IB Physics TI: Tools and inquiry worked solutions

\sigma=P/(AT^4). A watt is \mathrm{J\,s^{-1}}=\mathrm{kg\,m^2\,s^{-3}}; dividing by \mathrm{m^2} and by \mathrm{K^4} leaves \mathrm{kg\,s^{-3}\,K^{-4}}. B forgets to divide by the area, C uses the joule instead of the watt, D divides by the area twice.

1\,\mathrm{\mu m}=10^{-6}\,\mathrm{m} and 1\,\mathrm{nm}=10^{-9}\,\mathrm{m}, so 1\,\mathrm{\mu m}=10^3\,\mathrm{nm} and 0.532\,\mathrm{\mu m}=532\,\mathrm{nm}. The value also sits inside the visible band (400 to 700\,\mathrm{nm}, printed in the data booklet), which is a quick check. A and B shift the decimal point the wrong number of places, D multiplies by 10^4.

The seven base units are kg, m, s, A, mol, K and cd. The coulomb (\mathrm{A\,s}), the newton (\mathrm{kg\,m\,s^{-2}}) and the volt (\mathrm{kg\,m^2\,s^{-3}\,A^{-1}}) are derived, so B, C and D each contain one derived unit.

One year is about 3\times10^7\,\mathrm{s}, so 70 years is about 2\times10^9\,\mathrm{s}; at roughly one beat per second (a little faster in practice) the count is a few times 10^9, order of magnitude 10^9. A counts in minutes instead of seconds; C and D have no physical basis.

The mean is 9.88/4=2.47\,\mathrm{s}. The readings spread from 2.41 to 2.52\,\mathrm{s}, far more than the stopwatch's 0.01\,\mathrm{s} resolution, so the uncertainty comes from the spread: half the range is 0.055\,\mathrm{s}, rounded to one significant figure 0.06\,\mathrm{s}. A quotes only the resolution, C quotes the whole range, D keeps too many figures.

Percentage uncertainty =\dfrac{\Delta L}{L}\times100\%=\dfrac{0.5}{48.0}\times100\%=1.04\%\approx1\%. A quotes the absolute uncertainty as if it were a percentage, C doubles it, D divides 48 by 0.5 the wrong way and drops a factor of ten.

For a difference the absolute uncertainties add: \Delta\theta=(14.5\pm1.0)\,\mathrm{K}, so the percentage uncertainty is 1.0/14.5\times100\%=6.9\%. B uses only one reading's uncertainty, C adds the two percentage uncertainties of the readings (that rule is for products), A divides by the final temperature.

For a quotient the fractional uncertainties add: 0.1/6.0+0.005/0.250=0.0167+0.0200=0.0367, and 0.0367\times24\,\mathrm{\Omega}=0.88\,\mathrm{\Omega}\approx0.9\,\mathrm{\Omega}. A subtracts the fractions, B uses the current alone, D doubles the correct value.

Fractional uncertainties add for a quotient, and a power multiplies its fractional uncertainty by the size of the index: 2\%+2\times3\%=8\%. The minus sign of r^{-2} does not make the uncertainty smaller. A ignores the square, C cubes it, D multiplies the two percentages.

Averaging reduces random error: a reaction time that is sometimes long and sometimes short scatters the readings on both sides of the true value, and the scatter partly cancels in the mean. A, C and D shift every reading the same way (systematic errors), so the mean carries the same shift however many repeats are taken.

T\propto\sqrt m means T^2\propto m. Plotting T^2 (vertical) against m therefore puts the points on a line that passes through (0,0); its slope is the constant of proportionality. A gives a curve flattening out; C and D give curves because T grows with m, not with 1/m.

Check each line at the four x values. A gives 2.0,\,4.0,\,6.0,\,8.0: every one within 0.3 of the data, so the data are consistent with y\propto x. B misses at x=1 (2.5 against 2.1\pm0.3), C misses at x=3 (5.8 against 6.2\pm0.3), D misses at x=1 (1.7 against 2.1\pm0.3).

(a) East: 40\cos30^\circ=34.6\,\mathrm{N} (A1); north: 40\sin30^\circ=20.0\,\mathrm{N} (A1). (b) Add components: east 34.6\,\mathrm{N}, north 20.0+25=45.0\,\mathrm{N}; magnitude \sqrt{34.6^2+45.0^2}=56.8\approx57\,\mathrm{N} (A1); angle \tan^{-1}(45.0/34.6)=52^\circ north of east (A1). (c) Constant velocity needs zero resultant, so the third force is equal and opposite: 57\,\mathrm{N} at 52^\circ south of west (A1). (d) 56.8/10=5.7\,\mathrm{cm} (A1). Measure the angle from a stated reference line; an angle with no reference loses the direction mark.

(a) R=pV/(nT). pV has the unit \mathrm{Pa\,m^3=N\,m^{-2}\,m^3=N\,m=kg\,m^2\,s^{-2}} (M1), so R is in \mathrm{kg\,m^2\,s^{-2}\,K^{-1}\,mol^{-1}} (A1). (b) \alpha\theta is added to the number 1, so it has no unit; \alpha is in \mathrm{^\circ C^{-1}}, which equals \mathrm{K^{-1}} because a temperature difference of 1\,\mathrm{^\circ C} is 1\,\mathrm{K} (A1). (c) k_\mathrm{B} is in \mathrm{J\,K^{-1}=kg\,m^2\,s^{-2}\,K^{-1}} (M1); multiplying by K and dividing by kg leaves \mathrm{m^2\,s^{-2}}, whose square root is \mathrm{m\,s^{-1}} (A1). (d) The 3 is a pure number; the expression with 2 or 10 in its place would pass the same check (A1).

(a) Mean =72.08/4=18.02\,\mathrm{mm} (A1); half the range =(18.05-17.98)/2=0.035\approx0.04\,\mathrm{mm}, larger than the micrometer's 0.01\,\mathrm{mm}, so it is the uncertainty (A1). (b) V=\pi(18.02\times10^{-3})^2(40.0\times10^{-3})/4=1.020\times10^{-5}\,\mathrm{m^3} (M1); \rho=0.0864/1.020\times10^{-5}=8.47\times10^3\,\mathrm{kg\,m^{-3}} (A1). (c) \dfrac{\Delta\rho}{\rho}=\dfrac{0.1}{86.4}+\dfrac{0.1}{40.0}+2\times\dfrac{0.04}{18.02}=0.0012+0.0025+0.0044=0.0081 (M1); \Delta\rho=0.0081\times8.47\times10^3=69\approx0.07\times10^3, so \rho=(8.47\pm0.07)\times10^3\,\mathrm{kg\,m^{-3}} (A1). (d) The range is 8.40 to 8.54\times10^3\,\mathrm{kg\,m^{-3}}: it contains 8.53\times10^3 but not 8.73\times10^3 (A1).

(a) With W=0 the reading is 3.0\,\mathrm{mm} instead of 0: a zero error of +3.0\,\mathrm{mm} (A1); it shifts every reading by the same amount, so it is systematic (A1). (b) Averaging only reduces scatter about the mean; all five readings carry the same +3.0\,\mathrm{mm}, so the mean does too (A1). (c) Gradient =(43.0-3.0)\,\mathrm{mm}/4.0\,\mathrm{N}=10\,\mathrm{mm\,N^{-1}}, so k=4.0\,\mathrm{N}/0.040\,\mathrm{m}=100\,\mathrm{N\,m^{-1}} (A1). (d) No effect: the zero error moves the line up without changing its slope (A1). Subtract 3.0\,\mathrm{mm} from each reading before calculating an extension.

(a) The heater supplies energy at a constant rate, so mc\Delta\theta=Pt and \theta=\theta_0+\dfrac{P}{mc}\,t: linear in t (A1); gradient =P/(mc), the rate of temperature rise (A1). (b) (26.3-18.0)/300=0.0277\,\mathrm{K\,s^{-1}} (M1 A1; \mathrm{^\circ C\,s^{-1}} also accepted). (c) c=\dfrac{P}{m\times\text{gradient}}=\dfrac{48}{0.400\times0.0277}=4.3\times10^3\,\mathrm{J\,kg^{-1}\,K^{-1}} (A1). (d) The temperature of the water when the heater was switched on (A1). (e) Some of the 48\,\mathrm{W} warms the cup and leaks to the room, so the measured gradient is too small and c comes out too large: a systematic error (A1).

(a) Take logs of T=\dfrac{2\pi}{\sqrt{GM}}\,r^{3/2}: \log T=\tfrac32\log r+\log\dfrac{2\pi}{\sqrt{GM}} (M1), which has the form y=mx+c with m=1.5 and a constant intercept (A1). Changing the unit of T changes only the intercept. (b) \log(1.88\times10^9)=9.274; \log16.7=1.223 (A1). (c) \dfrac{1.223-0.248}{9.274-8.625}=\dfrac{0.975}{0.649}=1.50 (A1); this matches 1.5, so the data support T^2\propto r^3 (A1). (d) T=16.7\times86400=1.443\times10^6\,\mathrm{s} (M1); rearranging, GM=4\pi^2r^3/T^2, and with Callisto's values M=39.48\times(1.88\times10^9)^3\div\big(6.67\times10^{-11}\times(1.443\times10^6)^2\big)=1.9\times10^{27}\,\mathrm{kg} (A1).

(a) Independent: h (A1). Dependent: v, calculated as card length divided by gate time (A1). Controls, any two with reasons (A1 for both): the trolley's mass (friction and air resistance depend on it), the ramp's surface and angle (they set the friction and the path), the card and the position of the gate (they define the measured speed). (b) Five or more heights spread over as wide a range as the ramp allows, for example 0.10 to 0.50\,\mathrm{m}, each repeated three times (A1). (c) 0.0461\,\mathrm{s} is far from the other three, which agree within 0.0004\,\mathrm{s}; repeat that run and exclude the value only if the repeat agrees with the others, giving the reason in the report (A1). Mean of the rest =0.0410\,\mathrm{s}, so v=0.050/0.0410=1.2\,\mathrm{m\,s^{-1}} (A1). (d) v is calculated with a length that is too small, so every v is too small by the same fraction, 0.2/5.2\approx4\%: a systematic error that a graph of v^2 against h would show as a gradient that is too small (A1).

(a) 6/120=0.050\,\mathrm{s} (A1). (b) Average speeds: (19.6-11.0)\,\mathrm{cm}/0.050\,\mathrm{s}=1.72\,\mathrm{m\,s^{-1}} and (30.6-19.6)/0.050=2.20\,\mathrm{m\,s^{-1}}; these are the speeds at the middles of the intervals, 0.050\,\mathrm{s} apart, so a=(2.20-1.72)/0.050=9.6\,\mathrm{m\,s^{-2}} (M1 A1). (c) The ball is blurred over the exposure time, or it is closer to the camera than the rule, so its position on the rule depends on the viewing line (parallax) (A1). (d) Step 1: a=9.8\,\mathrm{m\,s^{-2}}, v=0+9.8\times0.050=0.49\,\mathrm{m\,s^{-1}} (M1). Step 2: a=9.8-2.0\times0.49=8.82\,\mathrm{m\,s^{-2}}, v=0.49+8.82\times0.050=0.93\,\mathrm{m\,s^{-1}} (A1). The drag term makes each step add less speed, which is how the model reaches a terminal speed (g/b=4.9\,\mathrm{m\,s^{-1}}).

(a) f=\dfrac v2\cdot\dfrac1L, so plot 1/L (A1); the theory predicts a straight line through the origin of gradient v/2. (b) 1/0.300=3.33\,\mathrm{m^{-1}} (A1); the fractional uncertainty is the same as that of L: 0.005/0.300=1.7\%, so \Delta(1/L)=0.017\times3.33=0.06\,\mathrm{m^{-1}} (A1). (c) v=2\times48.3=97\,\mathrm{m\,s^{-1}} (A1). (d) \Delta(\text{gradient})=(51.5-45.5)/2=3.0\,\mathrm{m\,s^{-1}} (M1); \Delta v=2\times3.0=6\,\mathrm{m\,s^{-1}} (A1); v=(97\pm6)\,\mathrm{m\,s^{-1}}, value and uncertainty to the same place (A1). (e) This experiment: 91 to 103\,\mathrm{m\,s^{-1}}; pulse method: 89 to 93\,\mathrm{m\,s^{-1}} (M1). The ranges overlap, so the results agree within their uncertainties, although the pulse method is far more precise (A1). (f) The intercept is much smaller than the uncertainty of one reading, so it is consistent with zero, as the theory predicts: no evidence of a systematic error in L or f (A1).

(a) \sin\theta=2.50/100.0=0.0250 (A1); percentage uncertainty =0.05/2.50+0.2/100.0=2.0\%+0.2\%=2.2\% (A1). (b) v=w/t=0.0250/0.0505=0.495\,\mathrm{m\,s^{-1}} (M1); a=v^2/(2s)=0.495^2/1.000=0.245\,\mathrm{m\,s^{-2}} (A1). (c) g=0.245/0.0250=9.80\,\mathrm{m\,s^{-2}} (A1). Percentage uncertainty: 2\times0.8\% (w) +0.4\% (s) +2\times1.0\% (t) +2.2\% (\sin\theta) =6.2\% (M1); \Delta g=0.062\times9.80=0.6\,\mathrm{m\,s^{-2}}, so g=(9.8\pm0.6)\,\mathrm{m\,s^{-2}} (A1). (d) 9.8 lies inside 9.2 to 10.4\,\mathrm{m\,s^{-2}}, so the result is consistent (A1). (e) The block height (2.0\%, tied with the doubled timing term): a taller block on a longer track halves its share, or a caliper reads h to \pm0.01\,\mathrm{cm}; for the time, a wider card gives a longer, relatively more precise gate time (A1).

(a) Ratios over successive 2\,\mathrm{min} intervals: 47.2/60.0=0.787 and 37.1/47.2=0.786 (M1); equal ratios in equal times are the signature of exponential decay (A1). (b) The value at 6\,\mathrm{min}: 31.6\,\mathrm{K}, against about 0.786\times37.1=29.2\,\mathrm{K} from the pattern (the ratios either side of it are 0.852 and 0.728). Check or repeat it and give the reason if it is left out of the fit (A1). (c) Gradient =\dfrac{\ln14.2-\ln60.0}{12-0}=\dfrac{2.653-4.094}{12}=-0.120\,\mathrm{min^{-1}}, so k=0.120\,\mathrm{min^{-1}} (M1 A1); halving time =\ln2/k=0.693/0.120=5.8\,\mathrm{min} (A1). (d) Each \Delta\theta is 0.5\,\mathrm{K} too large (M1). That is a small fraction at 60\,\mathrm{K} but 3.5\% at 14\,\mathrm{K}, so the late points lie above the true line: the graph curves upwards at large t and its gradient is less steep, giving k too small (A1). With the same thermometer for water and room, the offset would cancel.

(a) r/h: 0.29/0.40=0.725, 0.43/0.60=0.717, 0.58/0.80=0.725, 0.71/1.00=0.710, 0.86/1.20=0.717 (M1); the ratio is constant to within the reading uncertainty (about \pm0.01), so r\propto h (A1). A graph of r against h through the origin is the alternative test. (b) The energy left after the bounce is mgr out of mgh: fraction lost =1-0.72=0.28 (A1). (c) \pm1\,\mathrm{cm} is a fixed absolute uncertainty (M1), so the percentage is largest when r is smallest: 1/29=3.4\% at h=0.40\,\mathrm{m}, against 1/86=1.2\% at 1.20\,\mathrm{m} (A1). (d) Any one linked to reading r: more frames per second so the highest frame is closer to the true top; the camera at the height of the top of the bounce to remove parallax; repeat more drops at the small heights (A1). (e) The ball moves faster for larger h, so air resistance takes a larger fraction of its energy on the way down and up (A1).

(a) Convert the card width first: w=0.0400\,\mathrm{m}. Then v=w/t=0.500\,\mathrm{m\,s^{-1}}. For a quotient, \Delta v/v=0.0005/0.0400+0.002/0.080=0.0375, or 3.75\%.

(b) E_\mathrm{k}=\tfrac12(0.600)(0.500)^2=0.0750\,\mathrm{J} and \Delta E_\mathrm{p}=(0.600)(9.8)(0.020)=0.1176\,\mathrm{J}. Their ratio is 0.6378, or 64\%.

(c) Cancel the common mass before propagating: f=v^2/(2gh). Its fractional uncertainty is 2(3.75\%)+5.0\%=12.5\%. Thus \Delta f=0.0797 and f=(64\pm8)\%. The interval 56–72% excludes 100%, so the data resolve energy loss; negligible loss is not supported.

(a) VIt=mc\Delta T, so the temperature gradient s=\Delta T/t=VI/(mc) and c=VI/(ms).

(b) c=6.00(2.00)/(0.200\times0.0130)=4615\,\mathrm{J\,kg^{-1}\,K^{-1}}. The fractional uncertainty is 0.05/6.00+0.02/2.00+0.002/0.200+0.0005/0.0130=0.0668. Therefore \Delta c=308\,\mathrm{J\,kg^{-1}\,K^{-1}} and c=(4.6\pm0.3)\times10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}.

(c) The rounded range 4300–4900 does not include 4180. Energy reaching the surroundings or warming the container reduces the water’s temperature gradient, so VI/(ms) overestimates c. Insulate the vessel and add a lid to reduce heat transfer; account separately for container heating.

(a) In SI units, each product Ir^2 is 9.0\times10^{24}\,\mathrm{W}: 900(10^{11})^2, 400(1.5\times10^{11})^2 and 225(2.0\times10^{11})^2. A constant product supports I\propto r^{-2}.

(b) L=4\pi r^2I=4\pi(9.0\times10^{24})=1.13\times10^{26}\,\mathrm{W}. The fractional uncertainty is 2\Delta r/r+\Delta I/I=2(2\%)+5\%=9\%, corresponding to about 1.0\times10^{25}\,\mathrm{W}.

(c) For planets orbiting the same mass, T^2\propto r^3. Thus T_\mathrm{out}/T_\mathrm{in}=(2.0/1.0)^{3/2}=2.83. A log irradiance graph tests the radiation relation with slope -2; it measures no orbital periods. Testing the orbital law requires period data as well as radii.

(a) n=pV/(RT)=120/(8.31\times300)=0.0481\,\mathrm{mol}. Add fractional uncertainties: 0.02/1.20+0.01/1.00+1/300=0.0300, or 3.0\%.

(b) The temperature rises by 50\,\mathrm{K}. Hence \Delta U=\tfrac32nR\Delta T=\tfrac32(0.0481)(8.31)(50)=30\,\mathrm{J} to two significant figures. Equivalently, the initial nRT=pV=120\,\mathrm{J} and the fractional temperature increase is 50/300, giving the same result.

(c) The heater delivers E=VIt=(6.0)(1.0)(10)=60\,\mathrm{J}. The fraction in the gas is \Delta U/E=30/60=0.50.

(d) Some supplied energy warms the vessel or passes to the room. Measure the vessel’s temperature change and account for its energy gain, or add insulation and repeat the run to test the heat-loss explanation.