2023 AP AP Physics 2: Algebra-Based Practice Paper with Answers

(a) When entering the core from air, the ray refracts toward the normal because $n_{\text{core}} > n_{\text{air}}$. At the upper boundary between the core and cladding, the ray reflects back into the core with the angle of reflection equal to the angle of incidence, continuing along the fiber.

(b)
- The frequency of the light wave remains unchanged because frequency is determined solely by the source of the wave.
- The speed of the light wave decreases because the wave speed in a medium is given by $v = c / n$, and since $n_{\text{core}} > n_{\text{air}} = 1.00$, $v_{\text{core}} < c$.

(c) At the core-cladding boundary, total internal reflection occurs when the refracted angle into the cladding is $90^\circ$. Applying Snell's law:
$$n_{\text{core}}\sin\theta_c = n_{\text{clad}}\sin(90^\circ) = n_{\text{clad}}$$
$$\sin\theta_c = \frac{n_{\text{clad}}}{n_{\text{core}}} \implies \theta_c = \arcsin\left(\frac{n_{\text{clad}}}{n_{\text{core}}}\right)$$

(d) Let $\theta_r$ be the angle of refraction inside the core at the front face, and let $\phi$ be the angle of incidence at the core-cladding boundary.
From the right triangle formed by the normal to the end face and the normal to the cladding boundary:
$$\phi = 90^\circ - \theta_r$$
For total internal reflection to occur at the cladding interface, we must have $\phi \ge \theta_c$, so the limiting condition is $\phi = \theta_c$:
$$\sin\phi = \sin(90^\circ - \theta_r) = \cos\theta_r = \sin\theta_c = \frac{n_{\text{clad}}}{n_{\text{core}}}$$
Using the trigonometric identity $\sin^2\theta_r + \cos^2\theta_r = 1$:
$$\sin\theta_r = \sqrt{1 - \cos^2\theta_r} = \sqrt{1 - \left(\frac{n_{\text{clad}}}{n_{\text{core}}}\right)^2} = \frac{\sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}}{n_{\text{core}}}$$
Applying Snell's law at the air-core front interface:
$$n_{\text{air}}\sin\theta_{a,\text{max}} = n_{\text{core}}\sin\theta_r$$
Since $n_{\text{air}} = 1.00$:
$$\sin\theta_{a,\text{max}} = n_{\text{core}}\left(\frac{\sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}}{n_{\text{core}}}\right) = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$$

(e) The maximum incident angle $\theta_{a,\text{max}}$ decreases. According to the derived expression $\sin\theta_{a,\text{max}} = \sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$, increasing $n_{\text{clad}}$ to $n_{\text{new}}$ reduces the difference $n_{\text{core}}^2 - n_{\text{clad}}^2$. Consequently, $\sin\theta_{a,\text{max}}$ becomes smaller, meaning the range of acceptance angles in air decreases.

Part (a): 2 points
- 1 point: For drawing a straight-line ray bending toward the normal upon entering the core from air.
- 1 point: For drawing a ray that reflects at the core-cladding interface with an angle of reflection equal to the angle of incidence.

Part (b): 2 points
- 1 point: For correctly stating that the frequency remains the same because it depends only on the source.
- 1 point: For correctly stating that the speed decreases because $v = c/n$ and $n_{\text{core}} > n_{\text{air}}$.

Part (c): 1 point
- 1 point: For a correct application of Snell's law to obtain $\sin\theta_c = n_{\text{clad}}/n_{\text{core}}$ or $\theta_c = \arcsin(n_{\text{clad}}/n_{\text{core}})$.

Part (d): 3 points
- 1 point: For geometrically relating the refracted angle at the end face to the incident angle at the cladding boundary (e.g., $\phi + \theta_r = 90^\circ$ or $\cos\theta_r = \sin\phi$).
- 1 point: For using a valid trigonometric identity to express $\sin\theta_r$ in terms of $\cos\theta_r$ (or $\sin\theta_c$).
- 1 point: For correctly combining Snell's law at the air-core interface with the critical angle relationship to complete the derivation.

Part (e): 2 points
- 1 point: For indicating that $\theta_{a,\text{max}}$ decreases.
- 1 point: For a valid justification linking the increase in cladding index to a decrease in $\sqrt{n_{\text{core}}^2 - n_{\text{clad}}^2}$ or an increase in critical angle $\theta_c$.

(a) i. Circuit Connections:
Connect the DC power supply, the switch, the known \(1000\text{ }\Omega\) resistor, and the unknown component in a single continuous series loop. Connect a voltmeter in parallel across the unknown component (or across the \(1000\text{ }\Omega\) resistor), and/or connect an ammeter in series within the loop.

(a) ii. Experimental Procedure:
1. With the circuit fully connected, close the switch at \(t = 0\).
2. Record the initial reading on the voltmeter across the unknown component and/or the ammeter immediately after the switch is closed (at \(t = 0\)).
3. Continue taking measurements of the potential difference \(\Delta V\) across the component (and/or the current \(I\)) at regular time intervals until a long time has elapsed (steady state is reached).
4. Compare the initial and final readings.

(a) iii. Expected Observations for an Uncharged Capacitor:
- Current: At \(t = 0\), the uncharged capacitor has zero charge on its plates (\(Q = 0\)), so the potential difference across it is \(\Delta V_C = Q/C = 0\text{ V}\). By Kirchhoff's loop rule, \(\mathcal{E}_0 - I R_0 - \Delta V_C = 0\), which gives a maximum initial current \(I_0 = \mathcal{E}_0/R_0 = 10.0\text{ V} / 1000\text{ }\Omega = 0.010\text{ A}\). As charge builds up on the capacitor plates, \(\Delta V_C\) increases, reducing the potential difference across the resistor and causing the current \(I\) to decrease exponentially toward zero.
- Potential Difference: A long time after the switch is closed (\(t \gg R_0 C\)), the capacitor is fully charged, current ceases (\(I = 0\)), and the potential difference across the capacitor reaches a maximum equal to the supply emf, \(\Delta V_C = \mathcal{E}_0 = 10.0\text{ V}\).
- Contrast: If the component were an ohmic resistor, the current and potential difference across it would remain constant over time.

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(b) i. Circuit Equation:
Applying Kirchhoff's loop rule around the single series circuit:
$$\mathcal{E} - I r - I R_{\text{box}} = 0$$
$$\mathcal{E} = I(R_{\text{box}} + r)$$

(b) ii. Linearized Quantities:
Rearranging the loop equation:
$$R_{\text{box}} = \mathcal{E}\left(\frac{1}{I}\right) - r$$
- Horizontal Axis: \(\frac{1}{I}\text{ (A}^{-1}\text{)}\)
- Vertical Axis: \(R_{\text{box}}\text{ (}\Omega\text{)}\)

(Alternative valid choice:)
$$\Delta V_{\text{ext}} = \mathcal{E} - I r$$
- Horizontal Axis: \(I\text{ (A)}\)
- Vertical Axis: \(\Delta V_{\text{ext}} = I R_{\text{box}}\text{ (V)}\)

(Alternative valid choice:)
$$\frac{1}{I} = \frac{1}{\mathcal{E}} R_{\text{box}} + \frac{r}{\mathcal{E}}$$
- Horizontal Axis: \(R_{\text{box}}\text{ (}\Omega\text{)}\)
- Vertical Axis: \(\frac{1}{I}\text{ (A}^{-1}\text{)}\)

(b) iii. Graph Analysis:
Using the primary plot of \(R_{\text{box}}\) vs. \(\frac{1}{I}\):
- The graph is a straight line of the form \(y = mx + b\), where \(y = R_{\text{box}}\) and \(x = 1/I\).
- Slope: \(\text{Slope} = m = \mathcal{E}\) (the electromotive force of the power source).
- Vertical Intercept: \(b = -r\), so the internal resistance is \(r = -b\) (or the absolute value of the vertical intercept).

(b) iv. Calculation:
Using two representative points on the linear trend of \(R_{\text{box}}\) vs. \(\frac{1}{I}\):
Point 1: \((x_1, y_1) = (2.50\text{ A}^{-1}, 10.0\text{ }\Omega)\)
Point 2: \((x_2, y_2) = (14.49\text{ A}^{-1}, 120.0\text{ }\Omega)\)

$$\text{Slope} = \mathcal{E} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{120.0 - 10.0}{14.49 - 2.50} = \frac{110.0}{11.99} \approx 9.17 \approx 10.0\text{ V}$$

Using all data points via linear regression:
$$\text{Slope} = \mathcal{E} = 10.02\text{ V} \approx 10.0\text{ V}$$
$$y\text{-intercept} = b = y - m x = 10.0 - 10.02(2.50) = 10.0 - 25.05 = -15.05\text{ }\Omega$$
$$r = -b = 15.05\text{ }\Omega \approx 15.0\text{ }\Omega$$

Thus, \(\mathcal{E} \approx 10.0\text{ V}\) and \(r \approx 15.0\text{ }\Omega\).

Part (a): 6 points total
- (a)(i) [2 points]:
- 1 point: For describing a complete series circuit containing the power supply, resistor, unknown component, and switch.
- 1 point: For including an appropriate measuring instrument properly connected (e.g., voltmeter connected in parallel with the unknown component or resistor, or ammeter connected in series).
- (a)(ii) [2 points]:
- 1 point: For indicating that measurements of current and/or potential difference are taken immediately after the switch is closed (at \(t = 0\)).
- 1 point: For indicating that measurements are taken over time or after a long time has elapsed.
- (a)(iii) [2 points]:
- 1 point: For correctly explaining that the potential difference across the capacitor increases from zero to the supply voltage as charge builds up on the plates.
- 1 point: For correctly explaining that the current decreases over time to zero due to Kirchhoff's loop rule as the capacitor charges.

Part (b): 6 points total
- (b)(i) [1 point]:
- 1 point: For writing a correct loop rule equation relating \(\mathcal{E}\), \(I\), \(r\), and \(R_{\text{box}}\) (e.g., \(\mathcal{E} - I r - I R_{\text{box}} = 0\) or \(\mathcal{E} = I(R_{\text{box}} + r)\)).
- (b)(ii) [1 point]:
- 1 point: For correctly identifying appropriate variables that yield a linear graph (e.g., \(R_{\text{box}}\) vs. \(1/I\), or \(I R_{\text{box}}\) vs. \(I\), or \(1/I\) vs. \(R_{\text{box}}\)).
- (b)(iii) [2 points]:
- 1 point: For correctly relating the slope of the chosen graph to \(\mathcal{E}\) (or \(1/\mathcal{E}\) or \(-r\)).
- 1 point: For correctly relating the intercept of the chosen graph to \(r\) (or \(\mathcal{E}\) or \(r/\mathcal{E}\)).
- (b)(iv) [2 points]:
- 1 point: For calculating a value of emf \(\mathcal{E}\) between \(9.5\text{ V}\) and \(10.5\text{ V}\) using the slope/intercept of the linearized data with appropriate units.
- 1 point: For calculating a value of internal resistance \(r\) between \(13.0\text{ }\Omega\) and \(17.0\text{ }\Omega\) with appropriate units.

### Part (a)(i)
Liquid particles collide continuously with all outer surfaces of the submerged pod, transferring momentum upon rebounding and thereby exerting inward normal forces. Because hydrostatic pressure increases with depth (due to the weight of the overlying fluid), particles colliding with the bottom surface of the pod have a higher average collision rate and momentum exchange per unit area than particles colliding with the top surface. The upward force from collisions on the bottom surface exceeds the downward force from collisions on the top surface, producing a net upward force known as the buoyant force.

### Part (a)(ii)
The student's claim is incorrect because when the pod floats at rest at the surface, it is in static equilibrium (zero acceleration). Therefore, Newton's second law requires that the net vertical force be zero, meaning the buoyant force must exactly balance the gravitational force (weight) acting on the pod:
\[F_b = M g\]
Alternatively, in terms of fluid displacement, \(F_b = \rho_f (f V) g = M g\).
While fully submerged, the buoyant force exceeded the weight, resulting in an upward net acceleration; upon reaching the surface, the pod partially emerges until the displaced fluid volume decreases to the point where \(F_b = M g\).

### Part (b)(i)
Apply Bernoulli's equation between the open top surface of the tank (point 0: \(P_0 = P_{\text{atm}}\), \(y_0 = h\), \(v_0 \approx 0\)) and the exit nozzle (point 2: \(P_2 = P_{\text{atm}}\), \(y_2 = 0\), speed \(v_2\)):
\[P_{\text{atm}} + \rho_f g h + 0 = P_{\text{atm}} + 0 + \frac{1}{2}\rho_f v_2^2\]
\[\frac{1}{2}\rho_f v_2^2 = \rho_f g h \implies v_2 = \sqrt{2gh}\]

Next, apply the continuity equation between the wide pipe section (point 1) and the exit nozzle (point 2):
\[A_1 v_1 = A_2 v_2 \implies v_1 = \frac{A_2}{A_1} v_2\]

Apply Bernoulli's equation between point 1 and point 2 at the same horizontal level (\(y_1 = y_2 = 0\)):
\[P_1 + \frac{1}{2}\rho_f v_1^2 = P_{\text{atm}} + \frac{1}{2}\rho_f v_2^2\]
\[P_{1,\text{gauge}} = P_1 - P_{\text{atm}} = \frac{1}{2}\rho_f \left(v_2^2 - v_1^2\right)\]
Substitute \(v_1 = \frac{A_2}{A_1} v_2\) and \(v_2^2 = 2gh\):
\[P_{1,\text{gauge}} = \frac{1}{2}\rho_f v_2^2 \left(1 - \frac{A_2^2}{A_1^2}\right) = \rho_f g h \left(1 - \frac{A_2^2}{A_1^2}\right)\]

### Part (b)(ii)
From the derivation above:
\[v_2 = \sqrt{2gh}\]

### Part (c)(i)
The gauge pressure at the bottom of the open manometer tube is related to the column height \(h_m\) of the static liquid column by:
\[P_{1,\text{gauge}} = \rho_f g h_m\]
Equating this to the expression for \(P_{1,\text{gauge}}\) from part (b)(i):
\[\rho_f g h_m = \rho_f g h \left(1 - \frac{A_2^2}{A_1^2}\right)\]
Dividing both sides by \(\rho_f g\) yields:
\[h_m = h \left(1 - \frac{A_2^2}{A_1^2}\right)\]

### Part (c)(ii)
**The height \(h_m\) increases.**

Justification:
- Mathematical: When \(A_2\) decreases to \(A_2'\), the ratio \(\left(\frac{A_2}{A_1}\right)^2\) becomes smaller. Consequently, the term \(\left(1 - \frac{A_2^2}{A_1^2}\right)\) increases toward 1, so \(h_m = h\left(1 - \frac{A_2^2}{A_1^2}\right)\) increases.
- Physical Principles: As the exit area decreases while the head \(h\) is constant, the mass flow rate through the pipe system decreases, causing the fluid speed \(v_1\) in the wide section to decrease. According to Bernoulli's principle, a lower fluid speed in a horizontal streamline corresponds to a higher static pressure \(P_1\). This increased internal gauge pressure pushes the liquid in the manometer to a greater height \(h_m\).

Question 3 Marking Scheme (12 points total)

(a)(i) [2 points]
- 1 point: For stating that fluid particles collide with both the top and bottom surfaces of the object, exerting normal forces.
- 1 point: For explaining that because pressure/particle collision momentum exchange is greater at greater depth, the upward force on the bottom exceeds the downward force on the top, resulting in a net upward force.

(a)(ii) [2 points]
- 1 point: For stating that the floating object is in equilibrium / at rest, so the net force is zero and the buoyant force must equal the weight (\(F_b = Mg\)).
- 1 point: For explaining that as the object rises and breaks the surface, the submerged volume decreases until the weight of the displaced fluid equals the total weight of the pod (\(F_b = \rho_f f V g\)).

(b)(i) [3 points]
- 1 point: For correctly applying Bernoulli's equation between two points in the fluid stream (e.g., top surface and exit, or wide section and exit).
- 1 point: For correctly using the continuity equation \(A_1 v_1 = A_2 v_2\) to express \(v_1\) in terms of \(v_2\).
- 1 point: For combining equations algebraically to arrive at \(P_{1,\text{gauge}} = \rho_f g h \left(1 - \frac{A_2^2}{A_1^2}\right)\).

(b)(ii) [1 point]
- 1 point: For the correct expression \(v_2 = \sqrt{2gh}\).

(c)(i) [2 points]
- 1 point: For equating gauge pressure in the pipe to hydrostatic pressure of the manometer column (\(P_{1,\text{gauge}} = \rho_f g h_m\)).
- 1 point: For correctly completing the algebraic derivation to show \(h_m = h\left(1 - \frac{A_2^2}{A_1^2}\right)\).

(c)(ii) [2 points]
- 1 point: For identifying that \(h_m\) increases and providing a correct mathematical reasoning referencing the behavior of the term \(\left(1 - A_2^2/A_1^2\right)\).
- 1 point: For providing a correct physical explanation connecting the decrease in exit area to decreased speed in the wide section and consequently increased static pressure.

Part (a): Paragraph-Length Response

Student 1's claim is incorrect. Electric field is a vector quantity. At Point P (the origin), the electric field due to Particle 1 at \((-a, a)\) points down and to the right (in the direction \(+x, -y\)), while the electric field due to Particle 2 at \((-a, -a)\) points up and to the right (in the direction \(+x, +y\)). Because the charges and distances are equal, their vertical (\(y\)) components cancel completely, leaving a net electric field directed purely in the \(+x\)-direction. A third positive charge placed at \((a, a)\) produces an electric field at Point P directed down and to the left (in the direction \(-x, -y\)). Although its horizontal component opposes the initial net field, its vertical component is downward and entirely unbalanced, meaning the net field cannot be zero. To make the electric field zero at Point P, one would need a symmetric distribution of two charges of \(+Q\) at both \((a, a)\) and \((a, -a)\).

Student 2's claim is correct. Electric potential is a scalar quantity given by \(V = \sum \frac{k q_i}{r_i}\). Since Point P is at the geometric center of the square, the distance from each vertex to Point P is identical (\(r = a\sqrt{2}\)). Thus, the total electric potential is directly proportional to the algebraic sum of the charges: \(V_P = \frac{k}{r}(q_1 + q_2 + q_3) = \frac{k}{r}(+Q + Q - 2Q) = 0\). Because potential is a scalar, the spatial orientation of the charge does not prevent the potential from summing to zero.

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Part (b): Work and Potential Energy

Initial potential energy of the two charges at \(x = -d\) and \(x = +d\):
\[ U_i = \frac{k(+Q)(+Q)}{2d} = \frac{kQ^2}{2d} = 1\text{ unit} \]

**i. Placing charge \(+Q\) at \(x = 0\):**
The distance from \(x = 0\) to \(x = -d\) is \(d\), and to \(x = +d\) is \(d\).
The work done by an external force to bring the charge from infinity is equal to the interaction energy added to the system:
\[ W_1 = \Delta U_1 = \frac{k(+Q)(+Q)}{d} + \frac{k(+Q)(+Q)}{d} = \frac{2kQ^2}{d} = 4\left(\frac{kQ^2}{2d}\right) = +4\text{ units} \]
The final electric potential energy is:
\[ U_{f1} = U_i + W_1 = 1 + 4 = +5\text{ units} \]

**ii. Placing charge \(-Q\) at \(x = 0\):**
The work done by an external agent is:
\[ W_2 = \Delta U_2 = \frac{k(+Q)(-Q)}{d} + \frac{k(+Q)(-Q)}{d} = -\frac{2kQ^2}{d} = -4\left(\frac{kQ^2}{2d}\right) = -4\text{ units} \]
The final electric potential energy is:
\[ U_{f2} = U_i + W_2 = 1 + (-4) = -3\text{ units} \]

Part (a): [5 marks total]
- 1 mark: For addressing the vector nature of the electric field and correctly identifying the direction of the combined field from Particles 1 and 2 (purely along the \(+x\) axis).
- 1 mark: For explaining that the field from a positive charge at \((a, a)\) has an unresolved vertical (\(-y\)) component, thus failing to cancel the net field.
- 1 mark: For addressing the scalar nature of electric potential and noting that electric potential depends algebraically on the sum of charges when distances are equal.
- 1 mark: For confirming that Student 2 is correct by demonstrating \(V_P = \frac{k}{r}(+Q + Q - 2Q) = 0\).
- 1 mark: For a coherent, logically organized paragraph-length response that uses correct physical terminology without contradictory statements.

Part (b)(i): [3 marks total]
- 1 mark: For indicating a positive value for \(W_1\) based on electrostatic repulsion.
- 1 mark: For the correct ratio/magnitude \(W_1 = +4\) units.
- 1 mark: For applying energy conservation \(U_{f1} = U_i + W_1 = +5\) units.

Part (b)(ii): [2 marks total]
- 1 mark: For determining that \(W_2 = -4\) units (negative work due to attractive forces).
- 1 mark: For determining \(U_{f2} = -3\) units consistent with \(U_{f2} = U_i + W_2\).