Physics Textbook Website: Errata

Physics Textbook Website


"New Century Senior Physics - Knowledge, Processes and Reasoning"

- by Richard Walding, Greg Rapkins and Glenn Rossiter
Oxford University Press 1999

This page details the errors and omissions in the "Senior Physics - Knowledge, Processes & Reasoning" text (1997).
There are one or two corrections to the "New Century Senior Physics" (1999) text as well.

Please email the authors in the unlikely event that you find some more. We're only human.
Email to: or

Our editorial advisors were omitted from the acknowledgements page. Many thanks to:
  • John Hayward, Head of Physics at Somerville House (District Panel Chairman,Physics - Brisbane-South)
  • Ian Yarham, Head of Science at MacGregor State High School (State Panel Member and External Examiner, Physics)

    Chapter 1
    Page 11. About two-thirds of the way down the page it says "(made bigger by 1000)".
    This should read "(made bigger by 10000)". Sorry folks!
    Page 13. The second dot point in Multiplying and Dividing.
    Change 1.50 to 1.5 "This is rounded to 150 or 1.5 x 102 (2 sf)."
    Page 17. Question 18. The answer on page 698 should be 2.20 x 103 mm3 (or 2200 to 2SF), not 2350. The uncertainty of +/- 300 is okay (1 SF).
    Page 17. Question 20. The answer on page 698 should be 1980 m not 990 m. What a fool!
    Page 23. Question 33. Delete the ± out of line 4.

    Chapter 2
    Page 32. Question 10 (New Century Edition). Answers are: (a) 6 m/s, 0 m/s, 4 m/s, 0 m/s, -5 m/s; (b) 0 m/s; (c) 200m/60s = 3.3 m/s. Thanks to Geoff Fletcher and the Year 11 Physicists at Sheldon College, and Stephen Golding (Year 12 Brisbane Boys' College), for pointing that one out.
    Page 36. Question 18(d). Answer is 0s, 30 s and 50 s.
    Page 36. Question 19(d). Answer is 0s, 30 s and 60 s.
    Page 50. Question 43(b). Answer on page 698 should be 40 m; Q43(f) Answer should be -12.5 m s-2
    Page 50. Question 44(d). Acceleration should be -1.5 m s-2, not +1.5 m s-2. Answer on page 698 should be t = 6.9 s, not -6.9 s. Thanks to Anna Singleton for finding these. No prize - she's from Moreton Bay College.
    Page 50. Question 46(a). Answer should be 162.5 m. Thanks to Scott Kelly from Hendra Secondary College for letting us know.
    Page 54. Question 68(a). Answer is 5 - 10t1. Thanks to Linda Tseng for finding this one.
    Page 55. Question 75. The answer in the back should be 160 seconds.
    A thank-you Bernard McGrath and the gang at St Mary's College Toowoomba.(who found it on April 9th 2000) and to Tim Trudgian at St Paul's School Bald Hills (for reporting it on 9th May 2000). No Tim-Tams but we will send you a $1 Scratch-it" (one-quarter of any prize to you, one-quarter to each of the three of us).
    Page 707 (1999 edition): The answer to Q 75 is 160 seconds.

    Chapter 3
    Page 61. Question 7(b)- the answer has the wrong angle. It should be 39.8 degrees. Thank-you to Michael Dedajic for his keen eye.
    Page 61. Question 7(c)- the answer has the wrong units. It should be km/h.
    Page 65. The solution to the example has a mistake. The angle is found by taking the sin (not tan) of 5/8. This gives an angle of 39 degrees. The rest is okay.
    A big thank-you to Pia Scobie at Somerville House for finding this one. A $1 Scratch-it is on the way.

    Chapter 4
    Page 86. Question 17(c). The answer on page 699 should be 700 N + 35 N = 735 N.
    Page 91. One-third of the way down: where it says "Note: the tension in the string on the left has to equal...". Change the word 'left' to 'right'.
    Thanks to Brian McManus at Aquinas College, Southport, for finding that one. It survived the first edition without comment.
    Page 91. Question 20(ii). The answer on page 699 should be T = 15 N.
    Thanks to Glenn Scarffe (Griffith Uni). Whitman's Sampler on its way.
    Page 97. Question 23(c) in the "New Century" edition (1999)only. The question should ask for m not m.
    A big-thankyou to the pedants in Mr Keogh's class at Moreton Bay College
    Page 98. Question 27 in the "New Century" edition (1999)only. The question should ask for m not m.
    Page 100. Question 52. The answer should be 11.7 ms-1. The solution has been corrected on the CRP web site. A big thank-you to Paul Evans for this one.

    Chapter 5
    Page 103. Figure 5.2 should have the lengths of vv increasing as the object falls.
    Change the figure to look like this:

    Page 108. Activity 5.3.  (a) should read Horizontal range sh= u cos q t = R; velocity at maximum vertical height vv = u sin q + at/2 = 0. Thanks to Rod Duncan for pointing that out.

    Page 118. Question 15. The radius should be 30 cm not 30 m.
    Thanks to Lee McLennan and the gang at Charleville State High School for spotting this one. A Scratch-it is on the way.
    Page 700. Answer to Q50 should be: (a) 80 m; (b) 76 degrees; (c) Max height = 12.8 m, Horizontal distance = 102 m.
    Thanks to the Year 11s at Pine Rivers State High for your interest.

    Chapter 6
    Question 22 (b) - the answer is 6064 m/s not 5990 m/s.
    Thanks to Jeff Knowles again!

    Chapter 7
    There is a figure missing for Question 33 on page 163. It looks like this:

    Chapter 9
    Question 3 page 197 of New Century Senior Physics edition. Whoa! We slipped up here. The answer for the New Century edition should be (a) Horse A: 8400 J, Horse B: 6000 J; (b) Total work = 14400 J. Thank you to Dane Cavanagh of St Paul's College, Bald Hills for pointing that out. Your teacher should give you a VHA.

    Question 4 (c) page 198 of New Century Senior Physics edition. The answer is 280J not 320J. Thank you to Matthew Alexander from Marymount College, Burleigh Heads.

    Question 14 page 207. We've slipped up a bit here and we'll let our colleague, Brian McManus from Aquinas College, put you straight:
    The implication seems to be that the power of the aeroplane engine in straight and level flight is used solely to overcome drag caused by air resistance. That is simply not the case, of course. Most of the power or thrust is used to generate lift. I am also a pilot and the four forces in balance at equilibrium are lift, weight, thrust and drag. There is no direct relationship between constant forward speed and engine power at all, since various power settings will produce exactly the same constant forward speed depending on the configuration of the airframe such as undercarriage up or down, degree of flap extension and so on.

    Question 24 page 215. Figure 9.29 - the x-axis should read "displacement (m)"
    Mistake fround by Anna Singleton

    Chapter 10
    On page 228 Example 4: the second last line should read Tf = 5 + 25 and the answer should read Tf = 30oC.
    A big thank you to Matthew Buckley for pointing this out. No-one noticed that for 6 years.

    Chapter 11
    In the third box down on page 235, the solution should read 2000 N m-2

    The solution on the top of page 249 shows only the change in length. The final length would then be 199.862m (or 199.9m approximately). Strictly speaking, if you are using 3 dignificant figures, the answer would still be 200 m I guess). Maybe the question should have given the length as 200.0 m (4SF). We'll fix that in the next edition. Thank-you to Prue Harvey at Moreton Bay College for her attention to detail.

    Chapter 13
    Page 280. (New Century Edition only) Question 8: the answer to part (a) is 20 cm NOT 2 cm. Thank-you to Katie Park for her vigilance.
    Page 280. (New Century Edition only) Question 10: the answer to part (a) is 20 cm. Thank you to Kirsty MacNamara for finding this one.
    Page 280. Question 23 should read "what is the range of wavelengths of radio waves?" The answer on page 701 should be in metres (600 m to 10 m).
    Page 281. Question 27 - the answer on page 701 should read: (a) A down, B down; (b) A to left, B to right.
    Page 281. Question 29(b) - delete L from answer on p 701; Q 29(c) - add L to answer.
    Page 282. Question 35. The speed should be in cm s-1 (4.0 cm s-1).
    Thanks to Jeff Knowles for spotting these.

    Page 283. (New Century Edition) Question 31(a). The amplitude is 4 centimetres.
    Thanks to Prue Harvey for spotting this.

    Chapter 14
    Page 294 (or page 292 in the 1997 edition). Under the heading Changing the wavelength the third line in the first paragraph should read "...if the wavelength is equal to or greater than the opening."
    Thanks to Ryan Seo for spotting this.

    Chapter 16
    Page 710 (or page 702 in the 1997 edition). The answer to Q 43 is 439 Hz not 895 Hz. See the worked solutions page for details.

    Chapter 17
    Page 369. Question 21. Answer on page 702 should be f = 4 cm.
    Thanks to James Keogh's class for knowing it was virtual.

    Chapter 18
    Page 378. Question 9(c)and 9(d). The answers (page 702) should be (c) 38o; (d) 1.09

    Chapter 20
    Page 403. Activity 20.6. "Optologist" should be spelt orthoptist - a person who corrects disorders of vision by exercises.
    Thanks to Margaret-Mary Althaus of the QUT Optometry School for finding this one.

    Chapter 21
    Page 424. Question 8(b): Answer (page 703) is 1.5 x 10-4 N
    Page 426 (New Century Edition). Question 8(b): Answer is 1.5 x 10-4 N
    Page 424. Question 8(d): Answer (p 703) is 6.0 x 10-4 N
    Page 433. Question 21: Answer (p 703) is "on each, Q = 7 x 10-9 C"

    Chapter 22
    Page 467: the formula for efficiency in Q34(c) should be power out/power in. A big thatnk you to Prue Harvey (Moreton Bay College) for finding that one. A Tim-tam is on its way Prue.
    Question 8(c): Answer (p 703) is "assuming the conductor B is copper, R = 0.8 ohm".
    Question 10: Answer (p 703) is "voltmeter reads 11.8 V, ammeter reads 0.39A".
    Question 16: Answer (p 703) is 18.9 kWh or 2.9 hours
    Question 39: Answer (p 703) is R = 232 or 43 ohm.

    Chapter 23
    Question 17: The answer is 0.72 mA (7.2 x 10-4 A and not 72 mA)
    Thank you to Eve Hsing for finding this one.

    Chapter 24
    Question 6: Answer (p 703) is 1.4 V (the 4.6 V is the voltage across the resistor Rc).
    Thanks to Rosana Miles for spotting this one. A packet of Tim-Tams is on the way.

    Chapter 27
    Question 17 (1999 edition): Answer (p 712) is (a) 8.478 MeV, 2.826 MeV/nucleon; (b) 7.71 MeV, 2.57 MeV/nucleon. Thank you to the 2000 Year 12 Physics class at Moreton Bay College for finding that one.

    Chapter 28
    Page 584. The scales on the graph are wrong.
    On the y-axis, the 5 x 105 should be half way down next to the dotted line.
    On the x-axis, the times should be 0, 15, 30, 45, 60 and the unit should be 'hours' not days.
    Question 44. Answer in back of the 1997 edition of the book says Reaction II produces 9.61 x 1012 J/kg. This is wrong - the correct answer is 5.72 x 1014 J/kg) and Reaction II produces more energy per kg than Reaction I.
    Thank you to Clint Dempster for finding that one.

    Chapter 30
    Page 635: The arrows have been left out of the two diagrams of the trains at the bottom of the page.

    Figure 30.9(a) should be like the figure below (the arrows should be spaced the same both sides):

    Figure 30.9(b) should be like the figure below (the arrows should be closer to the left door than they are to the right door):

    Page 640: The second line under the heading "Relationship between frames" should read (t > to)
    Page 646 of 1999 edition: The last line under the heading "Relationship between frames" should not have the r in the equation that starts to = t/r ...
    The "r" was the typesetters mark-up to indicate a square root ("radical") symbol which is there as well.

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