CHALLENGE
Four thermometers as shown are placed in the Sun for 10 minutes. List them in order from
highest reading to lowest. Explain why.
ANSWER: Black, white, tight cotton, loose cotton. Black is a good absorber of heat, white
is poor. The loose cotton would stop the thermometers warming up but the tight cotton
would allow them to heat up a bit. This would make a good experiment.
Text box page 254
NOVEL CHALLENGE
You have three icecubes of the same mass. Which one will melt first? Why?
ANSWER: The hollow cube (has the biggest surface area so can dissipate heat the quickest).
Text box page 254
NOVEL CHALLENGE
How can you cook a hamburger thoroughly in the quickest time? Would you cook it on an open
grill (large heat, but some charring) or in a pan (small heat). Explain using physics
principles. Suggest to your Physics teacher that you have a endofterm BBQ and the school
pay for the hamburger patties. Good luck!
ANSWER: Cook it slowly so no charred meat is formed (charring acts as an insulator).
Text box page 255
NOVEL CHALLENGE
Is it better to try to cool water fast by leaving ice float or keeping the ice submerged.
Provide the physics principles behind this.
ANSWER: Allowing it to float. The cold water sinks and the warmer water at the bottom
rises. If the ice was held at the bottom, the cold water wouldn’t rise (by
convection).
Text box page 256
NOVEL CHALLENGE
Does dark water (e.g. tea) cool faster or slower than white water (e.g. milk in water)? Explain this before you try it. While you’re thinking about it, here’s another: does brown bread toast quicker than white bread?
Answer: Brown bread does char (toast) quicker than white because white bread has higher albedo and therefore reflects better. It is difficult to control variables as brown bread has more sugar, moisture and protein than white so you would have to produce just white bread but with some brown food colouring. Not sure about the white/black tea. Must try it.
Text box page 256
NOVEL CHALLENGE
Imagine you added equal volumes…etc.
ANSWER: the oil would rise to the higher temperature because it’s specific heat
capacity is lower (which means it takes less heat to change its temperature by 1° C than it does for an equal mass of water). They both have the
same thermal energy as equal amounts of heat energy was added from the hotplate (assuming
the beakers were identical).
Text box page 257
NOVEL CHALLENGE
80 mL of cold water is placed in a polystyrene cup and the cup is placed in a beaker of
hot water. Thermometers are placed in both containers. Predict the shape of the
temperatures versus time graph.
The experiment is repeated but a small cube of ice is place in the cold water. Now show
how the graph shapes will change.
Text box page 259
NOVEL CHALLENGE
The temperature around a bunsen burner is lower the further you are from it. But does the
temperature fall away evenly in all directions? Draw a bunsen flame as in the diagram
below and predict where points of similar temperature will be, keeping in mind that both
convection and radiation are operating. Join these points of equal temperature – they
are called ‘isotherms’ (Gk. ‘iso = ‘equal’; thermos =
‘heat’). Try it!
ANSWER: because of convection currents the region above the flame should be much hotter
than air at the same distance to the side of the flame.
Text box page 262
NOVEL CHALLENGE
Two houses are covered in snow on their roofs. In which house is the ceiling insulation
better: the one in which the snow melts quickly or slowly?
ANSWER: the one with the least snow because heat from inside the house is not getting
through the roof to melt the snow.
Text box page 262
NOVEL CHALLENGE
Cut a grape almost in half…etc.
ANSWER: Grape juice is an electrolyte and can conduct electricity from one half to the
other via the skin bridge. When the microwave oven is turned on, the waves heat each half
and also make charge oscillate across the skin bridge from one half to the other
(that’s just what microwaves do). This makes the bridge very hot and it soon catches
on fire so you now have electric current passing through flames. This causes the gas to
ionize and the gas itself begins to conduct electricity which causes a brilliant arc light
lightning. Cool! For further explanations of microwave oven physics see the "How
things work" webpage by Louis Bloomfield, Professor of Physics, University of
Virginia.
Text box page 271
NOVEL CHALLENGE
The June 2000 issue of the very prestigious New England Journal of Medicine reported that the average rate of jaw movement when gum chewing in USA is 100 Hz. What do you suppose they really meant?
Answer: probably meant 1 Hz or more likely 100 movements per minute.
Text box page
272
NOVEL CHALLENGE
A
boat catches on fire and the skipper jumps overboard and swims away. The
skipper hears an explosion while underwater and lifts his head out of the water
and hears another explosion. Bystanders said that there was only one explosion
but the skipper said there were two. Who was correct? Explain.
Answer:
sound travels faster in water (1482 m/s) compared to air (341 m/s). Hence, if he
swam 500 m, the sound would reach him through the water in 500/1482 = 0.34 s but
would take another 1.1 s longer to reach him through the air (t = 500/341 = 1.47
s).
Text box page
272
NOVEL CHALLENGE
Scientists monitor global warming by measuring ocean temperatures. They pass a 57 Hz signal at Heard Island in the southern Indian Ocean and time its arrival at points in major ocean basins of the world. How does speed vary with water temperature? Draw a graph.
Answer: The approximate change in the speed of sound with temperature is 4.0 m/s per 1°C but this is not linear. It also varies with Salinity (1PSU = 1.4 m/s) and depth (1km = 17 m/s). Visit: http://www.dosits.org/people/resrchxp/1.htm
Text box page
296
NOVEL
CHALLENGE
To calculate
the total number of nodal lines produced by two coherent point sources, let
q
= 90°,
calculate ‘n’ and multiply by 4 (because there are four lots of 90°
in a circle). Alternatively, calculate the angle between two nodal lines and
divide into 360°).
Each method gives a slightly different result. Why is this? I think it is
because the formulas are approximations for a case where the distance between
sources is very small and the “screen” is a long, long way away.
Text box page 321
NOVEL CHALLENGE
You are listening to a radio broadcast of a live orchestral concert in London 20 000 km
away. Would you hear it before or after a person at the rear of the concert hall 50 m away
from the orchestra? (Sound travels at about 330 m s^{1}).
ANSWER: Before. It would take 20000 x 10^{3}/3 x 10^{8} = 0.067 s to hear
it by radio but would take 50/330 = 0.15 s to hear it live.
Text box page 321
NOVEL CHALLENGE
Stealth bombers…etc.
ANSWER: Most radar absorbing materials are partially conducting plastic composites. As a
radar microwave penetrates these composites, the electric field in the wave drives charges
back and forth through the composites. Since they don’t conduct electricity well,
they turn the wave’s energy into thermal energy and thereby absorb it. Because there
will also be some reflection of the microwaves, the aircraft are designed to deflect the
reflected wave away from the radar transmitter so that it’s receiver won’t
detect the return wave. Stealth bombers rely on rectangularbased pyramid shaped cladding
which tends to present a very low profile to radar. We’ve got no idea of how to
improve the detection of a Stealth bomber otherwise we’d be millionaires and not
Physics teachers.
Text box page 322
NOVEL CHALLENGE
You can hear popcorn popping from the outside of a microwave oven…etc.
ANSWER: Microwaves are reflected off the metal grid on the door of the oven. This
reflection occurs because the wave’s electric field pushes charges inside the metal
walls and causes those charges to accelerate. These accelerating charges redirect (absorb
and reemit) the wave in a new direction – a mirror reflection. The waves actually
enter the holes of the metal grid on the front and penetrates a short distance before it
dwindles to insignificance. By the time the wave has travelled a millimetre through the
hole its power is halved. The rule is: if the electromagnetic wave’s wavelength is
significantly larger than the hole, it won’t pass through. Sound, on the other hand
passes through the walls because it causes the walls to vibrate and they transmit the
sound to the outside.
Text box page 323
NOVEL CHALLENGE
A microwave oven doesn’t heat evenly…etc.
ANSWER: 12.2 cm (l = v/f = 3 x 10^{8}/2.45 x 10^{9}
= 0.122 m = 12.2 cm).
Microwave ovens are about 36 cm wide which equals three wavelengths. A diagram should have
a standing wave 36 cm long with 5 nodes in between and about 6 cm apart. Hence, the
hotspots will be every 3 cm across the carousel. Try it for yourself. Get some 35 mm film
containers, put equal amounts of water in each and arrange them across the insides of the
oven. Take their starting temperature. Turn it on for 10 seconds and check the new
temperatures. Try it without the carousel. Careful if you heat too long. The water can
become superheated (over 100° C) and boil explosively when you
move it.
Text box page 331
NOVEL CHALLENGE
In the Song of the White Horse by David Belford, the lead soprano is required to
breathe in helium to reach the extremely high top note. Question: if you released some of
the helium into the middle of the orchestra, what would happen to the pitch of the
following instruments: wind, brass, strings, percussion?
ANSWER: The speed of sound in helium is 965 m s^{1} so the frequencies of the
standing waves in pipes would increase (f = v/2L). No change in pitch would occur for the
others. You could create have in a orchestra by releasing helium during a performance.
We’re only joking of course.
Text box page 337
NOVEL CHALLENGE
If you told a violinist that you’re a physicist and they should play the strings
about oneseventh of their length from the end what would they say? Measure where they
play – is it 1/7 th?
ANSWER: We’re not sure what they’d say – they’d probably tell you to
get lost  but this is where a good violinist would play. If they don’t, maybe
they’re not very good (but don’t say that either). What's the difference between
a violin and a viola? A: violas burn longer.
Text box page 338
NOVEL CHALLENGE
What is the squeaky sound when you are washing your hair with shampoo and what is the
similarity with the shrill sound beginners produce on a violin?
ANSWER: Longitudinal waves are produced instead of transverse ones. The stiffness of the
hair strand (the "string") longitudinally is very high so the frequencies are
also high. That's why teenage boys don't wash their hair very often.
Text box page 338
NOVEL CHALLENGE
You’ve dipped your finger in some wine (or metho) and run it around the rim of a
wineglass. A loud sound is produced. Why doesn’t it work if your finger is not
degreased with the wine?
ANSWER: Your finger has to pull the glass in a longitudinal direction so it has to stick a
bit. If it is not degreased, then it will slip.
Text box page 340
NOVEL CHALLENGE
An orchestra tunes up at the start of a concert but as the theatre warms up they have to
retune their instruments. Do they find that the pitch of their instruments rises or falls
as the theatre warms up. String musicians can change the tension of the strings. What do
wind musicians do?
ANSWER: Wind imusicians tune by adjusting the length of the pipe. E.g.. Clarinettists and
flautists can pull out at the barrel when sharp (which is usually the case). In cases of
flatness, faster air generally cures the problem, although adjusting the angle of blowing
can also fix the tuning in flutes (i.e.. the more it turns in, the flatter it is). Oboists
and bassoonists are able to adjust the way the double reed fits into the instrument.
(Answer by Ruimin Gao  gao@myplace.net.au).
Text box page 342
NOVEL CHALLENGE
If you hold the right tuning fork up to your mouth cavity you can cause the cavity to
resonate. Would you expect boys or girls to have the lower resonant frequency? Why?
ANSWER: Boys. Big cavity (in their mouth) therefore lower fundamental. People without
brains should have a different sound when tapped on the head compared to normal people.
Text box page 344
NOVEL CHALLENGE
When the physics laboratory is quiet, a dropped pin can be heard clearly at the back of
the room. Calculate the energy arriving to the ear of the person at the back. If you want
secondhand data, assume the pin has a mass of 0.2 gram and it is dropped from a height of
1.0 m (use GPE = mgh). Assume all GPE is tranformed into sound energy that radiates
outwards as a large spherical surface (A_{sphere} = 4p
r^{2}). Calculate the amount of energy per square centimetre at the back of the
room (say r = 5 m). Is it more or less than 10^{9} J/cm^{2}? Not much
huh?
ANSWER: Less
GPE = mgh = 0.2/1000 x 10 x 1 = 2 x 10^{3} J
V = 4p r^{2} = 4p 500^{2}
= 3.14 x 10^{6} cm^{2};
Energy/cm^{2} = 2 x 10^{3} J/3.14 x 10^{6} cm^{2} = 6 x
10^{10} J
Text box page 345
NOVEL CHALLENGE
Tacoma Narrows bridge
ANSWER: you should get various motions like standing waves on a string. In the "Urban
Legends" webpage (www.urbanlegends.com/science/bridge_resonance.html) they argue that
the motions are very complex. Too complex for me I'm afraid.
Text box page 348
NOVEL CHALLENGE
The frequency shift effect was first proposed by Doppler in 1842 but the first experiment
wasn’t done until French scientist BuysBallot had a go in 1845. He had a carriage
full of brass musicians go past him in a train as they blew a steady note. To study this
effect quantitatively, what sort of measuring devices would he need?
ANSWER: Something to measure speed (tape measure, watch); something to measure frequency
(nothing invented then – he would have to use a tuning fork or string and compare by
ear).
Text box page 357
NOVEL CHALLENGE
You walk towards a plane mirror at 1 m s^{1}. How fast does you image approach
you? The mirror now approaches you at 1 m s^{1}. How fast does your image
approach you now?
ANSWER: (a) 2 m/s; (b) 2 m/s.
Text box page 359
NOVEL CHALLENGE
In the April 1984 edition of New Scientist magazine, a report appeared on the work of
British inventor Charles deSelby… etc.
Was deSelby a big liar or what? What is wrong with his theory?
ANSWER: this is an April Fool’s joke in New Scientist.
Text box page 382
NOVEL CHALLENGE
In 1621, French …etc.
ANSWER: The best we can do is the following diagram. If the 12 o’clock position is 0° , then the rays to the right are at 78°
and 83° ; the bottom rays are at about 185°
and 225° but there is a lot of room for error if you use a
small protractor.
Text box page 385
NOVEL CHALLENGE
When you squeeze…etc.
ANSWER: The bubbles look like silver balls as light is totally reflected off the surface
of the bubble. They look great.
Text box page 388
NOVEL CHALLENGE
Have you ever taken a photo…etc.
ANSWER: It should be very hard to tell the difference. We’ll add some photos here one
day to show you.
Text box page 392
NOVEL CHALLENGE
You’ve made an air lens by taping two watchglasses together. This does nothing in air
but what will it do underwater? "Myopic people see better underwater" –
account for this.
ANSWER: Act as a diverging lens. Myopic people need diverging lenses to spread the rays
out a bit so they don’t focus in front of the retina. In water, the difference in
refractive index between water and the cornea is less than for air and the cornea so the
rays don’t bend as much. The rays will focus on the retina and not in front as they
do in air.
Text box page 394
NOVEL CHALLENGE
A concave mirror and a concave lens….etc.
ANSWER: The concave lens has a longer focal length as the difference in refractive indices
is no longer as great so the deviation is less. The mirror is unchanged.
Text box page 396
NOVEL CHALLENGE
Plot a graph of v (yaxis) against M (= v/u) on the xaxis using suitable data. Prove that
the intercept on the yaxis equals the focal length. But what does the slope of the line
equal? You’ll be surprised and delighted.
ANSWER: Proof:
1/f = 1/u +1/v
1 = f (1/u +1/v)
v = f (v/u + v/v)
v = f (v/u + 1)
v = f v/u + f, which is in the form:
y = m x + c, hence: the slope (m) = f, and the intercept (c) also = f.
Text box page 402
The word pupil comes from the Latin pupilla = ‘a doll’. When you look at
yourself in someone’s eye you see a small dolllike image of yourself. Now someone
was really creative with language. Quick now, is the image upright or inverted?
ANSWER: upright images are formed in a convex mirror.
Text box page 403
NOVEL CHALLENGE
Two eyes are better than one for depth perception. Would three be better than two. Why or
why not?
ANSWER: The brain relies on triangulation to estimate distance. Only two rays are needed.
Three eyes would make no difference.
Text box page 403
NOVEL CHALLENGE
You blink on average once every 5 seconds while you are awake. How many megablinks per
annum is this? When you go into a shopping centre your blink rate falls to one every 12
seconds. Propose a reason for this.
ANSWER: 6.3 megablinks/annum. The atmosphere (lighting, music, colours) is conducive to
making you feel relaxed so your blink rate slows down.
Text box page 403
NOVEL CHALLENGE
There are three forms of colour blindness: protanopia, deuteranopia and total.
Find out what they mean then propose a reason for the names based on the meaning of the
prefixes: pro = one, deut = 2.
ANSWER: Both refer to a defect called "redgreen colour blindness". It is a
genetic disorder and is sexlinked so it only afflicts males (about 1 in 10). Protanopia
means you have a defective perception of red. It is sometimes called red blindness.
Deuteranopia means you have a defective perception of green. ‘Pro’ means one
and refers to red as red is the first of the primary colours. ‘Deut’ refers to
green as green is the second of the primary colours.
Text box page 405
NOVEL CHALLENGE
In 1970 Dr Fyodorov of the Soviet Union treated a nearsighted man who had glass slivers
in his eye. After the operation his nearsightedness had been cured. Prospose what the
‘radial keratomy’ operation did.
ANSWER: Near (or short) sight results in a focal point in front of the retina. This can be
corrected by spreading the rays out a bit using a diverging (convex) lens. The radial
keratomy would have removed some of the surface of the lens making it less curved and thus
increasing the focal length.
Text box page 418
NOVEL CHALLENGE
In the mid1700s, French experimentalist François du Fay observed that a charged gold
leaf was attracted by some electrified substances and repelled by others. He called the
two types vitreous and resinous. Use Table 21.2 and you knowledge of what substances are
classified as vitreous and resinous to decide if the resinous rod would have a positive or
negative charge.
ANSWER: Glass is vitreous so it was the positive charge. Amber is resinous so it is
negative.
Text box page 419
NOVEL CHALLENGE
In 500 BC, Greek philosopher Thales of Miletus noticed that cork dusk was attracted to a
charged amber rod. It wasn’t until 1500 (almost 200 years later) that people noticed
that after a while the cork dust is repelled. Why is the dust repelled and why do you
suspect that people didn’t notice it earlier?
ANSWER: Greeks didn’t do experiments. Philosophers such as Plato and Aristotle said
the majority of men laboured "in order that the minority, the elite, might engage in
pure exercises of the mind – art, philosophy and politics". Experiments were not
pure exercises of the mind so they didn’t do them. They believed that manual labour
was for the slaves. Possibly also, they may have noticed how cork dust was repelled but as
it made no sense in terms of their theories of matter, they politely ignored it.
Text box page 423
NOVEL CHALLENGE
When sand falls through a plastic funnel on to a metal plate on a electroscope, the leaves
diverge. Explain this if you can.
ANSWER: The sand acquires a charge as a result of electrification by rubbing against the
plastic. According to the triboelectric series (page 418) glass, and hence sand, has a
lower affinity for electrons than polythene so the sand acquires a positive charge. Try it
and see. Let us know if it doesn’t become positive.
Text box page 426
NOVEL CHALLENGE
In his book, The Ascent of Science, Professor Brian Silver…etc.
ANSWER: 1.3 x 10^{3} N (not 1000 N).
Solution: moles of sugar molecules = m/M = 5/342 = 0.0146 mole;
1 molecule of sugar has 270 electrons (72 from C_{6}, 22 from H_{22} and
176 from O_{11});
Hence there are 0.0146 x 270 = 3.947 moles of electrons;
Which equals 3.947 x 6 x 10^{23} = 2.368 x 10^{24} electrons;
In this there are 2.368 x 10^{24}/10^{9} = 2.368 x 10^{15} billion
electrons;
If this many are taken from one cube to the other, the nett charge on each cube will be
2.368 x 10^{15} x 1.6 x 10^{19}C = 3.78 x 10^{4}C.
The force at 1000 m = kQ_{1}Q_{2}/d^{2} = 9 x 10^{9} x
(3.78 x 10^{4})^{2} / 1000^{2} = 1.3 x 10^{3}N.
Note: if the distance was 1 metre instead of 1 km, the force would be 1300N which is about
right.
Text box page 428
NOVEL CHALLENGE
Here’s a good idea we overheard at a UFO conference…etc.
ANSWER: F = 9 x 10^{9} (6 x 10^{23} x 1.6 x 10^{19})^{2}/1^{2}
= 8.3 x 10^{19} N (Whew, that’s big!).
Acceleration = F/m = 8.3 x 10^{19}/(100 x 10^{3}) = 8.3 x 10^{14}
ms^{2}. That’s big too – in one millionth of a second, the rocket would
be at the speed of light.
The Flaw is that you couldn’t have a mole of charge in a wheelbarrow – it would
fly apart. The only reason you can have a mole of hydrogen ions together is that there is
a mole of negative charge mixed in to neutralise it.
Text box page 458
NOVEL CHALLENGE
A writer in New Scientist magazine (July 1998) described how in 1938 he stayed in a
country house in England and helped wind up a 1 tonne steel ball suspended on a chain into
the roof space. During the evening, the ball was allowed to fall slowly, turning a
generator to keep the light glowing all night. He said that this was impossible as there
wasn’t enough gravitational potential energy in the ball to do this. Verify his claim
by working out how long a 60 W lamp would glow of the ball was raised 5.0 m. Assume the
energy conversion was 100% efficient (unlikely!). In actual fact it turns out that the
steel ball didn’t turn a generator but turned an enclosed 44 gallon drum partly
submerged in petrol. As the drum turned petrol evaporated and was burnt in a gas lantern.
ANSWER: GPE = mgh = 1000 x 10 x 5 = 5 x 10^{4} J
Time (t) = W/P = 5 x 10^{4} / 60 = 833 seconds (13.9 minutes). Not long –
hardly enough time for a sherry.
Text box page 459
NOVEL CHALLENGE
One that most people get wrong: 100W bulbs glow….etc.
ANSWER: The 40W bulb is brighter in series.
Solution: Firstly, calculate the resistance of each – that is the only thing that
doesn’t change (the current and voltage will differ when in series to when in
parallel).
P = VI = V^{2}/R; Hence R = V^{2}/P; R_{100watt} = 240^{2}/100
= 576 ohm; R_{40watt }= 240^{2}/40 = 1440 ohm;
In series, the total resistance is 576 + 1440 = 2016 ohm;
The current (I) will be V/R = 240/2016 = 0.119 A;
The power consumption and hence the brightness of each will be P = VI = I^{2}R;
P_{100watt} = 0.119^{2} x 576 = 8.2 watt;
P_{40watt} = 0.119^{2} x 1440 = 20.4 watt (and therefore brighter).
Text box page 459
NOVEL CHALLENGE
In an experiment to measure the efficiency of a microwave oven, 1 litre (1000 g) of water
was placed in a icecream bucket and its temperature measured . It was then cooked on
‘high’ for 2 minutes and it’s temperature measured again. Use the formulas
Q = mcD T and P = W/t to prove that the power output of the
microwave (P) = 34.8 x D T. Hint: let Q = W. If a 750 watt
microwave raised the temperature of 1 L of water by 20° C,
calculate the efficiency of the oven.
ANSWER: Part A: Q = W, therefore: mcD T = Pt; 1000 x 4.18 x D T = P x 120;
hence P = 1000 x 4.18/120 x D T = 34.8 D
T
ANSWER: Part B: For the water: Q = mcD T = 1000 x 4.18 x 20 =
83600 J
For the microwave oven: W = Pt = 750 x 120 = 90000 J
% efficiency = 83600/90000 x 100 = 92.9 %
Text box page 462
NOVEL CHALLENGE
Have you seen the new compact fluorescent ….etc.
ANSWER: Fluorescent is cheaper in the long run.
Solution: Consider usage over 8000 hours:
(a) Fluoro uses 5.76 x 10^{18}J = 160 kilowatt hours = $16. To this add the
initial cost ($20) = $36.
(b) Incandescent uses 2.16 x 10^{9} J = 600 kWh = $60. To this add 7 bulbs at 75
cents each ($5.25) = $65.25.
Text box page 461
NOVEL CHALLENGE
A 240 V jug rated at 2000 W takes 1 min 45 s to heat 2 cups (500 g) of water from 23° C to boiling (100° C). Calculate its
percentage efficiency.
If one litre of water was being boiled from the same temperature, propose whether the time
would be exactly twice or more or less than twice. Hmmm  think of the energy losses.
ANSWER: 76.6%
Q (water) = mcD T = 500 x 4.18 x 77 = 160930 J
W (jug element) = Pt = 2000 x 105 = 210000 J
% eff = W(out)/W(in) x 100% = 160930/210000 x 100 = 76.6%
ANSWER: If 1 litre was being boiled it should take less because you would still have the
same loss of heat in heating up the jug element. We tried it and got 105 seconds for 2
cups and 202 s for 4 cups. We'd love to test a lot of different volumes and plot a graph
but we're afraid it mightn't be linear and this would spoil a good story.
Text box page 480
NOVEL CHALLENGE
It is sometime said that the discharge of a capacitor is mathematically like the emptying
of a bucket of water with a hold in the bottom. Explain why the two discharge curves are
the same.
ANSWER: as the bucket empties the rate slows because there is a diminishing head of water
(less pressure). For a capacitor, as the charge flows out, the repulsive forces get weaker
and weaker so the rate also slows.
Text box page 510
NOVEL CHALLENGE
If you heat a iron bar attached to a magnet as shown, at a particular temperature (Curie
temperature) the bar falls off. Why might this be?
ANSWER: the heat disrupts the magnetic domains and the magnet is no longer a magnet.
Text box page 512
NOVEL CHALLENGE
Four ring magnets are placed on a wooden pole as shown. If the distance between the top
two is 10cm, calculate the other spacings?
ANSWER: Second spacing = 7.1 cm; first spacing = 5.8 cm.
Solution: let the magnets be numbered 1, 2, 3, 4 starting from the top. The force between
the magnets can then be labelled: F_{12}, F_{23}, F_{34}. Because
magnet 2 has to support the weight of 1 as well as 2, the force F_{23} must be
twice F_{12}. As the magnetic force is in an inverse square relationship with
distance: F_{12} µ 1/10^{2}; F_{23} µ 1/d^{2}; but F_{23} = 2 F_{12}, hence 1/d^{2}
= 2 ´ 1/10^{2}; 2d^{2} = 10^{2}; d =
7.1 cm.
Similarly for the bottom pair; magnet 3 has to support the weight of two other magnets; F_{34}
= 3 F_{12}, hence 1/d^{2} = 3 x 1/10^{2}; 3d^{2} = 10^{2};
d = 5.8 cm.
Text box page 518
NOVEL CHALLENGE
Which would reach the higher temperature…etc.
ANSWER: The magnet would because energy was needed to get the domains to align so it
contains stored (potential) energy. This would be release on dissolving but the
temperature change would be microscopic.
Text box page 519
NOVEL CHALLENGE
Which would reach the higher temperature when dissolved in acid: a steel spring…etc.
ANSWER: The compressed spring – because it has stored elastic potential energy which
is released on dissolution.
Text box page 519
NOVEL CHALLENGE
We read on the Internet that magnets are fed to cows…etc.
ANSWER: We’re not sure about how true this stuff is about cow magnets but if you want
to, check out the University of Minnesota’s web page at www.extension.umn.edu/Documents/J/O/JO1072.html.
If this has disappeared, search for "cow magnets" + Minnesota.
About the time to dissolve: you check it out and let us know.
Text box page 520
NOVEL CHALLENGE
Figure (a) shows a part of a field about…etc.
ANSWER: This is a tricky one because it very much depends on what field lines are meant to
represent. When Michael Faraday started using these lines in his description of magnetic
fields it was because he was poor at mathematics. But the use of lines became popular.
They represent the direction an isolated north pole would travel in the field. So they
don’t have a particular length – they curl around to meet the south pole or go
off infinitely into space. So figure (a) shows a part of the field but when the magnetic
field is turned off, the lines would still be there but would get fewer in number as the
field weakened. They would not get shorter or have a gap between them and the magnet. So
neither figures (b) or (c) are correct. The closest would be a figure with just one arrow.
When the field is off there should be no arrows.
Text box page 587
NOVEL CHALLENGE
Other things in nature have a halflife. For example t_{1/2} for human protein is 80 days; rat muscle 21 days and rat blood 6 days. In nonmammals halflife values are much lower as they have lower temperatures. How is this different to nuclear halflives?
Answer: nuclear halflives are not temperature dependant.
Text box page
597
NOVEL CHALLENGE
The time to hear a nuclear explosion was worked out by the Rand Corporation in 1968 (in the USA). The formula is: t = 5.8 x 10^{19} R^{10} W^{3} where t is the time in seconds to hear the explosion, W = the equivalent amount of TNT explosive in megatonne, and R = the distance from the explosion in metres. Calculate how much time would elapse before you could hear a 1000 megatonne bomb at 500 m.
Answer: 5.8 x 10^{19} x 500^{10} x 1000^{3} = 0.5664 s
Text box page 602
NOVEL CHALLENGE
A letter writer to the Courier Mail…etc.
ANSWER: Oh great! Radioactive waste all through space just waiting to be trapped into
earth orbit. No thanks!
Text box page 602
NOVEL CHALLENGE
Poet W H Auden wrote in his poem Marginalia…etc.
ANSWER: He meant that geologists and engineers would not challenge a dictator because they
are either gutless or too wound up in their own work. To substantiate this claim you would
have to document many geologists and engineers who stood by while tyrants were at work.
Maybe W H Auden was bashed up by a geologist for his loud mouth and bad poetry.
Text box page 603
NOVEL CHALLENGE
Two new heavy nuclides have been prepared recently. Work out what they are from the incomplete equations shown on p 603.
Answer: Rf257; Rf260
Text box page
607
NOVEL CHALLENGE
A person receives a dose of 0.3 mGy of slow neutrons, 6 mGy of gamma rays and 0.1 mGy of fast neutrons. How many millisievert is this?
Answer: H = 0.3 mGy x 5 + 6 mGy x 1 + 0.1 mGy x 20 = 9.5 mSv
Text box page 614
NOVEL CHALLENGE
A Crookes’ Radiometer consists of four paddles suspended on a needle point in a low
pressure glass container. One side of the paddle is painted black, the other side white.
When placed in the Sun it turns around. Explain whether the black side moves away from the
Sun or towards it (and why). It goes the opposite way near a block of dry ice (44° C). Most people (even scientists) gave the wrong explanation.
ANSWER: The black side moves away from the sun. The air is hotter next to the black side
so it expands and pushes the paddle away.
Text box page 637
NOVEL CHALLENGE
Can a shadow travel faster than light…etc.
ANSWER: A shadow maybe able to travel faster than light but so what? What’s actually
travelling? It’s not any object just the image of an object. It is not a violation of
the special theory of relativity because you cannot transmit information using a shadow.
This is one of the key points of the theory. Information can be transmitted by light, but
a shadow marks the absence of light so no information can be transmitted. It’s like
saying when you don’t speak the silence is travelling faster than sound. Silence
doesn’t transmit information (except when your mum gives you the cold shoulder –
her silence says a lot).
Text box page 645
NOVEL
CHALLENGE
You want to get to a star 100 light years away and have calculated you could do
it in 4.5 years if you travelled at 0.999c. The problem is that the human body
can’t stand accelerations greater than 5 ‘g’. How much time as measured by an
Earth observer would it take you to reach 0.999 c from rest at an acceleration
of 5 ‘g’? This next question may be too hard: estimate how much time it would
take in the spacecraft’s frame of reference?
Answer: t = v/a = 0.999 x 3 x 10^{8}/(5 x 10) = 5994000s = 69.375 days.
Don’t know how to do the second part as v is changing.
Text box page 647
NOVEL CHALLENGE
Imagine you’ve slid into a parallel universe in which the year is 1840. A US Marshall
is travelling by in a train at 0.75c and approaches two gunfighters about to have a duel.
Gunslinger A is closer to the Marshall. If the Marshall sees both men draw their guns at
the same time, who actually drew first in (a) the Marshall’s frame of reference; (b)
the gunslingers’ frame of reference.
ANSWER: (a) both the same as it says above; (b) B
Text box page 647
NOVEL CHALLENGE
A physicist driving a very fast sports car is booked by police for travelling through a
red traffic light. The physicist argues that because he was travelling fast with respect
to the light, the colour of the light had its wavelength altered and appeared green to
him. The judge said that he would let him off the charge of running a red light but would
fine him 1 cent for every m/s he was travelling over 100 km/h. How much was he fined?
ANSWER: $840 000 ($839 999.72)
f_{o} = 4.5 x 10^{14} Hz; f = 6 x 10^{14} Hz. Solve for v (= 0.28c
= 8.4 x 10^{7} m s^{1}). 100 km/h = 27.78 m s^{1}; hence he was
83999972 m s^{1} over the speed limit.
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