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

ERRATA
We've found a couple few more of errors in the text.
Chapter 1, p11: Subheading "Further Examples" (a) should be "radius of the Sun".
Thank you to
Joshua Noronha, St. Laurences College,
Brisbane.
Chapter 2, p33: 4th line from bottom. Answer to final part of example should be
negative (3.3 m s^{1}). Credit to Michael Schobbe, Year 11, Caboolture
SHS.
Chapter 2, p37 and p41. Table 2.7: the last column should have units for acceleration as ms^{2}. Same for acceleration in Table 2.10 p41. My thanks to students of Richard Baker, ELearning Leader, St Stephen's Catholic College, Mareeba, Qld.
Chapter 2, p50, Q37d. The correct answer is 715 km not 715 m. It was just a typo. My thanks to eagle eyed student Caitlin Ramsay in my Year 11 class at Moreton Bay College.
Chapter 2, p50, Q37f. The correct answer is 53846 hours not 55846. It was just a typo. My thanks to student Joshua Napier in Grade 11 at Meridan State College.
Chapter 2, p 50, Q40 about Brett Lee. Answer is 20
÷ (157.4 ×1000/(60×60)
= 0.46 s not 0.55 s as stated on p761. Credit to Alyssa Young, Year 11,
Caboolture SHS.
Chapter 2, p50,Q46(a) about the skateboard. The answer should be a displacement
of 225 m (area = 38 x 15/2  12x10/2 = 285  60 = 225 m). I misread the
question and worked out the time for 60s NOT 50s as required. My thanks to Kristoff Todd, Physics Teacher, Parklands Christian College, Park Ridge, Qld, and to Michael Leask and his students Joseph Weyer and Ryan Weston from Meridan State College. The answer for Q46(b) is probably closer to 395 m than 392.5. The line actually crosses the xaxis at 38 s not 37s as I originally worked out. Not that Q46(b) asks for DISTANCE not displacement.
Chapter 3, p57: Lines 4 & 5 should read "...but is heading northwest."
Spotted by Michael Leask.
Chapter 3, p 60. The
answer to Q4 should say that "height" is a vector (not a scalar) at it indicates
direction as well as magnitude. Thank you to Michael Leask, Caboolture SHS
Chapter 4, Q20: Answer
should be (a) (i) 3.04 m s^{2}; (ii) T = 104 N
Chapter 4, Q49: The angle was wrong. the answer is: Using trigonometry: the wind force F_{H} = F_{W} tan 37°
= 2.26 x 10^{3} N. Thank you to Bryan Corcoran for his eagle eye.
Chapter 4, Q59: Answer should be 15.3 m s^{1} (see worked solutions web
page).
Chapter 4, Q61: Answer is 953 N. Thank you to Brianna Towers for finding this.
Chapter 4, p5657. Physics teacher Rowan Barber points out: Looking at the example on the bottom of page 66 and the top of page 67 of New Century Senior Physics (2004) I am actually not sure how you got s= 37m for part (f). If one reads off the graph it looks like s = 40m, when t =15s. However if one substitutes t = 15s into the equation of best fit, s = 1.7t +16 s = 1.7 x 15 + 16 s = 25.5 + 16 s = 41.5 m. My response to Rowan (besides thanking him) is: "I think the question probably asked for the displacement at 12.5 s but a typo made it 15.0 s."
Chapter 4, p76. Q1(c) should be N22°W; Q1(d) should be S18°W. My thanks to Gaetano Caputa for alerting me to this.
Chapter 5, p112: Novel Challenge. The Acapulco cliff diver landed 5 m from the
base of the cliff, not 5 mm. Thank you to
Joshua Noronha, St. Laurence's College,
Brisbane.
Chapter 5, p 114. Activity 5.3. (a) should read Horizontal range s_{h}= u
cos θ × t = R; velocity
at maximum vertical height v_{v} = u sin θ + at/2 = 0. Thanks to Rod Duncan for
pointing that out.
Chapter 5, p129, Q26(d). The answer on page 763 for the horizontal range should
be in "m" not "m/s". Thank you to
Tristan Ockwell, St. Laurences College, Brisbane.
Chapter 6, p140. In Figure 6.2, the last number on the xaxis should be 10^{15} not 10^{15}. Thank you to Michael Leask, Science Teacher at Caboolture
State High School.
Chapter 6, p 137. Table 6.1 has wrong data in
the CaH and Ha
columns. The correct values are shown below. My thanks to Physics teaches Victor
Braun (for pointing out the error) and Mark Shields (for giving me the correct
values). Scratchies on their way chaps.
CaH  Ha  
396.9  656.3  
Virgo 
398.4 
658.8 
Perseus 
404.1 
668.2 
Hercules 
410.7 
679.1 
Pegasus II 
413.8 
684.3 
Ursa Major 1 
416.7 
689.1 
Gemini 
427.9 
707.5 
Ursa Major 2 
450.3 
744.7 
Hydra 
476.9 
788.7 
3C295 
579.5 
958.2 
Chapter 6, p143. Physics fact: the spin of 1998KY26 takes 10.7 minutes not
seconds. And the next line about the day being 0.09 seconds doesn't even make
sense. Delete it! Thank you again to Michael Leask at Caboolture State High
School.
Chapter 7, Q5, p172. The answer should be 9458 N not 9690 N as we have in
the back of the book. We (I) used the density of freshwater (1000 kg m^{3} instead of saltwater (1030 kg m^{3}). Thank you yet again to Michael Leask at Caboolture State High
School. We owe you a moche latte.
Chapter 8, Q16. Correct answer is 16.3 m s^{1}. I'm not sure how we got
that one wrong. A big thank you and two cans of Red Bull to the men with
the keen eyes from St Laurences' College, Brisbane: Joshua Noronha and
Duy Nguyen. Thanks again chaps!
Chapter 8, Q 14 (b), p197. The answer is 2.71 m/s not 1.86 m/s. The working is
as follows:
As the jet is backwards (v=35m/s) then (with bullet a, melon b and jet
c)
ma.ua =
(ma + mb  mc).v + mc.vc
0.01x545 = (0.01+30.3)v+0.3x35
5.45 + 10.5 = 2.71 m/s (not 1.86!?). A big thank you to the
Kennedy Assassination buff and Physics teacher Chris Smith (and his Yr 11
Physics class) at Sunshine Coast Grammar School.
Chapter 8, p206. In the solution to the example, the linear momentum should be
2.9 x 10^{3} not 2.9 x 103. Also, for the rotational momentum the
radius part should read (2.78 x 10^{3})^{2} not (2.78 x 10^{3})2.
That's not my fault  that's the typesetter stuffing things up afterwards. But,
I guess I should have checked. Thanks boys: that's Joshua Noronha & Ryan Schumacher from St Laurences' College, Brisbane again. This is getting embarrassing.
Chapter 10, Q40. Correct answer is 0.2 cm. See worked solutions. Credit to
Anthony Reid, North Lakes State College. We owe you a halfstrength decaf mocha
latte.
Chapter 11, page 264 Table 11.1. The units for alpha should be °C^{1} not m°C^{1} Thank you to Vic Braun for spotting that.
Chapter 11, page 267 Table 11.2. The units for beta should be °C^{1} not m°C^{1} Thank you again to Vic Braun for spotting that.
Chapter 11, page 266 Q 22. The answer in the back of the book has used the coefficient of expansion for steel (10 x 10^{6} °C^{1}) whereas the question clearly says "iron". Hence, the alpha value for iron should be used (12 x 10^{6} °C^{1}) which would given an answer of a 33°C temperature rise, and hence a final temperature (answer) or 43°C. My thanks to Physics teacher Adam Warren and his students from Glenala State High School (Brisbane) for this correction. However, another question often arises of whether the change in length for eaach 5 m rail is 1 mm or 2 mm. I believe it to be 2 mm based on this understanding:
In Case 1 the rail is seen to expand outwards from the centre of each 5 m rail. So it has to expand a total of 2 mm (1 mm from each end). The same goes for the rail on the right. Thus to close the 2 mm gap in the middle, each rail has to expand 2 mm. This gives and answer of 43°C for iron. If you say the rails are fixed at the outer ends (as in Case 2) then each 5 m rail has to expand just 1 mm, and this would given an answer of 26°C. I think Case 1 is correct.
Chapter 13, Q38. Correct answer is Open Pipe formula: L = nλ/4, where n = 1, 2, 3, 4, 5 (see webpage for
diagram). Big thank you to the eagle eyed Joshua Noronha (St Laurances).
Chapter 15, page 337, Q15. The answer is the back (1.1 x 10^{5}m
is wrong; it should be 8.281 x 10^{6} m. The solution is thus:
For a soap film in air, the refractive indices go LHL so the formula is 2d = m λ_{fil}_{m}
Where λ_{fil}_{m} is the wavelength of the blue light in the film, which equals the wavelength in
air/(refractive index of the soap film).
That is λ_{fil}_{m} = λ_{air}/1.33 =
450/1.33 = 338 nm.
2d = m x 338 where m = the number of the band starting with the first band as
m=0; thus for the 50th band m = 49.
2d = 49 x 338
d = 8281 nm or 8.281 x 10^{6} m
The answer in the back was wrong as the wavelength of light in air (450nm) was
used instead of the wavelength in the water medium (450/1.33).
Chapter 15, p326. Example, parts (c) and (d). The wavelength of 600nm found in part (b) has not been used in parts (C) and (D)  in fact 640nm has been used for some unknown reason (fat fingers?). The correct answer for (c) should be 2.25 x 10^{2} m; and the correct answer for (d) should be x = 4.5 mm, thus thickness = 9.0 mm. As well, the formula in the box for (d) has some typos: the numerator on the LHS should read (1½) x 600 x 10^{9}. My thanks again to Joshua Noronha (St Laurances). Goodness knows what he'll find in Chapter 16. I can hardly wait. My thanks to Rajika Subasinghe Arachchige (yet again) for correcting my corrections.
Chapter 16, page 364. Table 16.4 has a chainsaw listed with an absolute intensity of 10^{0.5}. This should read 10^{–0.5}. My thanks to Benjamin Gray from Brisbane State High School for spotting this mistake. I'll fix it in the next edition for sure. Thanks, Ben. Appreciated.
Chapter 16, p376. The formula at the top of the page should have a v_{s} in the denominator (on the bottom instead of v_{o}. What a stupid mistake. My thanks to ACT Physics student Rajika Subasinghe Arachchige.
Chapter 18, p403. Question 8(c) should be 38° not 38°C (as it is an angle not a temperature). Heartfelt thanks to my eagleeyed Year 11 student Claudia Bamford who wrote (in the form of a haiku it seems):
I have some excellent news!!
I found a mistake in the textbook. Written by yourself. 
Claudia also said: "I have attached a photo as evidence. Does this mean I get my name on your website???" Yes Claudia, a can of Golder Circle Creaming Soda will be yours soon. I'll make sure it is at 38°C.
Chapter 19, p 421. The text referring to Figure 19.8(c) under the subheading "Characteristics of the image" is wrong. Delete the words "between F and the lens"  this is clearly wrong. Thank you to Ron Grant for his eagle eye.
Chapter 19, p 423. Example calculation part (d): the image should be at 20 cm, not 10 cm. The error is in the last line. My thanks to one of Rowan Barber's physics students at Alexandra Hills TAFE, Redlands. Thanks guys.
Chapter 19 Q15, p426. It should say that the image was found on a screen.
Thanks Ron.
Chapter 19 Q27, p428. The answer in the back is wrong but the worked solution on
the web page is correct. Thanks Ron.
Chapter 21 Q26 (d) page 464. The questions also asks for velocity in the direction of the field. The answer would be v = u + at = 0 + 1.67 x 10^{8} x 8.3 x 10^{6} = 1386 m/s, downwards. I also worked out the resultant velocity and that came to 1833 m/s at an angle of 49° below the horizontal. My thanks to Cobe LeClerc from Meridan State College for pointing out the omission.
Chapter 22, Q9, p479. In the table where it has 4 V the corresponding current should be 0.025A not 0.25A. Well spotted by Issy Burdon. Thanks Issy  I'd send a Scratchit but times are tough.
Chapter 24 Q26, p545. The answer is 280 kohm not 280 ohms. Another one well spotted by Issy Burdon from St Margarets Anglican Girls School, Brisbane. Thanks Issy.
Chapter 26 Q32. The slope of the line of the graph is 1.32 hence B =m/LN = 1.32/(0.1x100) = 0.132 N/A (NOT 1.3 as the book states). Sorry about the error. My thanks to teacher Chris Smith, Head of Learning Area  Science, Sunshine Coast Grammar School, Queensland.
Chapter 27. Questions 18 to 28 have the answers in the back (page 768) as
Q1929.
Thank you to Kevin Fullbrook  Gold Coast Institute of TAFE for pointing that out.
Chapter 28, page 624  Predicting the type of decay. The
two dot points are in conflict with the graph on page 625. The graph is correct.
The dot points should read:
Chapter 27 p 615: Figure 27.22 is said to be for question 29, but is for
question 28. Likewise Figures 27.21 and 27.20 are said to be for questions 25
and 28 respectively, but in fact relate to questions 24 and 27.
Thanks to David Wiseman  Year 12 student at Toowoomba State High School  for
spotting that.
Chapter 28, page 625: the reaction at
the bottom of the page should have as the product C13 not C12. David Wiseman yet again.
Chapter 28, page 625.
Q9a has a wrong answer in the back of the book. It should read Na22 as in the
question. The rest is okay. Thanks to David Wiseman from Toowoomba SHS for spotting that too.
Chapter 28 p 629, first line under the N13 decay
equation. Should read "nitrogen" not "sodium". I think I had a brain freeze. Mr
Pedantic  David Wiseman  from Toowoomba SHS again.
Chapter 28 p 645: This is not really an error but to keep Mr Pedantic (David
Wiseman) happy I will acknowledge that the sentence could be written more
clearly. He said that his friend Steven Connell is disturbed with the statement
six lines from the bottom that says 'Radiation that
deposits one joule of energy per kilogram of tissue is called the absorbed dose.' David said "However, radiation that
deposits one joule per kilogram would have an absorbed dose of one gray. I think
the sentence should read something like 'The absorbed dose is the radiation
absorbed in joules per kilogram of tissue'". Thanks again David, I will make
the change in the next edition. You should get a job with Oxford University
Press as a proofreader (it would get you off my
back).
Chapter 28, p652, Q44. Should refer to Appendix 8 (Table of Masses) not Appendix 7. My thanks to my nitpicking Year 11 physics genius Mia Broedelet for finding this. Run out of cans of Creaming Soda so its a pat on the back for you I'm afraid.
Chapter 29, p661. Jayke Anderson from Caboolture State High School pointed out that the equation used for solution part (c), v=fλ, is missing λ in the rearrangement. I would have sent a $2 Scratchie if it wasn't for the Global Financial Crisis.