NOVEL CHALLENGE ANSWERS

This page has the answers to all the Novel Challenge questions in the text boxes in the margins.

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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 ping-pong 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 mm³. A typical large Chinese-made suitcase is 24” x 17” x 9” (600 mm x 400 mm x 230 mm) with a volume of 5.52 x 10⁷ mm³. Hence 862 balls will fit into this suitcase.

(d) How fast does grass grow? Answer: Blue couch: Between 20-50mm every 7-14 days in Spring/Summer; Green couch: between 15-25mm 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 100-page 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 m² at 0.080 kg per sq. metre = 1.8459 kg.

Covers (2) at 0.249 kg/m² = 0.0287 kg; spine 0.043m x 0.276m at 0.249 kg/m² = 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 mid-1600s, 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 band-aids 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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NOVEL CHALLENGE
A testable hypothesis: that students in the top half of my maths class will .....

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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). Fifty-four 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://humanics-es.com/stadiometer.htm

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NOVEL CHALLENGE

Humans have 10¹⁴ 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 10⁶ m.

Answer: 10¹⁴ x 0.01 x 10^-6 m = 1 x 10⁶m

Circumference = 2 x 3.14 x 6.38 x 10⁶ = 4.0 x 10⁷ 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 cm³; 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 Hungarian-German 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 25-35 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 T²⁰). How much faster is the rate at 6000 Kelvin than it would be at 5000 Kelvin?

Answer: Rate = 6000²/5000²⁰ = 1.2²⁰ = 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 10⁹ people. Say the average person approximates a rectangle 0.45 m x 0.35m (= 0.1575 m²). Total area = 1.015 x 10⁹ m² = 1015 km².
(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/champ-car5.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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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: v² = 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 t₁. Hence t₁ = s₁/v_OG
5. In this time R has moved a distance s₂. Hence t₁ = s₂/v_RG
6. Both are equal to t₁, hence: s₁/v_OG = s₂/v_RG
7. But s₁= 3 + s₂ thus (3 + s₂)/v_OG = s₂/v_RG
8. Rearranging:(3 + s₂)/s₂ = v_OG/v_RG
9 Let the distance O travels from the head of the column to the rear be s₃
10. Let the time for O to travel from head of column back to rear = t₂ (= s₃/v_OG)
11. In this time R has moved a distance s₄ in a time t₂ (= s₄/v_RG)
12. But s₄ = 4 - s₂
13. Substituting into equation in line 11: t₂ = (4 - s₂)/v_RG
14. But s₄ = 3 - s₃ = 4 - s₂. Hence s₃ = s₂ -1.
15. Substituting into line 10: t₂ = s₃/v_OG = (s₂ - 1)/v_OG
16. Hence t₂ = (s₂ - 1)/v_OG = (4 - s₂)/v_RG
17. Rearranging: v_OG/v_RG = (s₂ - 1)/(4 - s₂), but this equals (3 + s₂)/s₂ from line 8.
18. Hence (s₂ - 1)/(4 - s₂) = (3 + s₂)/s₂
19. Rearranging: (s₂ - 1)s₂ = (3 + s₂)(4 - s₂)
20. s₂² - s₂ = 12 - 3s₂ + 4s₂ - s₂²
21. 2s₂² - 2s₂ -12 = 0
22. Simplify: s₂² - s₂ - 6 = 0
23. Factorise: (s₂ - 3)(s₂ + 2) = 0. Hence s₂ = 3 (ignore -2)
24. If s₂ = 3, then s₁ = 6, s₃ = 2, s₄ = 1
THUS - the officer travels s₁ + s₃ = 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 t₁ + t₂, R has traveled 4 km, hence: v_RG = 4/(t₁+t₂)
28. Substituting for t₁ (equation 4) and t₂ (equation 15): (using actual values of s₁ and s₃):
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_OG² = 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 t₁ = 1.5 hr, t₂ = 0.5 hr. Total time 2 hours.

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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 arm-wrestled 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 human-sized 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 + (19-N) walks of 1.5km. Hence: 5 + N*4 + (19-N)*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.

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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.

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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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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 m³; mass = DV = 1020 x 1.1365 x 10^-3 = 1.159 kg.

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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.

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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.

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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?

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NOVEL CHALLENGE
For m₁ and m₂ to remain in the same positions relative to the cart, what force (F) has to be applied.
Answer: Assume no friction. F on m₁ = m₂g; a of m₁ = F/m = m₂g/(m₁+m₂). F applied has to accelerate cart and m₁ and m₂ at m₂g/(m₁+m₂). Hence: F = ma = (m₁+ m₂ + m_cart x m₂g/(m₁+m₂)).

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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 = 4/3pr³ = 1.767 m³; 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 free-fall (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 v-shaped, touring are rounded. The figure below shows the profiles of Dunlop’s mega-successful D207GP race tyre which has been cleaning-up 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 v-shaped 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 19-year 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_Wsinq = 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 high-g 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 push-off speed? Answer: 1.9 m s^-1.

Text box page  1126

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 at²

0 = v sinq t + 0.93t²

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 at² = v sin q x 1.64 + 5 x 1.64²;

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 by John Liu.
The Cartesian form of the parabolic equation for projectiles is
y= xtanA-g(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.(^ is "to the power of-")
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^2tanA-50g^2(secA)^2=v^4(sinA)^2. Rearranging terms,
v^4(sinA)^2-20gv^2tanA+50g^2(secA)^2=0.
The range, also from year 12 mathematics, is v^2sin2A/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+/-5Sqrt2)g/sin2A.
Substitute this value into range, we get
v^2sin2A/g=(20+/-10Sqrt2)g/sin2A x sin2A/g=(20+/-10Sqrt2). (The x in this statement is "multiplied by", not a variable.)
Because 20-10Sqrt2=5.86, this indictates 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+10Sqrt2=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 up-and-down 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 half-filled 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 side-by-side 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 (y-axis) 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 s-1, 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 10¹¹ m - 1.49 x 10¹¹ m = 6.29 x 10¹¹ m). For superior conjunction the distance apart is the sum of the two radii ( = 9.27 x 10¹¹ m). The time taken for a radio signal to travel between the planets ( t = s/v) is: t_inferior = 6.29 x 10¹¹ m / 3.0 x 10⁸ = 2097 s. Thus total time (out & back) = 4193 s.  For superior conjunction, t_superior = 9.27 x 10¹¹ / 3.0 x 10⁸ = 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 r³/T² compare to that of our Solar System?
ANSWER:     r³/T²     = (0.4 x 3 x 10⁸ x 365 x 24 x 3600)³ / (6 x 10⁶ x 365 x 24 x 3600)²
                                              = 5.4 x 10⁴⁶ / 3.6 x 10²⁸
                                              = 1.5 x 10¹⁸ m³/s²
This is about half the value for our Solar System (3.4 x 10¹⁸).

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: Hmmm! Friction. How would we know, we’re only high school physics teachers.

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 10⁶ m (=EC); AB = 1000 m; hence BC = 6.379 x 10⁶ 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 / T²)
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 10⁸ Pa
P (molars) = F/A = 1220/(0.008 x 0.008) = 1.91 x 10⁷ 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 m-3; 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 m-3 whereas the mantle has a density of 3300 kg m-3. 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 r³ - 4/3 p 4³ = 4/3 p (r³ – 4³)
m_steel = 4/3 p (r³ – 4³) x 7.8 grams;
V_ball = 4/3 p r³
When density = 1 it just floats; D = m/V, hence:
1 = 4/3 p (r³ – 4³) x 7.8 / 4/3 p r³;
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 10³ 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.

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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 m³, when it melts it produces 1 m³ 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 face-down; 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 face-up; 4. Polished surface will tend to roll and go face-up; 5. The edge with the ridge will not tend to roll and so it will go face-down; 6. The rubber edge will tend to bounce and go face-up. 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 year-olds and 97% of 50 – 94 year-olds 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 long-hand 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 regular-shaped 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 sprint-coast-sprint 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 = 1/2mv²; v = 3.46 m s^-1

Conservation of momentum: m_au_a + m_bu_b = m_av_a + m_bv_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_a² + ½ mv_b²

6m = mv_a² + 1/ mv_b² (Equation 2)

Substitute Equation 1 into Equation 2

6m = m(3.464 – ½ v_b)² + ½ mv_b²

6 = 12 – 3.464v_b + ¼ v_b² + ½ v_b²

¾ v_b² – 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 at²; 0.9 = 5t²; t = 0.424 s.

s_horiz(A) = v_at = 0.205 m

s_horiz(B) = v_bt = 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 t₁ seconds to empty. When repeated with just the big stopper removed it takes t₂ 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 t₁ (= 4/5 t₂)
The small hole has an area of p r²;
The big hole has an area of p (2r)² = 4p r²;
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</