Original GioPhysics extended practice
IB Physics · Extended Paper 1
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 180 minutes
- Questions
- 10
- Marks
- 132
- Answers
- Complete teacher key
Instructions
- Answer every question and show all reasoning.
- State assumptions and use appropriate significant figures and units.
- Support evaluative conclusions with the supplied evidence and the relevant physics.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Impulse and rebound from force data
A 0.400 kg trolley moves right at 1.50 m s⁻¹ before striking a buffer. The measured horizontal force on it has coordinates (0 ms, 0 N), (10 ms, −60 N), (25 ms, −60 N), and (45 ms, 0 N), joined by straight lines. Determine the impulse, final velocity and kinetic-energy change. Explain whether momentum of the trolley alone is conserved, identify where the missing momentum goes, and evaluate one limitation of integrating sampled force data.
Measured buffer force
The four measured force–time coordinates from the prompt are connected linearly; signed area represents impulse.
Read the exact data as a table
| Time / ms | Force / N |
|---|---|
| 0 | 0 |
| 10 | -60 |
| 25 | -60 |
| 45 | 0 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Projectile prediction with uncertainty
A ball is launched from ground level at speed (18.0 ± 0.3) m s⁻¹ and angle (35.0 ± 1.0)° above horizontal. Ignore drag and use g = 9.81 m s⁻². Calculate time of flight, maximum height and range from the central values. Estimate the percentage uncertainty in range using R = u²sin(2θ)/g and maximum-uncertainty addition, treating angular uncertainty in radians. Explain two reasons a measured range could be systematically shorter.
Open the interactive answer workspaceMuon survival and reference frames
Muons created 12.0 km above sea level travel downward at 0.998c. Their mean proper lifetime is 2.20 μs. Calculate the Lorentz factor, mean lifetime and mean travel distance in Earth's frame, and the atmosphere thickness in the muon frame. Estimate the survival fraction to sea level using exponential decay in either frame. Explain how the two descriptions agree without assigning an absolute time dilation.
Open the interactive answer workspaceConduction and radiation through a wall
A 12.0 m² wall consists of 0.100 m brick with thermal conductivity 0.72 W m⁻¹ K⁻¹ and 0.060 m insulation with conductivity 0.040 W m⁻¹ K⁻¹. Indoor and outdoor surface temperatures are 293 K and 273 K. Find the steady conductive power and interface temperature. If the outer surface has emissivity 0.85 and sees surroundings at 263 K, calculate its net radiative loss. Explain why simply adding that radiation to the conduction result may violate steady-state energy balance.
Open the interactive answer workspaceTest a simple-harmonic acceleration model
A motion sensor records acceleration against displacement for an oscillator: (−0.080 m, +5.12 m s⁻²), (−0.040 m, +2.56 m s⁻²), (0, 0), (0.040 m, −2.56 m s⁻²), and (0.080 m, −5.12 m s⁻²). Determine whether the data support SHM, find angular frequency and period, and calculate maximum speed and total energy for mass 0.250 kg. Evaluate how a constant acceleration-sensor zero offset would appear on the graph.
Acceleration against displacement
The five measured coordinates are listed in the prompt; their gradient, intercept and physical meaning are left for analysis.
Read the exact data as a table
| Displacement x / m | Acceleration a / m s⁻² |
|---|---|
| -0.08 | 5.12 |
| -0.04 | 2.56 |
| 0 | 0 |
| 0.04 | -2.56 |
| 0.08 | -5.12 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Resolve wavelength and coherence from fringes
Two narrow slits separated by (0.300 ± 0.005) mm are illuminated by monochromatic light. On a screen (2.00 ± 0.01) m away, the distance across 10 fringe spacings is (42.0 ± 0.5) mm. Determine fringe spacing and wavelength with percentage uncertainty. Explain why measuring many spacings helps, predict what happens when one slit is widened substantially, and distinguish loss of visibility from a change in fringe spacing.
Open the interactive answer workspaceElectric potential along a dipole axis
Charges +Q and −Q, with Q = 2.00 nC, lie at x = +0.0500 m and x = −0.0500 m respectively. At x = +0.200 m calculate electric potential and field. Find the work done by the electric field when a +1.00 nC test charge moves slowly from x = 0.200 m to x = 0.300 m. Derive the far-axis approximations for V and E in terms of dipole moment p = Qd and compare the exact potential at 0.200 m with the approximation.
Open the interactive answer workspaceInduction in a rotating coil
A 200-turn rectangular coil of area 3.00 × 10⁻³ m² rotates at 50.0 rad s⁻¹ in a uniform 0.400 T field. At t = 0 its area vector is parallel to the field. Derive flux linkage and emf as functions of time, calculate peak and rms emf, and find the first time emf reaches its positive maximum. If connected to a 60.0 Ω resistor, find average power and mechanical torque averaged over a cycle for constant speed.
Open the interactive answer workspaceInfer half-life and background from count data
A detector records total count rate from a radioactive sample: (0 min, 860 counts min⁻¹), (5 min, 510), (10 min, 300), (15 min, 195), and (20 min, 142.5). A separate long background measurement gives 90.0 counts min⁻¹. Determine the corrected half-life and decay constant, predict the total count rate at 30 min, and explain how Poisson counting uncertainty and an incorrect background estimate affect a logarithmic analysis.
Measured total count rate
Total count rate includes the separately stated 90 counts per minute background; correction is part of the analysis.
Read the exact data as a table
| Time / min | Total count rate / counts min⁻¹ |
|---|---|
| 0 | 860 |
| 5 | 510 |
| 10 | 300 |
| 15 | 195 |
| 20 | 142.5 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Photoelectric energy and measurement design
Light of wavelength 400 nm illuminates a metal with work function 2.20 eV. Calculate photon energy, maximum photoelectron kinetic energy, stopping potential and maximum electron speed. Predict what changes when wavelength becomes 300 nm at the same photon rate, and outline how stopping potential should be measured without confusing it with the voltage that reduces current merely to a small nonzero value. Use h = 6.626 × 10⁻³⁴ J s and electron mass 9.11 × 10⁻³¹ kg.
Open the interactive answer workspaceTeacher copyDetailed answer and marking guideShow 10 answers
Teacher copy
Detailed answer and marking guide
Accept equivalent physics expressed clearly. Award method credit where the reasoning is valid.
Impulse and rebound from force data
Answer: Impulse = −1.80 N s, final velocity = −3.00 m s⁻¹, and ΔK = +1.35 J. Trolley momentum is not conserved; the buffer and Earth receive the opposite impulse.
- Impulse is signed area under F–t. The first triangle is −½(0.010)(60) = −0.300 N s, the rectangle is −0.900 N s, and the final triangle is −0.600 N s.
- Total impulse is −1.80 N s. Initial momentum is (0.400)(1.50) = +0.600 kg m s⁻¹, so final momentum is −1.20 kg m s⁻¹.
- Final velocity is p_f/m = −1.20/0.400 = −3.00 m s⁻¹. Kinetic energy changes from 0.450 J to 1.80 J, so ΔK = +1.35 J.
- The trolley alone experiences an external impulse from the buffer and does not conserve momentum. In a larger trolley–buffer–Earth system, the buffer and Earth gain +1.80 N s, so total momentum remains conserved.
- Straight-line integration assumes no narrow force peak occurs between samples. A higher sampling rate and calibrated force zero reduce missing-area and offset errors; repeating impacts quantifies random variation.
Projectile prediction with uncertainty
Answer: Time = 2.10 s, maximum height = 5.43 m, range = 31.0 m. Estimated range uncertainty ≈ 4.6%, or about ±1.4 m. Drag and a launch point below the assumed level can shorten the range.
- Resolve velocity: u_y = 18.0sin35° = 10.32 m s⁻¹ and u_x = 18.0cos35° = 14.74 m s⁻¹.
- For equal launch and landing heights, T = 2u_y/g = 2.10 s. Maximum height is u_y²/(2g) = 5.43 m.
- Range is u_xT, equivalently u²sin70°/g = 31.0 m. These two routes provide an algebraic cross-check.
- For R ∝ u²sin2θ, fractional maximum uncertainty is approximately 2Δu/u + |2cot(2θ)|Δθ = 2(0.3/18.0) + 2cot70°(π/180) = 0.0460.
- Thus ΔR ≈ 1.4 m. Air drag reduces horizontal speed; a landing surface below or above the assumed equal level changes flight time. A launch-speed calibration bias or angle reference offset would also shift all trials systematically.
Muon survival and reference frames
Answer: γ = 15.82; Earth-frame lifetime = 34.8 μs and mean distance = 10.4 km; muon-frame atmosphere thickness = 0.759 km. Survival fraction ≈ exp(−12.0/10.4) = 0.316.
- Lorentz factor is γ = 1/√(1−0.998²) = 15.82. The muon lifetime is proper in its own rest frame.
- Earth observers measure dilated mean lifetime γτ₀ = (15.82)(2.20 μs) = 34.8 μs and mean travel distance vγτ₀ = 10.4 km.
- The survival probability over 12.0 km is exp[−t/(γτ₀)] with t = 12.0 km/(0.998c), giving exp(−12.0/10.4) = 0.316.
- In the muon frame, atmospheric thickness is length-contracted: L' = 12.0 km/15.82 = 0.759 km. Travel time L'/(0.998c) compared with τ₀ gives the same exponential factor.
- Earth describes a longer muon lifetime; the muon describes a shorter atmosphere. Lorentz transformations connect these frame-dependent intervals, so neither frame is an absolute privileged description and both predict the same arrival events.
Conduction and radiation through a wall
Answer: Thermal resistance = 0.1366 K W⁻¹; conductive power = 146 W; interface temperature ≈ 291.3 K if brick is inside. Net outer radiation ≈ 445 W. The specified outer-surface temperature cannot remain fixed without a matching energy source.
- Series thermal resistances are R_brick = L/(kA) = 0.100/[0.72(12.0)] = 0.0116 K W⁻¹ and R_ins = 0.060/[0.040(12.0)] = 0.125 K W⁻¹.
- Total resistance is 0.1366 K W⁻¹, so conductive power through the composite wall is ΔT/R = 20/0.1366 = 146 W, not 12.2 W.
- The brick temperature drop is P R_brick = (146)(0.0116) = 1.69 K. If brick faces indoors, the interface is therefore 293−1.69 = 291.3 K.
- Net radiation is εσA(T_s⁴−T_sur⁴) = 0.85(5.67 × 10⁻⁸)(12.0)(273⁴−263⁴) ≈ 445 W.
- The radiative loss exceeds the calculated conductive supply at the imposed temperatures, so the outer surface would cool unless convection or another source supplies the difference. A true steady-state solution must determine surface temperatures from simultaneous conduction, convection and radiation balances.
Test a simple-harmonic acceleration model
Answer: The linear negative gradient is −64.0 s⁻², so ω = 8.00 rad s⁻¹ and T = 0.785 s. With A = 0.080 m, v_max = 0.640 m s⁻¹ and E = 0.0512 J.
- SHM requires a = −ω²x: a straight acceleration–displacement graph through the origin with negative gradient. The supplied points satisfy this pattern.
- Gradient is (−5.12−5.12)/(0.080−(−0.080)) = −64.0 s⁻². Therefore ω = √64.0 = 8.00 rad s⁻¹.
- Period is T = 2π/ω = 0.785 s. Taking the largest measured displacement as amplitude gives v_max = ωA = (8.00)(0.080) = 0.640 m s⁻¹.
- The effective spring constant is k = mω² = (0.250)(64.0) = 16.0 N m⁻¹. Total energy is ½kA² = 0.0512 J.
- A constant acceleration zero offset adds the same value to every vertical reading. It preserves the gradient and inferred ω but shifts the line vertically so it no longer passes through the origin, creating an apparent nonzero acceleration at equilibrium.
Resolve wavelength and coherence from fringes
Answer: Fringe spacing = 4.20 mm; wavelength = 630 nm. Maximum fractional uncertainty ≈ 1.19% + 1.67% + 0.50% = 3.36%, so λ ≈ (630 ± 21) nm.
- Ten spacings occupy 42.0 mm, so fringe spacing s = 4.20 mm. Measuring across many fringes makes the same endpoint-reading uncertainty a smaller fraction of one spacing.
- For small angles, s = λD/d. Hence λ = sd/D = (4.20 × 10⁻³)(0.300 × 10⁻³)/2.00 = 6.30 × 10⁻⁷ m.
- Maximum fractional uncertainty is Δs/s + Δd/d + ΔD/D = 0.5/42.0 + 0.005/0.300 + 0.01/2.00 = 0.0336.
- Thus percentage uncertainty is 3.36% and absolute uncertainty is about 21 nm, giving λ = (630 ± 21) nm at matching precision.
- Widening one slit changes relative amplitudes and may add a diffraction envelope, reducing visibility because minima no longer cancel perfectly. The geometric fringe spacing λD/d remains essentially unchanged while coherent path separation and wavelength are unchanged.
Electric potential along a dipole axis
Answer: At 0.200 m, V = 47.9 V and E = 512 N C⁻¹ toward +x. At 0.300 m, V = 20.5 V, so field work = +2.74 × 10⁻⁸ J. Far away V ≈ kp/x² and E ≈ 2kp/x³; the potential approximation at 0.200 m is 44.9 V.
- At x = 0.200 m, distances are 0.150 m to +Q and 0.250 m to −Q. Potential is kQ(1/0.150−1/0.250) = 47.9 V.
- Both axial field contributions point toward +x there: E = kQ[1/(0.150)²−1/(0.250)²] = 512 N C⁻¹.
- At 0.300 m, V = kQ(1/0.250−1/0.350) = 20.5 V. Electric-field work is q(V_i−V_f) = 1.00 × 10⁻⁹(47.9−20.5) = 2.74 × 10⁻⁸ J.
- For x much larger than separation d, expand 1/(x∓d/2) to first order. Their difference is approximately d/x², so V ≈ kp/x² and E = −dV/dx ≈ 2kp/x³.
- Here p = Qd = 2.00 × 10⁻¹⁰ C m, giving V_approx = 44.9 V. Its roughly 6% underestimate shows that x/d = 2 is only moderately within the far-field regime.
Induction in a rotating coil
Answer: NΦ = 0.240cos(50t) Wb-turn; ε = 12.0sin(50t) V; ε_rms = 8.49 V; first positive maximum at 0.0314 s. Average power = 1.20 W and average opposing torque = 0.0240 N m.
- Flux per turn is Φ = BAcosωt. Flux linkage is NBAcosωt = 200(0.400)(3.00 × 10⁻³)cos(50t) = 0.240cos(50t) Wb-turn.
- Faraday's law gives ε = −d(NΦ)/dt = NBAωsinωt = 12.0sin(50t) V; the sign reflects Lenz's law for the chosen winding orientation.
- Peak emf is 12.0 V and sinusoidal rms emf is 12.0/√2 = 8.49 V. First positive maximum occurs when 50t = π/2, so t = 0.0314 s.
- With a pure 60.0 Ω resistance, average power is ε_rms²/R = 8.49²/60.0 = 1.20 W.
- At constant angular speed, average mechanical input equals electrical dissipation in the ideal model. Thus average torque magnitude is P/ω = 1.20/50.0 = 0.0240 N m and acts opposite rotation.
Infer half-life and background from count data
Answer: Corrected rates are 770, 420, 210, 105 and 52.5 counts min⁻¹; half-life from 10–20 min is 5.00 min, λ = 0.1386 min⁻¹, and predicted total at 30 min is about 103 counts min⁻¹.
- Subtract background from every total reading. Corrected rates are 770, 420, 210, 105 and 52.5 counts min⁻¹.
- The corrected rate halves from 210 to 105 in 5 min and again to 52.5 in the next 5 min, so the consistent later-data half-life is 5.00 min.
- Decay constant is λ = ln2/T_half = 0.6931/5.00 = 0.1386 min⁻¹.
- At 30 min, two more half-lives after 20 min, corrected rate is 52.5/4 = 13.1 counts min⁻¹. Add background to predict total 103 counts min⁻¹.
- Counts in a fixed interval have approximate standard uncertainty √N, so relative uncertainty grows as activity falls. A wrong background makes ln(corrected rate) curve, especially late when background is a large fraction; fit background and decay carefully rather than subtracting an arbitrary value.
Photoelectric energy and measurement design
Answer: At 400 nm: photon energy = 3.10 eV, K_max = 0.900 eV, stopping potential = 0.900 V, and v_max = 5.63 × 10⁵ m s⁻¹. At 300 nm, K_max and stopping potential rise to about 1.93 eV and 1.93 V; saturation current is unchanged for equal photon rate and efficiency.
- Photon energy is hc/λ = (6.626 × 10⁻³⁴)(2.998 × 10⁸)/(400 × 10⁻⁹) = 4.966 × 10⁻¹⁹ J = 3.10 eV.
- Einstein's equation gives K_max = hf−Φ = 3.10−2.20 = 0.900 eV, so stopping potential is K_max/e = 0.900 V.
- Convert kinetic energy to joules and use ½mv²: v = √[2(0.900)(1.602 × 10⁻¹⁹)/(9.11 × 10⁻³¹)] = 5.63 × 10⁵ m s⁻¹.
- At 300 nm photon energy is 4.13 eV, so K_max ≈ 1.93 eV and stopping potential ≈ 1.93 V. Equal photon rate and unchanged quantum efficiency give the same saturation electron rate, not greater current merely because each photon is more energetic.
- Reverse the collector voltage and extrapolate the photocurrent curve to zero current, correcting dark current and contact potential. Calling an arbitrary small-current voltage the stopping potential introduces detector-sensitivity bias.