An original Thinka practice paper modelled on the structure and difficulty of the May 2025 AP AP Physics 2: Algebra-Based paper. Not affiliated with or reproduced from AP.
Section II: Free-Response Questions
Answer all four questions. Show all work, starting with fundamental physics principles and reference equations. Spend approximately 25 minutes on Q1, 30 minutes on Q2, 25 minutes on Q3, and 20 minutes on Q4.
4 Question · 40 marks
Question 1 · frq
10 marks
A very long, straight wire is fixed along the line \(y = 0\) in the \(xy\)-plane and carries a steady current \(I_0\) in the \(+x\)-direction. A particle with positive charge \(+q\) and mass \(m\) is located at point \(S\) with coordinates \((0, -d)\) and is initially moving with speed \(v_0\) in the \(+x\)-direction. The surrounding medium is a vacuum.
A.
i. Indicate the direction of each of the following quantities at point \(S\). Choose from: Into the page, Out of the page, *\(+x\)-direction*, *\(-x\)-direction*, *\(+y\)-direction*, or *\(-y\)-direction*. - The direction of the magnetic field \(\vec{B}\) produced by the wire at point \(S\). - The direction of the magnetic force \(\vec{F}_B\) exerted on the charged particle by the magnetic field at point \(S\).
ii. An external uniform electric field \(\vec{E}\) is applied in the region so that the net force on the charged particle at point \(S\) is zero. Derive an expression for the magnitude of the required electric field \(E\) in terms of \(I_0\), \(q\), \(m\), \(d\), \(v_0\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Also, state the direction of the electric field \(\vec{E}\).
B.
The electric field is removed and the particle is removed. A small circular conducting loop of radius \(r_0\) and resistance \(R\) is placed at rest in the \(xy\)-plane with its center at \((0, -2d)\). The current in the long wire now decreases steadily over time.
Indicate whether the induced current in the circular loop is clockwise, counterclockwise, or zero.
____ Clockwise
____ Counterclockwise
____ There is no induced current in the loop.
Justify your answer using physics principles such as Faraday's law and Lenz's law.
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Worked solution
Part A (i) - By the right-hand rule for a long straight wire carrying current in the \(+x\)-direction, the magnetic field below the wire (at \(y < 0\), including point \(S\)) points out of the page (in the \(+z\)-direction). - The magnetic force on a moving charge is \(\vec{F}_B = q(\vec{v} \times \vec{B})\). With \(\vec{v}\) in the \(+x\)-direction and \(\vec{B}\) out of the page (\(+z\)-direction), the cross product points in the **\(-y\)-direction** (downward).
Part A (ii) - Begin with Newton's first law / condition for zero net force: \[ \sum \vec{F} = \vec{F}_E + \vec{F}_B = 0 \] \[ F_E = F_B \] - Substitute the formulas for electric and magnetic forces: \[ qE = q v_0 B \implies E = v_0 B \] - The magnetic field at distance \(d\) from a long straight current-carrying wire is: \[ B = \frac{\mu_0 I_0}{2\pi d} \] - Substitute \(B\) into the electric field equation: \[ E = v_0 \left(\frac{\mu_0 I_0}{2\pi d}\right) = \frac{\mu_0 I_0 v_0}{2\pi d} \] - Since \(\vec{F}_B\) is directed in the \(-y\)-direction, the electric force \(\vec{F}_E\) must point in the \(+y\)-direction to cancel it. Because the charge \(+q\) is positive, the electric field \(\vec{E}\) must be directed in the **\(+y\)-direction.
Part B** - Counterclockwise. - The magnetic field produced by the wire at the location of the loop is directed out of the page, so the magnetic flux through the loop is directed out of the page. - As the current \(I\) decreases, the magnitude of the magnetic field and thus the magnetic flux out of the page decreases. - According to Lenz's law, the induced current must generate an induced magnetic field that opposes this change, which means the induced magnetic field must point out of the page. - By the right-hand rule for loops, a magnetic field directed out of the page is produced by a counterclockwise induced current.
Marking scheme
Part A(i): 2 points - Point A1: For correctly identifying that the magnetic field is directed out of the page at point S. - Point A2: For correctly identifying that the magnetic force is directed in the -y-direction (or consistent with the magnetic field direction indicated).
Part A(ii): 5 points - Point A3: For stating a fundamental physics principle or equilibrium equation, such as \(\sum \vec{F} = 0\), \(F_E = F_B\), or \(qE = qvB\). - Point A4: For writing a correct expression for the magnetic field of a long wire: \(B = \frac{\mu_0 I_0}{2\pi d}\). - Point A5: For setting the electric force equal to the magnetic force: \(qE = q v_0 B\). - Point A6: For arriving at the correct final expression: \(E = \frac{\mu_0 I_0 v_0}{2\pi d}\). - Point A7: For correctly stating that the electric field is in the +y-direction.
Part B: 3 points - Point B1: For selecting 'Counterclockwise'. - Point B2: For indicating that the external magnetic field / magnetic flux through the loop due to the wire is directed out of the page and is decreasing. - Point B3: For correctly applying Lenz's law to conclude that the induced magnetic field must point out of the page to oppose the decrease, resulting in a counterclockwise current.
Question 2 · Translating Between Representations (TBR)
12 marks
A sample of a monatomic ideal gas is sealed inside a thermally conducting vertical cylinder of cross-sectional area \(A\) by a movable, frictionless piston of mass \(M_p\). The top of the piston is open to the atmosphere at pressure \(P_{\text{atm}}\). A light vertical string is attached to the center of the top of the piston, runs over an ideal frictionless pulley, and supports a suspended counterweight of mass \(m_c\), where \(m_c < M_p\). The cylinder is immersed in a large water bath maintained at a constant temperature \(T_0\). The system is in thermal and mechanical equilibrium, the piston is at rest, and the gas occupies volume \(V_0\).
A. On the dot shown below representing the piston, draw and label the forces exerted on the piston. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
$$\bullet$$
B. Derive an expression for the internal energy \(U\) of the gas in terms of \(M_p\), \(m_c\), \(A\), \(V_0\), \(P_{\text{atm}}\), and fundamental constants. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. The string attached to the counterweight is cut, and the piston slowly moves downward while the container remains in the water bath at constant temperature \(T_0\). The piston comes to rest at a new equilibrium state at time \(t_f\). On the axes provided below, sketch the expected relationship between the pressure \(P\) and volume \(V\) of the gas for the thermodynamic process that the gas undergoes during this compression. Draw an arrow on your sketch to represent the direction of the thermodynamic process.
$$\begin{array}{r|l} P & \\ \uparrow & \\ & \\ & \\ 0 & \hline \longrightarrow V \end{array}$$
D. With the string still severed, the temperature of the water bath is changed to a new constant temperature \(T_{\text{new}}\). When the gas reaches thermal equilibrium with the bath, it expands until it once again occupies the original volume \(V_0\). Indicate whether \(T_{\text{new}}\) is greater than, less than, or equal to \(T_0\).
______ \(T_{\text{new}} > T_0\)
______ \(T_{\text{new}} < T_0\)
______ \(T_{\text{new}} = T_0\)
Briefly justify your answer by referencing at least one feature of your answers to parts A, B, or C.
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Worked solution
Part A: The piston is in static equilibrium (\(\sum \vec{F} = 0\)). The forces acting on the piston are: 1. Downward gravitational force: \(F_g = M_p g\) 2. Downward atmospheric force: \(F_{\text{atm}} = P_{\text{atm}} A\) 3. Upward tension force from the string: \(F_T = m_c g\) 4. Upward force exerted by the enclosed gas: \(F_{\text{gas}} = P_{\text{gas}} A\)
Part B: Begin with Newton's second law for the piston in equilibrium: $$\sum F_y = 0$$ $$F_{\text{gas}} + F_T - F_g - F_{\text{atm}} = 0$$ Substitute the expressions for the forces: $$P_{\text{gas}} A + m_c g - M_p g - P_{\text{atm}} A = 0$$ $$P_{\text{gas}} A = P_{\text{atm}} A + (M_p - m_c)g$$ $$P_{\text{gas}} = P_{\text{atm}} + \frac{(M_p - m_c)g}{A}$$
For a monatomic ideal gas, the internal energy is related to pressure and volume by: $$U = \frac{3}{2} n R T$$ Using the ideal gas equation \(P V = n R T\): $$U = \frac{3}{2} P_{\text{gas}} V_0$$ Substituting the expression for \(P_{\text{gas}}\): $$U = \frac{3}{2}\left(P_{\text{atm}} + \frac{(M_p - m_c)g}{A}\right)V_0$$
Part C: Because the cylinder remains in the constant-temperature water bath (\(T = T_0 = \text{constant}\)), the process is isothermal. By the ideal gas law, \(P V = \text{constant}\), which gives \(P \propto \frac{1}{V}\). The graph is a smooth, concave-up hyperbola connecting a state of larger volume and lower pressure to a state of smaller volume and higher pressure. The arrow points upward and to the left (toward decreasing volume).
Part D: Selection: \(T_{\text{new}} > T_0\) Justification: From the force balance in Part A and derivation in Part B, when the string is cut, the tension force \(F_T\) is removed, so the downward force on the piston increases from \(M_p g + P_{\text{atm}} A - m_c g\) to \(M_p g + P_{\text{atm}} A\). For the piston to remain in equilibrium at the restored volume \(V_0\), the new gas pressure must be: $$P_{\text{new}} = P_{\text{atm}} + \frac{M_p g}{A} > P_{\text{gas}, 0}$$ According to the ideal gas law (\(P V = n R T\)), because the volume is the same (\(V = V_0\)) but the pressure has increased (\(P_{\text{new}} > P_0\)), the final temperature must be greater than the initial temperature: \(T_{\text{new}} > T_0\).
Marking scheme
Part A: 3 points • Point A1: For drawing and labeling the downward gravitational force (\(F_g\)) and downward atmospheric force (\(F_{\text{atm}}\)) on the piston. • Point A2: For drawing and labeling the upward tension force (\(F_T\)) on the piston. • Point A3: For drawing and labeling the upward force exerted by the gas (\(F_{\text{gas}}\)) on the piston.
Part B: 4 points • Point B1: For starting with a fundamental equation from the reference sheet (e.g., \(\sum F = 0\), \(U = \frac{3}{2}nRT\), and/or \(PV = nRT\)). • Point B2: For writing a correct force balance equation consistent with part A: \(P_{\text{gas}}A + m_c g - M_p g - P_{\text{atm}}A = 0\). • Point B3: For correctly solving for the absolute gas pressure \(P_{\text{gas}} = P_{\text{atm}} + \frac{(M_p - m_c)g}{A}\). • Point B4: For correctly substituting \(P_{\text{gas}}\) into \(U = \frac{3}{2}P_{\text{gas}}V_0\) to obtain \(U = \frac{3}{2}\left(P_{\text{atm}} + \frac{(M_p - m_c)g}{A}\right)V_0\).
Part C: 3 points • Point C1: For drawing a curve connecting a point in the lower-right region to a point in the upper-left region. • Point C2: For drawing a curve that is concave up (isothermal curve). • Point C3: For drawing an arrow along the curve directed from right to left (toward lower volume / higher pressure).
Part D: 2 points • Point D1: For selecting \(T_{\text{new}} > T_0\). • Point D2: For a valid justification referencing features from Part A, B, or C showing that the removal of the counterweight increases the equilibrium pressure, so restoring the gas to the original volume \(V_0\) requires a higher temperature via \(PV = nRT\).
Question 3 · frq
10 marks
In Experiment 1, a student is given a spool of thin, uniform cylindrical metal wire of unknown resistivity \(\rho\). The student is tasked with designing an experiment to determine the resistivity \(\rho\) of the metal. The student has access to a variable DC power supply, a switch, an ammeter, a voltmeter, a meterstick, a micrometer caliper, and connecting wires of negligible resistance.
A. Describe a procedure for collecting data that would allow the student to determine the resistivity \(\rho\) of the metal wire. In your description, include the specific measurements to be made and any steps necessary to reduce experimental uncertainty.
B. Describe how the collected data could be analyzed to determine the resistivity \(\rho\). Include references to appropriate equations and to relationships between measured and known quantities.
C. In Experiment 2, the student uses a micrometer caliper to measure the diameter of a different cylindrical wire sample to be \(D = 0.40\text{ mm}\) (giving a cross-sectional area of \(A \approx 1.26 \times 10^{-7}\text{ m}^2\)). The student cuts five segments of this wire with various lengths \(L\), and measures the electrical resistance \(R\) of each segment. The data collected are recorded in Table 1 below.
Table 1 Length \(L\text{ (m)}\)Resistance \(R\text{ (}\Omega\text{)}\)0.403.30.806.51.209.41.6012.92.0015.8 i. Indicate two quantities, either measured quantities from Table 1 or calculated quantities, that could be graphed on the vertical and horizontal axes to produce a straight line that could be used to determine \(\rho\). Vertical axis: _______________ Horizontal axis: _______________
ii. On a grid, create a graph of the quantities indicated in part C (i). Clearly label the axes, including units and scales as appropriate, and plot the data points.
iii. Draw a best-fit line for the data graphed in part C (ii).
D. Using the best-fit line that you drew in part C (iii), calculate an experimental value for the resistivity \(\rho\) of the metal.
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Worked solution
Part A: 1. Use the micrometer caliper to measure the diameter \(d\) of the wire at multiple different locations along its length to reduce uncertainty, and calculate the average diameter to find cross-sectional area \(A = \pi (d/2)^2\). 2. Measure a specific length \(L\) of the wire using the meterstick. 3. Connect the wire segment in series with the variable DC power supply, a switch, and the ammeter. Connect the voltmeter in parallel across the measured length \(L\) of the wire. 4. Close the switch, adjust the power supply to set a current, and record the potential difference \(\Delta V\) and current \(I\). 5. Repeat measurements for several different lengths \(L\) (or repeat current/voltage measurements at several settings for each length to reduce random error).
Part B: From Ohm's law, the resistance of each wire segment is given by \(R = \frac{\Delta V}{I}\). The resistance of a cylindrical conductor is related to its geometry and resistivity by \(R = \rho \frac{L}{A}\). Plotting \(R\) versus \(L\) yields a straight line with slope \(m = \frac{\rho}{A}\). The resistivity is then calculated from the slope and the cross-sectional area using \(\rho = m \cdot A = m \cdot \pi \left(\frac{d}{2}\right)^2\).
Part C: i. Vertical axis: \(R\) (in \(\Omega\)); Horizontal axis: \(L\) (in \(\text{m}\)) (or vice versa). ii. & iii. Plotting \(R\) on the vertical axis (scaled from 0 to 18 \(\Omega\)) and \(L\) on the horizontal axis (scaled from 0 to 2.2 \(\text{m}\)): Plot points: (0.40, 3.3), (0.80, 6.5), (1.20, 9.4), (1.60, 12.9), (2.00, 15.8). Draw a straight best-fit line passing through the points.
Part D: Select two points on the best-fit line, for example \((0.20\text{ m}, 1.6\ \Omega)\) and \((1.80\text{ m}, 14.2\ \Omega)\): \(\text{slope} = \frac{14.2\ \Omega - 1.6\ \Omega}{1.80\text{ m} - 0.20\text{ m}} = \frac{12.6\ \Omega}{1.60\text{ m}} = 7.875\ \Omega/\text{m}\). Since \(R = \frac{\rho}{A} L\), \(\text{slope} = \frac{\rho}{A}\). \(\rho = \text{slope} \times A = (7.875\ \Omega/\text{m}) \times (1.26 \times 10^{-7}\text{ m}^2) \approx 9.9 \times 10^{-7}\ \Omega\cdot\text{m}\).
Marking scheme
Part A (2 points): • Point A1: For describing a procedure that includes measurements of the wire's dimensions (diameter/cross-sectional area and length) and electrical quantities (current and potential difference, or direct resistance). • Point A2: For indicating an appropriate method to reduce experimental uncertainty (e.g., measuring the wire diameter at multiple positions, repeating trials for multiple lengths/voltages).
Part B (2 points): • Point B1: For stating the fundamental relationship between resistance, resistivity, length, and cross-sectional area: \(R = \rho L / A\). • Point B2: For relating measured electrical quantities to resistance (e.g., \(R = \Delta V / I\)) and explaining how resistivity \(\rho\) is determined from the graph's slope (e.g., \(\rho = \text{slope} \times A\)).
Part C (4 points): • Point C1: For identifying appropriate quantities that yield a linear graph (e.g., \(R\) vs. \(L\) or \(L\) vs. \(R\)). • Point C2: For labeling both axes with appropriate quantities, units, and linear scales covering more than half the grid. • Point C3: For correctly plotting the data points from Table 1. • Point C4: For drawing a single straight best-fit line that reasonably reflects the trend of the data.
Part D (2 points): • Point D1: For correctly calculating the slope of the best-fit line using two points on the line (not data points unless they lie on the line) and relating the slope to \(\rho\) via \(\rho = \text{slope} \times A\). • Point D2: For a calculated value of \(\rho\) in the acceptable range of \(9.0 \times 10^{-7}\ \Omega\cdot\text{m}\) to \(1.1 \times 10^{-6}\ \Omega\cdot\text{m}\) with correct units.
Question 4 · frq
8 marks
Monochromatic light of frequency \( f \) is incident on a clean metal emitter inside an evacuated phototube. The work function of the metal is \( \Phi \), where \( hf > \Phi \). An adjustable electric potential difference is established between the emitter and collector to determine the stopping potential \( \Delta V_s \) required to reduce the photocurrent to zero.
A student makes the following claim: *"If the frequency of the incident light is doubled to \( 2f \), the stopping potential \( \Delta V_s \) needed to stop the photoelectrons will also be exactly doubled."*
A. Indicate whether the student's claim is correct or incorrect. Without manipulating equations, justify your answer by describing how the energy of each incident photon and the maximum kinetic energy of the emitted photoelectrons change when the frequency is doubled.
B. Derive an expression for the stopping potential \( \Delta V_s \) in terms of \( f \), \( \Phi \), and fundamental physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Then, determine an algebraic expression for the stopping potential \( \Delta V_s' \) when light of frequency \( 2f \) is used.
C. Indicate whether the algebraic expression derived in part B is or is not consistent with your qualitative answer from part A. Briefly justify your answer using your derived expression.
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Worked solution
Part A: - Claim: Incorrect. - Justification: The energy of each incident photon is proportional to its frequency, so doubling the frequency doubles the energy delivered by each photon to an electron. The maximum kinetic energy of an emitted photoelectron equals the photon energy minus the fixed work function of the metal emitter. Since the same constant work function is subtracted from double the original photon energy, the maximum kinetic energy of the emitted electrons more than doubles. Consequently, the stopping potential required to bring the fastest electrons to rest must increase by more than a factor of two.
Part B: - Start from conservation of energy / the photoelectric equation: \[ K_{\max} = E_{\text{photon}} - \Phi \] \[ K_{\max} = hf - \Phi \] - Relate maximum kinetic energy to stopping potential \( \Delta V_s \): \[ K_{\max} = e \Delta V_s \] - Substitute and solve for \( \Delta V_s \): \[ e \Delta V_s = hf - \Phi \implies \Delta V_s = \frac{hf - \Phi}{e} \] - For incident light with doubled frequency \( 2f \): \[ \Delta V_s' = \frac{h(2f) - \Phi}{e} = \frac{2hf - \Phi}{e} \]
Part C: - Consistency: The expression derived in part B is consistent with the answer in part A. - Justification: Rewriting \( \Delta V_s' \): \[ \Delta V_s' = \frac{2(hf - \Phi) + \Phi}{e} = 2\Delta V_s + \frac{\Phi}{e} \] Since \( \Phi > 0 \) and \( e > 0 \), the term \( \frac{\Phi}{e} > 0 \). Thus, \( \Delta V_s' > 2\Delta V_s \), confirming quantitatively that doubling the frequency more than doubles the stopping potential.
Marking scheme
Part A (3 points): - Point A1: For stating that the claim is incorrect (or making a statement consistent with the justification). - Point A2: For identifying that doubling the frequency doubles the energy of each incident photon. - Point A3: For reasoning that because the work function is constant, subtracting the same work function from twice the photon energy results in the maximum kinetic energy (and stopping potential) increasing by more than a factor of 2.
Part B (3 points): - Point B1: For starting the derivation with a fundamental principle or reference equation: \( K_{\max} = hf - \Phi \) and \( K_{\max} = e\Delta V_s \). - Point B2: For deriving a correct expression for \( \Delta V_s = \frac{hf - \Phi}{e} \). - Point B3: For determining the correct expression for the new stopping potential \( \Delta V_s' = \frac{2hf - \Phi}{e} \).
Part C (2 points): - Point C1: For stating that the derived expression is consistent with the qualitative reasoning in part A. - Point C2: For demonstrating from the algebraic expression that \( \Delta V_s' = 2\Delta V_s + \frac{\Phi}{e} > 2\Delta V_s \) (or noting that the non-zero negative intercept in \( \Delta V_s(f) \) implies non-proportional scaling).
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