Supervisions: Worked Solutions to 2018 Past Papers

Here are numerical answers, hints, some heavier hints that only appear when the text is highlighted, and selected images and videos of worked solutions to the 2018 past papers in IA NST Maths, IB Physics A, and IB Physics B. Do let me know if you think you've found any errors! (I'd afraid there will probably be one or two.)

Note that where there are video clips, they don't have everything written that you'd write in an answer---I've made use of the fact that I can say things as well to avoid writing everything! You'd need to include more words, and intermediate steps in some cases, in an exam.

To get most benefit from the process,

People often like to read worked solutions as soon as they get stuck. I've relegated the explanation of why this is a mistake to the bottom of this page as it had become a little polemical and unkind. (But it is, for all that, correct.) There is also somewhat friendlier advice on the advice page.

IA NST Mathematics, 2018 Paper 1

1

Answers: (a) x = y(log(10))/(log(2)); (b) 79.

2

Answers: (a) x = -5, -1/2, or 1; (b) (-3 - √5)/2 < x < (-3 + √5)/2.

3

Answers: φ=0 with t=-1, and φ=π/2 with t=0.

4

Answers: (a) 1/9; (b) exp(sin2(x)) + C.

5

Answers: (a) 2π/3; (b) B(3√3)/(4π).

6

Answers: (a) √3; (b) cos-1(1/3).

7

Answers: (a) 1.094; (b) 9/2.

(a) Use the first few terms of a binomial expansion.

Solution.

8

Answers: x=π/4, y=exp(π/4)/√2.

9

Answers: (a) (-1/2,-1); (b) 3/2.

10

Answers: (a) (1,4); (b) (0,-2).

Solution.

11Z

Answers: (a) |z|=√2/2, arg(z)=5π/12, |w|=√((log(2))2/4 + 25π2/144), arg(w) = π - tan-1(5π/(6 log(2))); (b) z=iπ(n - 1/4) where n is any integer.

(a) Recall the rules for modulus and argument of a product, to get z. The log is obtained from the modulus / argument form. (b) Write tanh in terms of exponentials. (c) Just use De Moivre's theorem with n=5, and take the real part. This question is rather straightforward. But watch the quadrant of w in (a).

Solution for (b).

12W

Answers: (a)(ii) (√π/2)(√2 - 1); (a)(iii) √π(√2 - 1); (b)(i) (4π/3)abc; (b)(ii) 0.

(a)(iii) Get the new limits for I from the diagram. They are from y=0 to y=∞ on the outside then from x=y to x=2y on the inside. Now integrate by parts to get I in terms of the integral you need to find. And you already know what I is from (a)(ii). (b)(i) It can be done in Cartesians, but the limits are inconvenient. It's easier to use modified spherical co-ordinates, re-scaling by a, b, and c. (b)(ii) The answer above is hint enough; we're looking for a symmetry property. Specifically that if we split it into two integrals added up, each is odd in one of the Cartesian variables.

Solution to (a); solution to (b).

13Y

Answers: (a) y = log(x)/x + C/x; (b) y = exp(3x)/4 + C exp(-x); (c) 5x2/2 + 4xy - 2y4 = C; (d) log(y) = 2√(x/y) + C.

(a) With a little shuffling this is linear and takes an integrating factor. (b) Likewise. (c) Cross-multiply, take the RHS over to the LHS, then recognize what you've got as an exact differential. (You may not think of this as a technique for solving ODEs, but occasionally it comes in useful.) (d) This takes one of the standard substitutions, u = y/x.

14X

Answers: (a) 4xy = 1; (c) (-√3/2,-√3/2), (-1/2,1/2), (0,0), (1/2,-1/2), (√3/2,√3/2); (d) Respectively minimum, maximum, saddle point, maximum, minimum.

Solution including approximate sketch.

15T

Answers: (b) 1/a0 + (a1/a02)x + (a12/a03 - a2/a02)x2; (c) 1 + (1/6)x4; (d) log(2) + (1/2)x-1 + (3/8)x-2.

(b) It's possible to find the derivatives of 1/f(x) in terms of those of f(x) to get the coefficients in the series for the former. Or alternatively substitute the entire series into a binomial expansion. (c) Treat dividing by the 1/2 power as multiplication by the -1/2 power. (d) Use the laws of logarithms to turn this into something plus log(1 + something small), where the something small consists of negative powers of x.

Solution to (a) and (d).

16Y

Answers: (a) yi = x0 tan(θi); (b) g(θ) = 1/π and G(θ) = θ/π + 1/2 for θ between -π/2 and π/2, with both 0 outside this range except that G(θ) is 1 for θ > π/2; (c) F(y) = 1/2 + tan-1(y/x0)/π; (f) x0 log(3)/π; (g) x0 log(3)/π.

(g) This is a geometric series as (f) gave us a mean proportional to x0.

Solution.

17Z

Answers: (a) (1/2)log(1 - cot2(x)) + C; (c) En+1 = (2n-1)En/(2n) hence E4 = 5π/32.

(a) Substitute u = tan(x) then use partial fractions. (b) This is just integration by parts. (c) It's probably easier not to use trig. substitutions (until evaluating E1 as π) but instead in the integrand for En multiply top and bottom by an extra factor of (1 + x2), then write the integral as two integrals added up, the first of which is En+1 and the second of which can be wrestled into a multiple of En if you remember that the x2 is x times x when setting up integration by parts. This is much harder than (b)!

Solution to (c).

18R

Answers: (a) 0, 6, 0, 0; (b) -6 with (1,0,-1)/√2, 0 with (1,2,1)/√6, 12 with (1,-1,1)/√3; (c) (1,2,1) [or any nonzero multiple thereof].

(a) Recall how the determinants of the transpose and of the square relate to the original determinant; there's no need to square the whole matrix! (c) Follow the instructions.

Solution to (c).

19T*

Answers: (a)(ii)(1) No for m=0, yes for m=1 and m=2; (a)(ii)(2) No for m=0 and m=1, yes for m=2; (b)(i) diverges; (b)(ii) converges.

(b)(i) Use the ratio test (or just look at what the term itself does as n goes to ∞). (b)(ii) There are several methods, but perhaps find a way to use a binomial expansion of the square root when n is large (which will require taking out a factor from inside the square root so that you leave behind 1 + a2/n4), then look at the powers of n in the terms you've got. A more elegant (but perhaps harder-to-spot) method is to multiply top and bottom of the term by the factor √(n4 + a2) - n2.

20S*

Answers: (a) tan-1(α)/α.

(a) After differentiating the first result, put in α=√3 to derive the result required. (b) Follow the instructions. The solution for E is in the first instance π log(1 + α) + C but we can find that C is zero by looking at what E does, in the original definition, when α=0. Hence, with α=1, we get the result required.

IA NST Mathematics, 2018 Paper 2

1

Answers: (a) (0,1,2)/√5; (b) 2/√5.

2

Answers: First nπi then (2m+1)πi where n and m are any integers.

Solution.

3

Answer: 2 cosh(φ) and 2 sinh(φ) [or write out the exponentials].

4

Answer: x2 - (3/2)x4.

I'd expand the numerator by substituting x2 into the log(1 + z) series, then multiply by the binomial expansion of (1 + x2)-1.

5

Answer: log(x exp(-t)).

Solution.

6

Answers: (a) cos(f)(df/dr)(x/r, y/r, z/r); (b) (df/dr)z/r.

Solution.

7

Answers: (a) y = x exp(-x); (b) y = 3/(3 + x3).

(a) Find an integrating factor, and don't forget to multiply the RHS by it! (b) Take the (xy)2 term to the other side and use separation of variables.

8

Answers: (a) 0; (b) π2.

Solution.

9

Answers: (-√(3/2),0) and (√(3/2),0).

10

Answers: (a) 1/2; (b) π/2.

To find α we need a quick integral, but the mean is obvious from symmetry so doesn't require integration.

11S

Answers: (b) √15; (c) (b× + 2b - 3)/5.

(a) The trick is to use the vector triple product identity on the quadruple product, first treating the (a×b) as a single vector, then the (c×d), then equating the two formulae. (b) The first bit is bookwork. Following instructions, we get possible values of µ as -1 and 1; the method could miss µ=0, but that gives a bigger l in any case. (c) First dot the equation with to discover that x·n̂ is -1. Then cross with the equation, tidy up with the vector triple product identity, combine with the original equation to eliminate b×, then substitute in the value of x·n̂, and solve.

Solution to (c).

12X

Answers: (a) sin(φ); (b) cos(φ)/r; (c) fy = gr sin(φ) + gφ cos(φ)/r; (d) fyy = grrs2 + 2gsc/r - 2gφsc/r2 + grc2/r + gφφc2/r2, hence ∇2f is grr + gr/r + gφφ/r2; (e)(i) f=2xy and f=(2y,2x); (e)(iii) 3√2.

This question is rather long and full of detail. The method in the "hard" step in part (d), of applying a differential operator to itself, is the same as that for one of the question sheet questions, showing that ∇2 in two dimensions is invariant under a rotation of co-ordinates; I'm not going to reproduce it here.

13R

Answers: (a)(i) F gives 2 and G gives 0; (a)(ii) F gives 2π and G gives 0; (a) F isn't conservative, but G is, equalling (x2y2); (b)(i) from (1,0,0) to (0,1,2π/3); (b)(ii) 5/6.

(b)(ii) This is an integral with respect to scalar arclength, which is a little unusual. We get everything in terms of t, and ds in terms of dt by using ds2 = dx2 + dy2 + dz2. (We should find ds is 10 dt.)

14V

Answers: (a)(i) Ω = {G∩x0, G∩x1, G∩x2, D∩x0, D∩x1, D∩x2}; (a)(ii) pg0; (a)(iii) pg0 + (1-p)d0; (a)(iv) (1-p)d1/(pg1 + (1-p)d1); (a)(v) p(1-g0)/(p(1-g0) + (1-p)(1-d0)); (b)(iii) 1/(1+g0) [or just 1/2 if we allow ourselves to modify g0, the question being not quite clear].

Answer for (bi) and (bii).

15Y

Answers: (a)(i) y = (A cos(x) + B sin(x))exp(x) + x2 + 2x + 1; (a)(ii) y = A exp(-x) + B exp(-2x) + x exp(-x); (b) y = kx log x + Cx.

(b) Having obtained the given equation, put v = du/dx. Solving the equation for v results in v=k/x, then integrate to get u, then substitute back.

16W

Answers: (a)(i) 6a·b; (a)(ii) 0; (a)(iii) -2a·b; (a)(iv) a×b; (b)(i) |a×b|2/2; (b)(ii) 2π(a·b)R3.

(a)(i) This follows from the value of ∇·r (,,which is 3). (a)(ii) This follows from the value of ×r (which is 0). (a)(iii) Use the vector triple product identity, then work in components for the bit you haven't found before. (a)(iv) See (a)(iii). (b)(i) Note that as the triangle contains the origin, r is within the triangle, so it is orthogonal to the element of vector surface, meaning that the only nonzero contribution comes from the first part of F, which is in the same direction as the element of vector surface in any case. (bii) You may want to consider using a vector calculus theorem here, specifically the divergence theorem, the divergence of F being conveniently constant.

Solution to (b).

17R

Answers: (c)(ii) -1 and 6; (c)(iii) for µ=-1, any non-zero multiple of (1,1,0), and for µ=6, any non-zero multiple of (18,-3,-7); (d) λ=7 and r is (x, x-1, 2) for any x, a straight line.

(c)(i) Consider whether the three vectors form a basis. (d) Consider whether b is in the plane (and we know it is only a plane) spanned by the three c vectors.

18T

Answers: (b) the a and b turn out to be the Fourier coefficients (given by the usual formulae) for f(t); (c) the a are all zero, and bn is zero for even n and 4/(πn) for odd n, leading to the required form; (d) p=2.

(b) Substitute the K formula into the yf formula, use the cosine subtraction formula on the cos(nθ - nt) that results, and equate coefficients. (c) Remember that you can split the integral into the integral from -π to 0 plus the integral from 0 to π, and this may be useful if the definition of the function changes for positive and negative t, as it does here.

19W*

Answers: (a)(i) one stationary point, (-2/(a2-4), -a/(a2-4), (4-3a2)/(a2-4)2); (a)(ii) maximum if |a| < 2, saddle if |a| > 2, no stationary point if |a|=2.

(a)(ii) Starting from f = axy + x - x2 - y2, obtain the determinant of the Hessian as 4 - a2, with the second-order partial with respect to x always negative, so the condition depends only on a. (b) The Lagrange multiplier method will give you, for each variable nk, the correct exponential factor with the k-dependence, provided that you identify one of your Lagrange multipliers with the β in the question. So far it's like the questions related to the Boltzmann distribution from the question sheet. But then we need to go further, finding the other Lagrange multiplier by using the first constraint (which will need the sum of a geometric series). This is still just four marks! Then applying the other constraint allows us to relate E and β as required. Summing that sum is made easier if you notice that it's equivalent to applying a certain mathematical operation to the previous sum that you did, namely differentiation with respect to β. (This is a hard question. Unless you already know second-year thermodynamics, in which case it's easy.)

Solution to (b).

20Y*

Answers: (a)(i) ηξ/4 + g(ξ) + h(η) where g and h are arbitrary functions; (a)(ii) a=b=d=0 and c=-1/2.

(a)(ii) This is more easily done independently of (i). The boundary conditions must apply for all x, giving a and d from the condition on u and then b from the condition on its derivative. Finally c comes from the equation itself. (b) Find the first derivatives of z with respect to x and y by implicit partial differentiation of the implicit equation. The differentiate again. We should get 2xz/(z2 + x)3 and -2z/(z2 + x)3 for the second x- and y- derivatives respectively, hence the result.

Solution to (ai): page 1, page 2.

IB Physics A, 2018 Paper 1

A1

Answers: 6Li is a fermion and 7Li is a boson.

Count the subatomic particles (nucleons and electrons)---they all have spin 1/2!

A2

Answer: 8.5×10-10 m.

The maximum wavelength corresponds to the smallest energy gap, which is between the ground and first excited states of this particle in a box.

Solution.

A3

Answers: eigenvalues -ħ/2 and ħ/2 with eigenfunctions (α + β)/√2 and (α - β)/√2 respectively.

This is most straightforwardly set up as a matrix eigenvector / eigenvalue problem. Using α = (1,0) and β = (0,1) the Sx operator becomes ħ/2 times the matrix with 0's in both diagonal positions and 1's in both off-diagonal positions which can be obtained by writing Sx in terms of the ladder operators.

A4

Answers: 125.7 kg m2, 0.52%.

We can work with relative errors and the special case of combining errors when the formula just consists of multiplication and powers.

Solution.

A5

Answer: capacitance 100 nF.

The examiners' notes make clear that they wanted a passive filter here---one resistor and one capacitor, as studied in IA---and didn't approve of all of the op-amps that they saw. They even point a moral about the selective revision that they suspect has led students to serve up chips with everything. It don't think that this is fair, but then examiners are necessarily further from ideal than op-amps.

B6

Answers: (b) A exp(ik0x) + B exp(-ik0x), C exp(ik1x) + D exp(-ik1x), F exp(ik2x) where ħ2k02/2m = E, ħ2k12/2m = E - V1, ħ2k22/2m = E - V2 with the usual boundary conditions; (c) E(E - V2) = (E - V1)2; (d) F = (ik1/k2)exp(-ik2a) A.

(a) This is bookwork. (b) This is standard and straightforward. (c) We should get here using the boundary conditions: first decide we need B=0, then perhaps eliminate A and F, and finally C-D and C+D to get the condition on the k which becomes the condition on the energies. But note that it is the expected condition on the k; we've seen quarter-wave layers before! (d) Finding J using the definition as in (a) is straightforward when, as here, we have single plane waves.

B7

Answers: (b) C = √(ma/2)/ħ for both, energies ħ√(a/(2m)) and 3ħ√(a/(2m)), the number of nodes; (c) wavefunction zero at the origin, energy the same as the second energy from (b); (d) 1/π.

(b) Just substitute in to the equation, and equate coefficients. (c) We already have the appropriate solution from the previous part. (d) This is the modulus squared of the inner product of the new ground state with the old ground state: we'll need the normalization factors for each first.

Solution for diagram and (d).

B8

Answers: (a) eigenfunctions A exp(inφ) + B exp(-inφ) for n an integer ≥ 0, energies ħ2n2/(2I); (b) angular momentum -2ħ, 0, 2ħ with energies 2ħ2/I, 0, 2ħ2/I and probabilities 1/6, 2/3, 1/6 respectively; (c) (1 + cos(2φ)exp(-2iħt/I))/√(3π).

(b) We can avoid having to take inner products here, because it's straightforward to write cos2φ in terms of the eigenfunctions of angular momentum (which are exp(inφ) for any integer n). (c) Multiply the terms in the wavefunction, when expressed in terms of the stationary states, by the appropriate time-dependent exponential phase factors coming from the energies.

Solution: page 1, page 2, page 3, page 4.

B9

Answers: (d) <p>t/m and -<d(V(x))/dx>.

This is essentially bookwork. As a question it is too easy because too familiar.

D12

Answers: (b) something like from n=1 to 5 or from n=1 to 6; (c) error in N0 is 1.4 (absolute error) which makes it much more accurate than the counts in the table, whose errors are the square roots of their values; (d) b=1.0 gives a χ2 of 6.34 (from the first five or six points) which is much smaller and more believable than that for b=1.1; (e) add a constant background.

In (b) we should really plot log(count/N0) against n, producing a straight line through the origin whose slope gives minus b. The last few points lie off the line, and we are supposed just to make some reasonable guess of how many points are well described by it. (c) Use Poisson statistics, as these are counts, discrete events occurring independently in continuous time at some underlying average rate. The standard deviation is just the square root of the mean. (d) This is just computation, providing that you have the correct errors (the square roots of the theoretical counts, ideally, though the square roots of the measured counts are almost as good). We will need to restrict ourselves to the first five or six points because we know from our graph that the model doesn't work beyond that, whatever we do with b. (e) A constant background is the obvious explanation on physical grounds, as well as looking like it can explain the behaviour of the tail of the data.

D13

IB Physics A, 2018 Paper 2

A1

Answers: 2.53×10-7 and 5.06×1010 m.

This question, like so many section A questions, uses Δθ = 1.22λ/D, and λ = hc/E, and small angles.

A2

Answer: 0.447 mm.

There's a formula for this, radius of nth zone = √(nλR), where n=2 in this case, and with it we need 1/R = 1/a + 1/b if a and b are the source-to-aperture and aperture-to-screen distances.

A3

Answers: air-to-body 0.00098 and water-to-body 0.99767

This is a familiar example. Recall that the power transmission coefficient isn't just the square of the amplitude transition coefficient. We can get it from one minus the power reflexion coefficient (and the power reflexion coefficient really is just the modulus squared of the amplitude reflexion coefficient as we're in the same medium) or remember (or derive!) the full formula for the power transmission coefficient itself.

A4

Solution.

A5

Answers: 0.21 carriers per atom---with the usual sign convention it is -0.21 electrons, so it should be understood as 0.21 holes, though this is really the surplus left after partial cancellation of electrons and holes with different effective masses.

The Hall coefficient is -1/ne, where n gives the number density of charge carriers, which we then compare to the number density of atoms.

B6

Answers: (e) as it depends only on γ, the dissipative term, which here is R/L, and we can optimize by reducing R as far as possible.

This question is almost all bookwork, enlivened only by the confusion over factors of m in the original paper. (It's not quite perfectly fixed even in the final version we now have; the γ/m in (a) should presumably just be γ.)

B7

Answers: (c) 4×10-3; (d) same angular separation; (e) still same angular separation for fine fringes but envelope function zeros separated by 2×10-2 (twice this across central maximum of envelope).

Nothing new or interesting happens in this question.

Sketch for the last bit of (e).

B8

Answers: (c) minima at x=λ/2, 3λ/2, etc,; (d) first minimum in I at 147.3 nm, spacing of minima in I 294.7 nm, first zero in visibility at 144.7 µm, spacing of zeros in visibility 289.4 µm. (e) 6.25 mm

The first few parts are bookwork. (d) We can show that the intensity is proportional to 1 + cos((k1 + k2)x/2)cos((k1 - k2)x/2) to get the expected fringes with oscillating visibility. Note that x is twice the distance moved by the mirror, and in (d), unlike (c), we plot against the latter. (e) Use the fact that the fringe visibility (as a function of τ=x/c) goes as the Fourier transform of the power spectrum (as a function of ω=2πc/λ).

Solution.

B9

Answers: (b) √(Tρ); (e)(i) if the incident wave is in the region with density ρ1, ρ2 > ρ1, giving a phase change of π; (e)(ii) ρ2 is 0 or ∞; (e)(iii) ρ2 = ρ1.

More bookwork. It's hard to see quite what was wanted in (c): the formula given works for the power carried along the string because at each value of x the string position varies sinusoidally with time, just like an oscillator. The examiners' notes, assuming that (c) and (d) in the notes should be taken to mean (b) and (c) as printed in the question, indicate that we should use impedance somehow in (c), but neither the original derivation for oscillators, nor the extension to strings, requires a linear force / velocity relationship. Of course we often assume one, to get power in terms of velocity squared and impedance, but that's not what we're doing here. (Clearly in (c) there should also be a dot between the vectors for the proper notation.) (e) Write down the formulae for r and R as part of your explanation of what is going on, in terms of the Z's hence the ρ's, though you should be familiar with all of these cases and limits anyway.

C10

Answers: (b) density of states mV√(2mε)/(π2ħ3), Fermi energy ħ2(3π2n)2/3/(2m); (e) 0.299 J K-1mol-1, but there are phonons.

This is largely bookwork; in (c) we have to assume that the density of states is sufficiently slowly-varying over a range within kT of µ that µ itself can be taken to be constant.

C11

Answers: (c) 0.00136 m s-1, 2.71 µm.

Also largely bookwork. In (c) the current density nevdr equals 13A/(cross-sectional area of wire).

Solution for (d).

IB Physics B, 2018 Paper 1

A1

We need the superposition of the fields due to plane conductors: try drawing them out individually (on the same diagram) and then doing a vector sum. The fields in each region should be uniform, but those in the 45° region should be stronger (i.e. with field lines more closely spaced).

Solution.

A2

Answer: 3Ar/(2rz4).

Consider the flux of fluid through the surface of a cylinder of radius r and length δz, on-axis.

Solution.

A3

For all sorts of reasons. Don't try this at home. But clearly the intent of the question is that there may not be enough upthrust to support the boat if the gas bubbles reduce the mean density of the water too much.

A4

Answer: 2.84×1013 Hz.

There's a formula for this.

A5

This is in the lecture notes.

B6

Answers: the last part is a factor of 3.

For (a) and (b) we can first find the solution in the potential φ by trying a linear combination of the two solutions to Laplace's equation found earlier, then finding the constants associated with each by applying boundary conditions: for large r we must get the solution on the upper plate, and at r=a we must get φ=0.

Solution (for later parts of question).

B7

Answers: IΔxΔy(0,0,1); at Cambridge magnitude 58.7 µT, direction 69.4° below the horizontal (within a plane that includes the vertical, and due North).

(a) This is fairly quick: use small-angle approximations to get the contributions from the four sides, which are all equal. (b) More work is needed here. The sides of length Δy only approximately cancel out in their effect on B; the difference between them has a term first order in Δx, producing a formula for B which is precisely 2 times the one we want. There's then a contribution from the sides of length Δx: because of the small angles this time the term itself is of order Δy, but both contribute equally, producing a formula for B which is half as big as the one we want, and in the opposite direction. The sum of the contributions is then the answer. For the last part, split the magnetic dipole moment at the centre of the Earth into a component in the direction away from Cambridge, and another component perpendicular to that. The former gives the downward component of the B-field at Cambridge, 2B0 sin 53°, and the latter the Northward component, B0 cos 53°.

Solution: page 1, page 2, page 3, page 4, page 5 (where the last result should have the direction specified as in the answers above).

B8

Answers: A = E02√(σ/(8µ0ω)).

Much of this is bookwork. The examiners point out (and they are right here) that we should justify which root of i to take. When finding the time-averaged Poynting flux, recall that if E and H are the usual complex quantities (such that the actual fields are Re(E) and Re(H)) then the time average of the product of the fields is (1/2)Re(E*H).

B9

Answers: speed is 1/√(LC) which in this case is c/√ε; d=0.325 mm; force is 4.08×10-4 N.

The capacitance is straightforward. For the inductance find the (necessarily uniform) B-field using an Ampèrian loop which closes a long way away, and get the flux from there. The next few parts are standard. Find d by recalling that there is no reflexion if we match impedances, i.e. make our √(L/C) equal the 5Ω. Find the force as EQ, where Q is related to V via the known capacitance, so we can get it in terms of V2, which comes from the power. (We can implicitly use r.m.s. voltage and mean power and force as everything is in phase.)

C10

Answers: (a) stable unless I3 is intermediate between I1 and I2, in which case unstable; (b) always stable, giving rise to steady precession.

As far as the formula for p2 this is bookwork and substituting in. Note that a negative p2 is needed for complex exponentials, oscillatory behaviour, and hence stability, which gives the solutions to (a) and (b). For the last part, consider the space of the angular momentum vector within the body frame. Conservation of energy constrains us to be on an ellipsoid in that space, where the axes depend on the moments of inertia about the principal axes. But then conservation of angular momentum constrains us to a sphere of fixed radius. The intersection is little, disjoint rings close to a rotation about the axes with the greatest or least moment of inertia, giving stability. But there's a much richer space to explore close to a rotation about the axis of intermediate moment of inertia.

C11

Answer: 1.29×109 m.

(a) Use small changes. (b) Use small angles. Then for the corotating object, note that ΔF=mω2a for the centrifugal effect, and that we can relate this ω2 to GM/R3 in the orbit. For the asteroid equate, at radius a anywhere in the body, the force per unit volume pulling in due to gravity, and the tidal force per unit volume.

C12

Answers: (a) 0.071°; (b) 0.0829 J; (c) 0.543 J.

The calculation of C has been done in an examples sheet question. Briefly, we need to find the strain angle at a distance r from the axis, which is rφ/L, then the sheer stress, then the torque per unit area, Gφr2/L, then the integral of that over the circle (area element in plane polars) to get the total torque, which is C. (a) The power transmitted by the shaft is Cω, allowing us to find C hence φ. (b) Energy is Cφ/2. (c) We just need the moment of inertia of the cylinder to find this. For thin slices, the original C formula wants the small change in φ across the slice if L is dx, so becomes C = (πa4G/2)(dφ/dx). Then look at the first order change in that. This becomes the net torque on the small element and gives rise to angular acceleration in the last part.

C13

Answers: a = dρg/η, b = ρg/2η; ρgd3/3η; just replace g with g sin(θ) throughout.

This should be reasonably straightforward. It's not clear why there are so many marks for tilting the plate. The component of g in the x-direction introduces a pressure gradient in the fluid, but that doesn't affect the viscous steady-state solution at all. It's just a case of replacing g with the z-component of it.

IB Physics B, 2018 Paper 2

A1

I'd answer this by sketching the (electric) field pattern associated with a TE01 mode and pointing out that we can't do this with magnetic fields because it breaks the continuity of the perpendicular component of B at the boundary (a condition that ultimately comes from ∇·B = 0).

A2

Answer: 14.9 kW.

This should just work. Remember to use absolute temperatures (i.e. in Kelvin.)

A3

Answer: V = -n2mA2/nx2-2/n/2 + C.

Find the acceleration, and get it in terms of x rather than t. Then -dV/dx = ma becomes a differential equation for V in terms of x.

Solution.

A4

Answers: something like 900nm, but it's quite sensitive to what you guess for the molecular size.

Using p=nkT (the ideal gas law) we get a number density n of 2.4×1025m-3. So far so accurate. Now we want λ = 1/(nσ√2), where σ is πd2, but d is the diameter of a molecule, which we have to guess. Given air is mostly N2, 1Å (which you might choose anyway on order-of-magnitude grounds) is about right for d. But this is disturbingly close to chemistry.

A5

"Thermal effects can be ignored" means T = 0 K in effect, so states are full if the energy for an electron filling them is less than µ, and empty if it is greater. We must share the interaction energy equally amongst the electrons, and draw separate graphs for U<0 and U>0.

Solution.

C10

Answers: dF = -SdT + γdA; work 2.2 µJ, heat 4.4 µJ; β = -dγ/dT; -6.3×10-5K, with internal energy increasing by the amount of work done, almost exactly the same amount of work as before (2.2 µJ).

(a) This just needs the Maxwell relation found above. (b) Here we need to use the Maxwell relation that comes from the original dU, along with the observation that partial dS/dT at constant A is positive as it is CA/T. (c) Similarly split the dS from the original expression for dU into a dT and dA part, and use the form of CA quoted in (b) as well as the Maxwell relation as in (a). In the final parts, reversible adiabatic changes have dS = 0.

Outline of solution.

C11

This is mostly bookwork. At the end, with one dimension fewer, we have a quadrant rather than an octant in k-space, hence dN/dν is proportional to ν rather than ν2, giving the result.

C12

Answers: T(a) = Tair + Ha/(2γ).

(a) H is dissipated power per unit volume, which is JE = J2/σ. (b) For a steady state we need to solve ∇2T = -H/κ, using just the radial-derivative part of the Laplacian, and the boundary condition at the surface. In the last part the heat flux per unit area, γ(T(a) - Tair), has to equal -κ(dT/dr) at r=a, which is Ha/2 using the solution from (b). (I'm working with H to avoid writing the large block of constants.)

Bonus Question: IB Physics B, 2015 Paper 1

C12

Answers: M = E(1-σ)/((1-2σ)(1+σ)); ωH = (a2/b)√(3πM/4mh), ωV = (a2/b)√(3πG/4mh); σ=0.4919.

To find M, write out all three of the relationships between the e's and the τ's, then eliminate τ2 and τ3. The calculation of TV is identical to that of C in question C12 above (and to an examples sheet question). To calculate TV, find in terms of the displacement y from the axis the vertical strain, yθ/h in the small-angle approximation, the vertical stress (via M), and the torque per unit area about the axis, My2θ/h. Then put y = r sin(α) where α is the plane polar angle (since we're using θ and φ already for other things) and integrate over the area of the circle. The ω's come from √((restoring torque per unit angle)/(moment of inertia)) in each case, where the moment of inertia for H is mb2/3 and for V is 2mb2/3. Thus the ratio that is 7.9 turns out to be √(M/G) and we can find σ, which is just under 1/2 as we'd expect for something like rubber.

Reading Worked Solutions Too Early

Many students think that, given limited time, they shouldn't squander too much of it attempting questions that they don't know how to do. They observe that, if they don't work out what to do fairly quickly, they'll have try a lot of things, drawing diagrams and writing equations and following through methods that probably won't work. They risk trying to construct mental models for things that don't yet make sense to them, only to find that they never could make sense. These students think that looking at a worked solution fairly early on will cut out a part of the process as inefficient as it is unpleasant, freeing them to learn useful, valid methods at a greater rate.

These students are wrong.

You can only learn to solve problems by trying to solve problems; you will only learn content securely by trying to apply it to something new; and you won't appreciate what is going on in a worked solution, let alone remember the method subsequently, unless you've explored the problem as best you can first.

Suppose that you had to train for a swimming race. You might well look at advanced swimmers in order to see how it's done. It would be no bad thing to seek feedback from a coach on peculiarities of your stroke. You might even read a book about the technique of swimming. But none of this will help very much unless you are prepared to spend many hours at the tedious and exhausting business of swimming as fast as you can until you can do it that bit faster. Each hour in the pool makes a barely perceptible change to your swimming speed. But if you don't do it you're hardly training for the race at all. And so it is with mathematics. If you don't have the time or patience to try the questions that you don't initially know how to do, before looking at someone else's solution, then you don't have the time or patience needed to get better at mathematics.

Footnote on Physics A Paper 1 Question A5

I shouldn't think that the examiners dispute that a low-pass filter can be built with an op-amp. They just seem to feel that the inclusion of an unnecessary IC makes the circuit over-complicated, therefore worse, and that students wouldn't have done that if they'd understood about passive filters. But depending on how we interpret the requirement of a fixed input resistance, an active filter may be necessary, and there's the vexed question of what exactly they mean by "ideal": if they mean that it has zero output impedance, like the ideal op-amp, which students might reasonably guess, then again we need an op-amp. It shouldn't take guesswork to find out what they mean by "ideal low-pass filter", because the term does have a standard meaning in the field of filter design: that the thing has a perfect step-function cut-off in the frequency domain. But they don't mean that, because it can't be done. So who knows?.