AP Physics 2 FRQ Practice Guide

FRQs are 50% of your score (90 points out of 180 FRQ points). The rubrics reward clarity and justification—not depth. You have six FRQs, 15 minutes each: be strategic and systematic.

The six FRQ archetypes

Every year, College Board recycles these patterns:

1. Experimental design: Design a lab to find an unknown (e.g., density via buoyancy). Must include controlled variables and measurement method.
2. Multi-step calculation: 3–4 part problem (e.g., circuit analysis, then power dissipation, then energy).
3. Conceptual explanation: "Explain why…" or "Describe how…" Requires claim + evidence + reasoning.
4. Diagram interpretation: Read a PV diagram, circuit, free body diagram, or ray diagram; extract data.
5. Phenomena analysis: Describe motion or behavior using physics principles (e.g., loop-the-loop, magnetic deflection).
6. Design & justification: "Propose a method to [measure/calculate/demonstrate]…" then justify why it works.

Claim-Evidence-Reasoning (CER) template

When asked to explain or justify:

Claim: State the answer in 1 sentence.

The object floats because the buoyant force equals its weight.

Evidence: Cite the equation, principle, or data.

$F_b = \rho_{\text{liquid}} V_{\text{displaced}} g$ and $F_g = mg$. If $\rho_{\text{liquid}} = 1000 \, \text{kg/m}^3$, $V_{\text{displaced}} = 0.01 \, \text{m}^3$, then $F_b = 100 \, \text{N}$. For equilibrium, $m = 10 \, \text{kg}$.

Reasoning: Connect evidence back to the claim.

Since the upward buoyant force equals the downward weight, the net force is zero and the object is in equilibrium—neither sinking nor rising.

Rubric breakdown: Often 1 pt claim, 1 pt equation, 1 pt calculation, 1 pt reasoning. Miss reasoning? Lose 1 point. Skip equation? Lose 1–2 points.

Circuit diagram rules

Always include:

1. Nodes (dots or circles at junctions).
2. Battery polarity (long line = $+$, short line = $-$).
3. Current direction arrows on each branch (conventional current, out of $+$ terminal).
4. Resistor labels ($R_1$, $R_2$, etc.) with values.
5. Simplified equivalent if combining series/parallel.

Avoid:

• Unmarked junctions (ambiguous series vs parallel).
• Ambiguous battery placement.
• Forgetting total equivalent resistance label.

Example:

[Battery 12V] ─────┬─── [R₁ 6Ω] ───┬───[Ground]
                   │               │
                   └─── [R₂ 6Ω] ───┘
                   (parallel, total 3Ω)

Then label: $R_{23} = 3\Omega$, $I_{\text{total}} = 4 \, \text{A}$, $I_{R_1} = 2 \, \text{A}$ (half the total, by current divider).

Ray diagram rules (lenses)

3-ray method:

1. Ray 1: Parallel to axis → refracts through focal point $F$.
2. Ray 2: Through lens center → passes straight.
3. Ray 3: Through $F$ → refracts parallel.

Where 3 rays meet = image location.

Diverging lens: all images are virtual, upright, reduced.

Checklist:

• ☐ Horizontal optical axis.
• ☐ Focal points on both sides, distance $f$ from lens.
• ☐ Object (arrow) on left, image on right (real) or left (virtual).
• ☐ All 3 rays drawn, converging/diverging at image.
• ☐ Image height estimated from diagram.

Worked Example 1: Circuits (Kirchhoff's Laws)

Prompt: A two-loop circuit has battery 12V and resistors 4Ω, 6Ω in loop 1; battery 8V and resistor 3Ω in loop 2. Loops share a common node. Find current through the 6Ω resistor.

Solution:

Step 1: Draw circuit & label nodes, currents.

Loop 1:     Loop 2:
[12V]—[4Ω]  [8V]—[3Ω]
   |         |
   |    [6Ω] |
   └────●────┘
      Node X

• $I_1$ = current loop 1 (through 4Ω, 6Ω).
• $I_2$ = current loop 2 (through 3Ω).
• $I_3 = I_1 + I_2$ = total current through 6Ω (by KCL at node X).

Step 2: Apply KVL to each loop.

Loop 1 (clockwise): $12 - 4I_1 - 6(I_1 + I_2) = 0$ $12 = 10I_1 + 6I_2 \quad \text{...(1)}$

Loop 2 (clockwise): $8 - 3I_2 - 6(I_1 + I_2) = 0$ $8 = 6I_1 + 9I_2 \quad \text{...(2)}$

Step 3: Solve simultaneously.

From (1): $I_1 = \frac{12 - 6I_2}{10}$

Substitute into (2): $8 = 6 \left(\frac{12 - 6I_2}{10}\right) + 9I_2$ $80 = 6(12 - 6I_2) + 90I_2$ $80 = 72 - 36I_2 + 90I_2$ $8 = 54I_2$ $I_2 = 0.148 \, \text{A}$

Back-substitute: $I_1 = \frac{12 - 6(0.148)}{10} = 1.11 \, \text{A}$

Current through 6Ω: $I_3 = I_1 + I_2 = 1.11 + 0.148 = 1.26 \, \text{A}$

Rubric (4 points):

• 1 pt: correct circuit / node labels.
• 1 pt: KVL equations correct.
• 1 pt: correct algebra.
• 1 pt: final answer with units.

Worked Example 2: Optics (Lens Equation & Ray Diagram)

Prompt: A converging lens ($f = 20 \, \text{cm}$) has an object at $d_o = 30 \, \text{cm}$. (a) Find image distance. (b) Draw ray diagram. (c) Describe the image.

Solution:

(a) Lens equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{20} = \frac{1}{30} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$ $d_i = 60 \, \text{cm}$

Image is 60 cm from lens, on opposite side (real image).

(b) Ray diagram:

• Object: arrow at 30 cm, height 5 mm.
• Ray 1 (parallel): refracts through $F$ at 20 cm.
• Ray 2 (center): straight through lens.
• Ray 3 (focal): enters through $F$, exits parallel.
• All 3 converge at 60 cm.

(c) Image description:

Magnification: $m = -\frac{d_i}{d_o} = -\frac{60}{30} = -2$

• Type: Real (can project).
• Size: 2× object height (magnified).
• Orientation: Inverted (negative $m$).
• Location: 60 cm from lens, opposite side.

Rubric (4 points):

• 1 pt: lens equation setup & algebra.
• 1 pt: $d_i = 60 \, \text{cm}$ correct.
• 1 pt: ray diagram (3 rays meet at image, labeled).
• 1 pt: description (real, inverted, 2× magnified).

Worked Example 3: Thermodynamics (PV Diagram & First Law)

Prompt: A gas undergoes: A→B isobaric (2 atm, 1 L → 2 L), then B→C isochoric (to 1 atm). Find $W$, $Q$, $\Delta U$ for each. ($n = 0.05 \, \text{mol}$, diatomic gas $C_V = \frac{5}{2}R$).

Solution:

A→B (isobaric, $P = 2 \, \text{atm} = 202650 \, \text{Pa}$):

Work done on the gas (the gas expands, so it does $+203$ J of work and $W$ is negative): $W = -P \Delta V = -202650 \times (0.002 - 0.001) = -202.65 \, \text{J}$

Temperature change (from $PV = nRT$): $T_A = \frac{PV_A}{nR} = \frac{202650 \times 0.001}{0.05 \times 8.31} = 488 \, \text{K}$ $T_B = \frac{202650 \times 0.002}{0.05 \times 8.31} = 976 \, \text{K}$ $\Delta T = 488 \, \text{K}$

Internal energy change: $\Delta U = nC_V \Delta T = 0.05 \times \frac{5}{2} \times 8.31 \times 488 = 1012 \, \text{J}$

First Law: $Q = \Delta U - W = 1012 - (-203) = 1215 \, \text{J}$

B→C (isochoric):

Work: $W = 0$ (no volume change).

Temperature at C: $T_C = \frac{101325 \times 0.002}{0.05 \times 8.31} = 488 \, \text{K}$ (back to $T_A$).

$\Delta U = nC_V \Delta T = 0.05 \times \frac{5}{2} \times 8.31 \times (488 - 976) = -1012 \, \text{J}$

$Q = \Delta U - W = -1012 - 0 = -1012 \, \text{J}$

Summary:

Rubric (4 points):

• 1 pt: process type, correct work formula.
• 1 pt: $\Delta U$ using $nC_V \Delta T$.
• 1 pt: First Law to find $Q$.
• 1 pt: both legs or correct total.

Common FRQ point-leaks

1. Forgetting to draw / label diagrams. Rubrics award points for the diagram itself.
2. Skipping intermediate steps. Graders give partial credit only if logic is shown.
3. Work sign errors. Isobaric expansion: $W > 0$. Compression: $W < 0$.
4. Units missing on final answers.
5. Misinterpreting the prompt. "Find and explain" = calculation + justification, not just answer.
6. Rounding too early. Keep 3–4 sig figs through calculation.

FRQ prep schedule

• Weeks 1–3: 1–2 FRQs per week, timed.
• Week 4: Full set (all 6) under exam conditions (90 min).
• Night before: Review 1 sample FRQ in each of your 3 weakest units.

Revisit the 3-day or 7-day cram plan → if you need a quick unit refresher before tackling these FRQs.