One question per chapter, 15 in all, taken from the course's corrected exercises. Answer the ones you can do and skip those from a chapter you have not seen yet: each answer is checked right away, and the solution opens after.
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These chapters come from an earlier course, and this one relies on them from the first weeks.
201-SN3-RE · Differential equations
None of these equations comes in its textbook form. Before any computation, rewrite it until you recognize its TYPE: separable dxdy=f(x)g(y), linear y′+P(x)y=Q(x), homogeneous y′=F(xy), or linear if x is viewed as a function of y. Then solve through the given point: the point (x0,y0) means y(x0)=y0.
a) dxdy=e2x+y+ey, through the point (0,0). Compute y(−21) and the right endpoint of the interval on which the solution exists.
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Answer
a)Separable: y=−ln(23−x−2e2x), y(−21)≈−0.597, right endpoint ≈0.396
a) The trap is in the exponent: e2x+y=e2xey, not e2x+ey. Factor out ey: dxdy=ey(e2x+1), a SEPARABLE equation. e−ydy=(e2x+1)dx, so −e−y=2e2x+x+C. At the point (0,0): −1=21+C, C=−23. Hence e−y=23−x−2e2x and y=−ln(23−x−2e2x). At x=−21: 23+21−2e−1=2−2e1≈1.816, so y≈−0.597. The solution exists only as long as the argument of the logarithm is positive: 3−2x−e2x>0. This function is strictly decreasing, equals 1 at 0 and vanishes between 0.39 and 0.40; by bisection, the right endpoint is about 0.396. As you approach it, e−y→0 and y→+∞: the solution blows up in finite time, although the right-hand side looked harmless.
203-SN1-RE · Measurement, uncertainty and graph analysis in the lab
In the mechanics lab, you measure the length of an object with a ruler graduated in millimetres, then the period of a pendulum with a hand-held stopwatch. Lab convention: the reading uncertainty of an instrument with divisions is half of the smallest division; that of a stopwatch started by hand is 0.2 s, the reaction time.
a) What is the absolute reading uncertainty of the ruler, in cm?
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Answer
a)ΔL=0.05 cm
a) The smallest division is 1 mm, so ΔL=0.5 mm =0.05 cm. Each end of the object is read to within half a division.
A vertical spring of constant k=250 N/m is fixed to the ceiling, its other end hanging free. A 2.00 kg block is hung from it. Take g=9.80 m/s2 and neglect the mass of the spring.
b) Now repeat, hooking the block on while the spring is at its natural length, then releasing it from rest. Find the MAXIMUM extension reached. Compare with the result of a) and state the general ratio.
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Answer
b)xmax=2mg/k=15.7 cm, exactly double
b) Here we must conserve energy, not balance forces. The block starts from rest and stops at the lowest point, so ΔK=0: all the gravitational potential energy lost has been stored in the spring. mgxmax=21kxmax2, whence xmax=k2mg=0.157 m, or 15.7 cm. That is EXACTLY double the equilibrium extension, and the result is general: xmax=2xeq, whatever the mass and the stiffness. A load released suddenly stresses a structure twice as much as a load lowered gently, which is a basic principle of engineering design.
Chapter 1 · Geometric optics and optical instruments
One single equation governs lenses and mirrors. What changes from one case to the next is not the formulas but the SIGNS, and that is where the whole chapter is won or lost.
We have a converging lens of focal length 20.0 cm.
a) An object is placed 30.0 cm from the lens. Calculate the image position and the magnification, then describe the image completely.
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Answer
a)di=+60.0 cm, m=−2.00: real, inverted, enlarged
a) di1=f1−do1=20.01−30.01=603−2=601, so di=+60.0 cm. The magnification is m=−30.060.0=−2.00. The image is therefore REAL since di>0, INVERTED since m<0, and ENLARGED twice since ∣m∣=2. It forms 60 cm on the other side of the lens and could be caught on a screen: this is the projector configuration.
Chapter 2 · Lab measurements, uncertainties and graphical analysis
On a grating spectrometer, a line appears on both sides of the direct beam. Its position on the graduated circle is read on the LEFT and then on the RIGHT, and the deflection angle is HALF THE DIFFERENCE of the two readings.
Readings: θL=(139∘12′±2′) and θR=(221∘00′±2′). The diagram below shows the spectrometer seen from above.
c) Calculate the uncertainty on θ, in arcminutes and then in degrees.
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Answer
c)Δθ=2′=0.0333∘
c) The two readings are independent and each carries 2′: a difference adds the absolute uncertainties, Δ(2θ)=2′+2′=4′. Dividing by two, Δθ=24′=2′, that is 0.0333∘, or 5.82×10−4 rad. Result: θ=(40.900±0.033)∘, a relative uncertainty of only 0.08%.
Chapter 3 · Simple harmonic motion and mechanical waves
A 0.50 kg block is attached to a horizontal spring of constant k=200 N/m and slides without friction. It is pulled 8.0 cm from its equilibrium position and released from rest at t=0.
d) Calculate the speed of the block as it passes 4.0 cm from equilibrium, and check that it agrees with conservation of energy.
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Answer
d)v≈1.39 m/s; at half the amplitude, THREE QUARTERS of the energy is still kinetic
d) v=ωA2−x2=20(0.080)2−(0.040)2=200.0048≈1.39 m/s. Energy: E=21kA2=21(200)(0.0064)=0.64 J; at x=0.040 m, Ep=21(200)(0.0016)=0.16 J, so Ek=0.48 J and v=0.502(0.48)≈1.39 m/s: the two routes agree. At half the amplitude, three quarters of the energy is still kinetic, not half: energy goes as the SQUARE of the displacement.
Chapter 5 · The Doppler effect: moving source, moving observer and radial component
A train blows its whistle at f0=450 Hz and travels at 70.0 m/s on a straight track. An observer O stands d=60.0 m from the track. Speed of sound: v=340 m/s.
Only the projection of the velocity onto the line of sight counts, vr=vscosθ, where θ is the angle between the velocity of the train and the direction from the train to the observer. We therefore write f′=f0v−vscosθv, and we neglect the propagation time of the sound.
b) The train passes the point of the track closest to O. What frequency does O hear then, and why?
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Answer
b)f′=450 Hz exactly: the radial component vanishes
b) At the closest point, the velocity is perpendicular to the line of sight: θ=90°, cosθ=0, vr=0. At that instant the train is neither approaching nor receding from O; its distance goes through a minimum and its derivative is zero. So f′=f0=450 Hz exactly. No shift at the moment of passing: this is the most counterintuitive result of the chapter, and the one that exams like to ask about.
A 500 nm laser passes through a single slit of width 0.10 mm. The screen is 3.0 m away.
The diagram below is not to scale: the slit is greatly enlarged.
b) Calculate the width of the central maximum on the screen.
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Answer
b)First minimum at 15 mm, central maximum 30 mm wide
b) First minimum: sinθ=aλ=0.10×10−3500×10−9=5.0×10−3 (small angle). Position: y=Ltanθ≈(3.0)(5.0×10−3)=15 mm. The central maximum extends from one minimum to the other: 30 mm wide.
Data: h=6.63×10−34 J·s, c=3.0×108 m/s, 1 eV =1.6×10−19 J. The diagram below places these three kinds of radiation.
a) Calculate the energy of a photon of green light (500 nm), in joules and then in electron-volts.
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Answer
a)E=λhc≈4.0×10−19 J, that is ≈2.5 eV
a) E=λhc=500×10−9(6.63×10−34)(3.0×108)≈4.0×10−19 J, that is 1.6×10−193.98×10−19≈2.5 eV. (Shortcut: 5001240=2.48 eV.) Visible photons all lie between 1.8 and 3.1 eV: the same scale as chemical bonds, which is why visible light drives photosynthesis and vision.
Chapter 8 · Quantum physics: photons, line spectra and duality
A hydrogen-like ion has a nucleus of charge +Ze and a single electron: He⁺ (Z=2), Li²⁺ (Z=3). The Bohr model applies to it with En=−n213.6Z2 eV and rn=Zn2a0. Take hc=1240 eV·nm.
a) Calculate the ionization energy of He⁺ from its ground state.
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Answer
a)54.4 eV
a) Eion=13.6×22=54.4 eV. The electron is attracted by a double charge, at half the distance: it is bound four times as strongly.
The following atomic numbers are recalled: Th 90, Pa 91, U 92, Ra 88, Rn 86, C 6, N 7, Na 11, Ne 10, Co 27, Ni 28, K 19, Ar 18, Ca 20. The diagram below shows the decay of cobalt-60.
a) State the two conservation laws that are enough to balance any nuclear equation, then complete: 92238U→?+24He and 88226Ra→?+24He.
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Answer
a)A and Z are conserved: thorium-234 and radon-222
a) Conservation of the mass number A and conservation of the charge Z: the sum of the A values and the sum of the Z values are the same on both sides. These two lines are enough, there is nothing else to memorize. 92238U→90234Th+24He: 238=234+4 and 92=90+2. 88226Ra→86222Rn+24He: radium indeed gives radon, which incidentally explains why this gas appears in the basements of granitic regions.
A spaceship of proper length L0=120 m and proper width 18.0 m travels at v=0.600c toward a star located 6.00 light-years from Earth, a distance measured by Earth astronomers. The figure, not to scale, shows the ship at rest with its proper dimensions, and the same ship measured from Earth, whose dimensions are to be found.
a) Calculate the length and width of the ship as measured from Earth. What does the crew on board measure with their own tape measure?
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Answer
a)96.0 m by 18.0 m seen from Earth; 120 m by 18.0 m on board
a) γ=1−0.3601=0.8001=1.250. Length seen from Earth: L=γL0=1.250120=96.0 m. Width seen from Earth: 18.0 m, UNCHANGED, because contraction acts only along the direction of motion. On board, the crew still measures 120 m by 18.0 m: in their own frame the ship is at rest, so they read the proper length, and nothing has changed for them. Neither party is wrong.
Conduction through a flat wall obeys Fourier's law: Φ=ekAΔT, which is written Φ=RΔT with the thermal resistance R=kAe. A window of 1.20 m² separates a room at 20 °C from the outdoors at −10 °C. Glass: k=0.80 W/(m·K), thickness 4.0 mm. Still air: k=0.026 W/(m·K).
a) Calculate the thermal resistance of a single pane.
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Answer
a)R=4.17×10−3 K/W
a) R=kAe=0.80×1.204.0×10−3=4.17×10−3 K/W. The thickness must be in metres.
Chapter 12 · Greenhouse effect, albedo and radiative forcing
The luminosity of the Sun is L=3.828×1026 W and the mean Earth-Sun distance is d=1.496×1011 m. Radius of the Earth: R=6.371×106 m. The mean albedo of the Earth is α=0.30. The diagram below, not to scale, shows the situation.
a) Calculate the solar constant S, the flux received on a surface perpendicular to the rays at the distance of the Earth.
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Answer
a)S=1361 W/m²
a) The power of the Sun is spread over a sphere of radius d: S=4πd2L=4π×(1.496×1011)23.828×1026=1361 W/m².
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