National Science Olympiad Problems

Physics (NSEP / F=ma Level)

1 Atwood Machine Variant Mechanics Medium

A modified Atwood machine consists of a pulley of mass $M$ and radius $R$ (solid disk, moment of inertia $I = \frac{MR^2}{2}$) with masses $m_1$ and $m_2$ ($m_1 > m_2$) hanging from a massless, inextensible string that does not slip on the pulley.

Find the acceleration of the system.
Find the tensions on each side of the pulley.
If $m_1 = 5\;\text{kg}$, $m_2 = 3\;\text{kg}$, and $M = 2\;\text{kg}$, calculate numerical values.

Hint

The no-slip condition gives $\alpha = a/R$. For $m_1$: $m_1 g - T_1 = m_1 a$. For $m_2$: $T_2 - m_2 g = m_2 a$. For pulley: $(T_1 - T_2)R = I\alpha = \frac{MR}{2}a$. Adding: $a = \frac{(m_1 - m_2)g}{m_1 + m_2 + M/2}$. With given values: $a = \frac{2g}{9} \approx 2.18\;\text{m/s}^2$. $T_1 = m_1(g - a) = 5(7.62) = 38.1\;\text{N}$. $T_2 = m_2(g + a) = 3(11.98) = 35.9\;\text{N}$.

2 Satellite Orbit Transfer Gravitation Hard

A satellite in a circular orbit of radius $r_1 = 7000\;\text{km}$ around Earth needs to transfer to a circular orbit of radius $r_2 = 42000\;\text{km}$ (geostationary orbit) using a Hohmann transfer.

Calculate the velocity at each circular orbit.
Calculate the two velocity changes ($\Delta v_1$ and $\Delta v_2$) needed for the Hohmann transfer.
Calculate the transfer time.
Why is the Hohmann transfer the most fuel-efficient two-impulse transfer?

(Use $M_{\text{Earth}} = 5.97 \times 10^{24}\;\text{kg}$, $G = 6.674 \times 10^{-11}\;\text{N}\!\cdot\!\text{m}^2/\text{kg}^2$)

Hint

Circular orbit velocity: $v = \sqrt{GM/r}$. $v_1 = \sqrt{GM/r_1} \approx 7.55\;\text{km/s}$. $v_2 = \sqrt{GM/r_2} \approx 3.07\;\text{km/s}$. Transfer ellipse: semi-major axis $a_t = \frac{r_1 + r_2}{2} = 24500\;\text{km}$. At periapsis: $v_p = \sqrt{GM\!\left(\frac{2}{r_1} - \frac{1}{a_t}\right)} \approx 10.15\;\text{km/s}$. $\Delta v_1 = v_p - v_1 \approx 2.60\;\text{km/s}$. At apoapsis: $v_a = \sqrt{GM\!\left(\frac{2}{r_2} - \frac{1}{a_t}\right)} \approx 1.69\;\text{km/s}$. $\Delta v_2 = v_2 - v_a \approx 1.38\;\text{km/s}$. Transfer time $= T/2 = \pi\sqrt{a_t^3/(GM)} \approx 5.3\;\text{hours}$. Hohmann is optimal because any other transfer ellipse has a larger semi-major axis, requiring more energy by the vis-viva equation.

3 Coupled Oscillators Waves Hard

Two identical pendulums (length $L$, mass $m$) are connected by a weak spring (constant $k$) at their midpoints ($L/2$ from the pivot).

Find the two normal mode frequencies.
If pendulum A is displaced and released from rest while B is at equilibrium, describe the subsequent motion (beat phenomenon).
Calculate the beat frequency and the time for complete energy transfer from A to B.

Hint

Normal modes: (1) In-phase: both swing together, spring unstretched. $\omega_1 = \sqrt{g/L}$. (2) Out-of-phase: they swing oppositely. The spring force adds a restoring torque: $\omega_2 = \sqrt{g/L + k/(2m)}$. General motion of A: $\theta_A(t) = A\cos(\omega_{\text{avg}}t)\cos(\Delta\omega\,t/2)$, which is beats. Beat frequency $= \frac{\Delta\omega}{2\pi} = \frac{\omega_2 - \omega_1}{2\pi}$. Complete energy transfer time $= \frac{\pi}{\Delta\omega}$.

Chemistry (NSEC / USNCO Level)

4 Buffer System Design Acid-Base Medium

You need to prepare 1.00 L of a buffer solution with pH = 5.00 and buffer capacity of 0.10 M using acetic acid ($\text{p}K_a = 4.74$) and sodium acetate.

Calculate the required concentrations of acetic acid and sodium acetate.
How many moles of NaOH can this buffer neutralize before the pH rises above 5.50?
If you only had $\text{Na}_2\text{HPO}_4$ ($\text{p}K_{a2} = 7.20$) and $\text{NaH}_2\text{PO}_4$ ($\text{p}K_{a2} = 7.20$), explain why this would be a poor choice for pH 5.00 buffer.

Hint

Henderson-Hasselbalch: $\text{pH} = \text{p}K_a + \log\!\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$. $5.00 = 4.74 + \log(r)$, so $r = [\text{A}^-]/[\text{HA}] = 10^{0.26} = 1.82$. Buffer capacity $\beta \approx C_{\text{total}} = [\text{HA}] + [\text{A}^-] = 0.10$. So $[\text{HA}] = 0.0355\;\text{M}$, $[\text{A}^-] = 0.0645\;\text{M}$. For pH rise to 5.50: new ratio $= 10^{0.76} = 5.75$. Let $x$ mol NaOH added: $\frac{0.0645 + x}{0.0355 - x} = 5.75$. $x = 0.0167\;\text{mol}$. Phosphate buffer at pH 5.00 is poor because pH is far from $\text{p}K_{a2} = 7.20$, giving ratio $= 10^{-2.2} \approx 0.006$, which means virtually no $\text{HPO}_4^{2-}$ — negligible buffering capacity.

5 Thermochemistry Calculation Thermodynamics Medium

Using the following standard enthalpies of formation:

$\text{CO}_2(g)$: $-393.5\;\text{kJ/mol}$
$\text{H}_2\text{O}(l)$: $-285.8\;\text{kJ/mol}$
$\text{C}_2\text{H}_5\text{OH}(l)$: $-277.7\;\text{kJ/mol}$
$\text{CH}_3\text{CHO}(l)$: $-192.3\;\text{kJ/mol}$

Calculate the standard enthalpy of combustion of ethanol.
Calculate $\Delta H$ for the oxidation of ethanol to acetaldehyde: $\text{C}_2\text{H}_5\text{OH}(l) + \tfrac{1}{2}\text{O}_2(g) \to \text{CH}_3\text{CHO}(l) + \text{H}_2\text{O}(l)$
Explain why bond energy calculations might give a slightly different answer.

Hint

Combustion: $\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \to 2\text{CO}_2 + 3\text{H}_2\text{O}$. $\Delta H = [2(-393.5) + 3(-285.8)] - [-277.7 + 0] = -787.0 - 857.4 + 277.7 = -1366.7\;\text{kJ/mol}$. Oxidation to acetaldehyde: $\Delta H = [-192.3 + (-285.8)] - [-277.7 + 0] = -478.1 + 277.7 = -200.4\;\text{kJ/mol}$. Bond energies differ because: (1) they are averages over many compounds, (2) they refer to gas-phase species, ignoring intermolecular interactions and phase change enthalpies.

6 Redox Balancing Challenge Electrochemistry Hard

Balance the following redox reaction in acidic solution: $$\text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{H}_5\text{OH} \to \text{Cr}^{3+} + \text{CO}_2$$

Balance using the half-reaction method.
If 25.00 mL of 0.0200 M $\text{K}_2\text{Cr}_2\text{O}_7$ is required to oxidize the ethanol in a blood sample, calculate the blood alcohol mass (in mg).
This is the basis of the breathalyzer test. Explain why the color change from orange to green is observed.

Hint

Reduction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$. Oxidation: $\text{C}_2\text{H}_5\text{OH} + 3\text{H}_2\text{O} \to 2\text{CO}_2 + 12\text{H}^+ + 12e^-$. Multiply reduction by 2: $2\text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{H}_5\text{OH} + 16\text{H}^+ \to 4\text{Cr}^{3+} + 2\text{CO}_2 + 11\text{H}_2\text{O}$. Moles dichromate $= 0.025 \times 0.020 = 5.0 \times 10^{-4}\;\text{mol}$. Moles ethanol $= 5.0 \times 10^{-4}/2 = 2.5 \times 10^{-4}\;\text{mol}$. Mass $= 2.5 \times 10^{-4} \times 46.07 = 11.5\;\text{mg}$. Orange $\text{Cr}_2\text{O}_7^{2-}$ ($\text{Cr}^{6+}$, $d^0$) is reduced to green $\text{Cr}^{3+}$ ($d^3$) which has characteristic d-d transitions.

Biology (NSEB Level)

7 Genetics — Dihybrid Cross with Epistasis Genetics Hard

In Labrador retrievers, coat color is determined by two genes: the E gene (pigment deposition) and the B gene (pigment color). E is epistatic to B.

$B\_E\_$ = Black, $bbE\_$ = Brown (chocolate), $\_\_ee$ = Yellow (regardless of B)

Cross a black lab ($BbEe$) with a yellow lab ($Bbee$). What are the expected phenotypic ratios?
Two black labs produce a yellow puppy. What are the possible genotypes of the parents?
If you cross two dogs that are both $BbEe$, what phenotypic ratio do you expect? How does this differ from a standard 9:3:3:1 dihybrid ratio?

Hint

(1) $BbEe \times Bbee$: B locus gives $1BB:2Bb:1bb$. E locus gives $1Ee:1ee$. Combined: $B\_Ee$ (3/8 black), $B\_ee$ (3/8 yellow), $bbEe$ (1/8 brown), $bbee$ (1/8 yellow). Phenotypes: 3 black : 1 brown : 4 yellow = 3:1:4. (2) Yellow puppy ($ee$) means both parents carry $e$: both are $B\_Ee$. Since both are black ($B\_E\_$), they are either $BBEe$ or $BbEe$. (3) $BbEe \times BbEe$: 9 $B\_E\_$ (black), 3 $bbE\_$ (brown), 3 $B\_ee$ (yellow), 1 $bbee$ (yellow). So 9 black : 3 brown : 4 yellow. This is a 9:3:4 ratio (recessive epistasis), modified from the standard 9:3:3:1.

8 Cellular Respiration Efficiency Biochemistry Medium

Complete oxidation of one molecule of glucose through aerobic respiration:

Calculate the total ATP yield from glycolysis, pyruvate oxidation, the citric acid cycle, and oxidative phosphorylation. Use the modern estimates (2.5 ATP per NADH, 1.5 ATP per FADH$_2$).
The free energy of glucose combustion is $-2870\;\text{kJ/mol}$. If each ATP hydrolysis yields $30.5\;\text{kJ/mol}$, calculate the thermodynamic efficiency of aerobic respiration.
Why is the actual efficiency lower than the theoretical maximum? List at least three reasons.

Hint

(1) Glycolysis: 2 ATP + 2 NADH. Pyruvate oxidation: 2 NADH. Citric acid cycle: 2 GTP + 6 NADH + 2 FADH$_2$. Totals: 4 ATP/GTP, 10 NADH (= 25 ATP), 2 FADH$_2$ (= 3 ATP). Grand total: ~30-32 ATP (depends on shuttle system used for cytoplasmic NADH; malate-aspartate gives 2.5 ATP/NADH, glycerol-3-phosphate gives 1.5). Using malate-aspartate: 32 ATP. (2) Efficiency $= \frac{32 \times 30.5}{2870} = \frac{976}{2870} \approx 34\%$. (3) Reasons: proton leak across inner mitochondrial membrane, cost of transporting ATP/ADP/P$_i$, heat dissipation, not all NADH enters ETC at same point, maintenance of proton gradient for other processes.

9 Population Genetics — Hardy-Weinberg Evolution Medium

In a population, 16% of individuals show the recessive phenotype for a particular trait controlled by a single autosomal gene with two alleles.

Assuming Hardy-Weinberg equilibrium, calculate the allele frequencies and the frequency of heterozygous carriers.
If the population size is 10,000, how many individuals of each genotype do you expect?
Describe five conditions required for Hardy-Weinberg equilibrium and how violations of each would change allele frequencies.

Hint

(1) $q^2 = 0.16$, so $q = 0.4$, $p = 0.6$. Heterozygote frequency $= 2pq = 2(0.6)(0.4) = 0.48 = 48\%$. (2) $AA = p^2 \times 10000 = 3600$. $Aa = 4800$. $aa = 1600$. (3) Five conditions: (i) No mutation — mutations introduce new alleles. (ii) Random mating — non-random mating changes genotype (not allele) frequencies. (iii) No selection — differential survival/reproduction changes allele frequencies directionally. (iv) Large population — small populations experience genetic drift (random changes). (v) No gene flow — migration introduces/removes alleles.

Astronomy (IOAA Level)

10 Stellar Luminosity and HR Diagram Astrophysics Hard

A star has apparent magnitude $m = +3.5$ and is at a distance of 50 parsecs.

Calculate the absolute magnitude $M$.
If the Sun has $M = +4.83$, calculate the luminosity of this star in solar luminosities.
The star has a surface temperature of 8500 K (compared to the Sun's 5778 K). Calculate its radius in solar radii.
Classify this star on the HR diagram and predict its spectral type.

Hint

(1) $m - M = 5\log(d/10) = 5\log(50/10) = 5\log 5 = 5(0.699) = 3.495$. So $M = 3.5 - 3.5 = 0.0$. (2) $M_{\text{star}} - M_{\odot} = -2.5\log(L/L_{\odot})$. $0.0 - 4.83 = -2.5\log(L/L_{\odot})$. $L/L_{\odot} = 10^{4.83/2.5} = 10^{1.932} \approx 85.5\,L_{\odot}$. (3) $L = 4\pi R^2 \sigma T^4$. $\frac{R}{R_{\odot}} = \sqrt{L/L_{\odot}} \times \left(\frac{T_{\odot}}{T}\right)^2 = \sqrt{85.5} \times \left(\frac{5778}{8500}\right)^2 \approx 9.25 \times 0.462 \approx 4.28\,R_{\odot}$. (4) This is an A-type subgiant/giant star, spectral type approximately A2-A3, positioned above the main sequence on the HR diagram.

11 Kepler's Laws and Exoplanet Detection Celestial Mechanics Medium

A star of mass $1.2\,M_{\odot}$ shows periodic radial velocity variations with amplitude $K = 50\;\text{m/s}$ and period $P = 3.5\;\text{years}$.

Assuming a circular orbit and edge-on inclination ($i = 90°$), estimate the minimum mass of the orbiting planet.
Calculate the orbital radius of the planet.
If the planet transits the star (radius $= 1.3\,R_{\odot}$) and causes a 1.2% dip in brightness, estimate the planet's radius.

Hint

(1) From Kepler's third law: $a^3 = (M_{\text{star}}/M_{\odot}) \times P^2(\text{yr})$ in AU. $a^3 = 1.2 \times 3.5^2 = 14.7$, $a = 2.45\;\text{AU}$. Planet mass: $m_p \sin i = M_{\text{star}} \times K \times \left(\frac{P}{2\pi G M_{\text{star}}}\right)^{1/3}$. Simplified: $m_p \approx K \times \left(\frac{M_{\text{star}} P}{2\pi G}\right)^{1/3} \approx 2.3\,M_{\text{Jupiter}}$. (2) $a = 2.45\;\text{AU} = 3.67 \times 10^{11}\;\text{m}$. (3) Transit depth $= (R_p/R_{\text{star}})^2 = 0.012$. $R_p = R_{\text{star}}\sqrt{0.012} = 1.3 \times 6.96 \times 10^8 \times 0.1095 = 9.9 \times 10^7\;\text{m} \approx 1.42\,R_{\text{Jupiter}}$.

12 Cosmological Redshift Cosmology Hard

A distant galaxy has a measured redshift of $z = 0.85$. The hydrogen-alpha emission line (rest wavelength 656.3 nm) is observed.

Calculate the observed wavelength of H-alpha from this galaxy.
Using Hubble's law ($H_0 = 70\;\text{km/s/Mpc}$) and the non-relativistic approximation, estimate the distance.
Now use the relativistic Doppler formula to find the recession velocity. By what percentage does the non-relativistic approximation overestimate the velocity?

Hint

(1) $\lambda_{\text{obs}} = \lambda_{\text{rest}}(1 + z) = 656.3 \times 1.85 = 1214.2\;\text{nm}$ (infrared). (2) Non-relativistic: $v = cz = 0.85c = 2.55 \times 10^5\;\text{km/s}$. $d = v/H_0 = 2.55 \times 10^5/70 = 3643\;\text{Mpc}$. (3) Relativistic: $1 + z = \sqrt{\frac{1+\beta}{1-\beta}}$. $(1.85)^2 = \frac{1+\beta}{1-\beta}$. $3.4225 - 3.4225\beta = 1 + \beta$. $\beta = \frac{2.4225}{4.4225} = 0.5478$. $v = 0.548c$. Non-relativistic gives $0.85c$. Overestimate: $\frac{0.85 - 0.548}{0.548} \approx 55\%$. At $z = 0.85$, the non-relativistic approximation is quite poor.

Cross-Disciplinary Problems

13 Radioactive Decay and Carbon Dating Nuclear / Archaeology Medium

A wooden artifact from an archaeological site has a $^{14}\text{C}$ activity of 8.2 disintegrations per minute per gram of carbon. Living wood has an activity of 15.3 dpm/g. The half-life of $^{14}\text{C}$ is 5730 years.

Calculate the age of the artifact.
Why is carbon dating unreliable for samples older than ~50,000 years?
A sample of bone contains both original carbon and contaminating modern carbon (from groundwater). If 5% of the carbon is modern contamination, by how much would the apparent age underestimate the true age for a 20,000-year-old sample?

Hint

(1) $A = A_0 e^{-\lambda t}$. $\lambda = \frac{\ln 2}{5730} = 1.21 \times 10^{-4}\;\text{yr}^{-1}$. $8.2 = 15.3\,e^{-\lambda t}$. $t = -\frac{\ln(8.2/15.3)}{\lambda} = \frac{\ln(1.866)}{\lambda} = \frac{0.6245}{1.21 \times 10^{-4}} \approx 5160\;\text{years}$. (2) After ~50,000 years (~9 half-lives), activity drops to ~0.2% of original — comparable to background radiation, making measurement unreliable. (3) True activity at 20,000 yr: $A = 15.3 \times e^{-\lambda \times 20000} = 15.3 \times 0.0891 = 1.36\;\text{dpm/g}$. With 5% contamination: $A_{\text{measured}} = 0.95(1.36) + 0.05(15.3) = 1.29 + 0.77 = 2.06\;\text{dpm/g}$. Apparent age $= \frac{\ln(15.3/2.06)}{\lambda} = 16{,}400\;\text{years}$. Underestimates by ~3,600 years.

14 Biophysics — Osmotic Pressure Physics / Biology Hard

Red blood cells are placed in three different NaCl solutions at $37°\text{C}$. The intracellular fluid has an osmolarity of approximately 300 mOsm/L.

Calculate the osmotic pressure of normal saline (0.9% NaCl, MW = 58.44 g/mol) using the van't Hoff equation ($\pi = iMRT$, where $i$ is the van't Hoff factor). Is this isotonic with blood?
If cells are placed in 0.45% NaCl, calculate the osmotic pressure difference and predict what happens to the cells.
A patient needs an IV drip of 5% glucose (MW = 180 g/mol, $i = 1$). Is this approximately isotonic? Calculate to verify.

Hint

(1) 0.9% NaCl = 9 g/L. Molarity $= 9/58.44 = 0.154\;\text{M}$. NaCl dissociates: $i = 2$ (approx). Osmolarity $= 2 \times 0.154 = 0.308\;\text{Osm/L} \approx 308\;\text{mOsm/L}$. $\pi = iMRT = 2 \times 0.154 \times 8.314 \times 310 = 793\;\text{kPa} \approx 7.8\;\text{atm}$. This is approximately isotonic (close to 300 mOsm/L). (2) 0.45% NaCl: 154 mOsm/L. This is hypotonic. Water enters cells by osmosis, cells swell and may lyse (hemolysis). Pressure difference: $\Delta\pi = RT\Delta c \approx 3.9\;\text{atm}$. (3) 5% glucose: $50/180 = 0.278\;\text{M}$, $i = 1$. Osmolarity = 278 mOsm/L. Close to isotonic (within 7% of 300). Approximately isotonic — suitable for IV use.

15 Environmental Science — Greenhouse Effect Physics / Chemistry / Environment Olympiad Level

Earth receives solar radiation with intensity $S = 1361\;\text{W/m}^2$ (solar constant). Earth's average albedo is $\alpha = 0.30$.

Calculate the equilibrium surface temperature of Earth assuming no atmosphere (Stefan-Boltzmann law). Compare with the actual average temperature of 288 K.
Model a simple one-layer atmosphere that is transparent to visible light but absorbs and re-emits all infrared radiation. Calculate the surface temperature in this model.
$\text{CO}_2$ concentration has risen from 280 ppm (pre-industrial) to 420 ppm. Using the logarithmic forcing relation $\Delta F = 5.35\ln(C/C_0)\;\text{W/m}^2$, calculate the radiative forcing. If the climate sensitivity is $0.8\;\text{K}$ per $\text{W/m}^2$, estimate the expected warming.

Hint

(1) Energy balance: $\pi R^2 S(1-\alpha) = 4\pi R^2 \sigma T^4$. $T = \left[\frac{S(1-\alpha)}{4\sigma}\right]^{1/4} = \left[\frac{1361 \times 0.7}{4 \times 5.67 \times 10^{-8}}\right]^{1/4} = 255\;\text{K}$. Actual 288 K — the 33 K difference is the greenhouse effect. (2) One-layer model: atmosphere at $T_a$ radiates $\sigma T_a^4$ up and down. Surface: $\sigma T_s^4 = \frac{S(1-\alpha)}{4} + \sigma T_a^4$. Atmosphere: $2\sigma T_a^4 = \sigma T_s^4$. So $T_s = 2^{1/4} \times 255 = 303\;\text{K}$. (3) $\Delta F = 5.35\ln(420/280) = 5.35 \times 0.405 = 2.17\;\text{W/m}^2$. $\Delta T = 0.8 \times 2.17 = 1.73\;\text{K}$ warming.