An original Thinka practice paper modelled on the structure and difficulty of the May 2024 AP AP Physics 2: Algebra-Based paper. Not affiliated with or reproduced from AP.
部分 II: Free-Response Questions
Answer all four questions. Questions 1 and 4 are short free-response questions worth 10 points each (approx. 20 minutes each). Questions 2 and 3 are long free-response questions worth 12 points each (approx. 25 minutes each). Show all work.
4 題目 · 44 分
題目 1 · free-response
10 分
In an experiment demonstrating wave-particle duality, a beam of electrons is accelerated from rest through a variable electric potential difference \(\Delta V\). The accelerated electrons pass through a thin crystalline foil of interatomic spacing \(d\) and strike a fluorescent screen positioned a fixed distance \(L\) behind the foil. The electrons form circular diffraction rings on the screen.
The diameter \(D\) of the first-order diffraction ring is measured for different trials. The relationship between the ring diameter and the de Broglie wavelength \(\lambda_e\) of the electrons is given by \(D \approx \frac{2L\lambda_e}{d}\).
The results of the first two trials are recorded in the table below. In Trial 3, a beam of neutral neutrons (mass \(m_n = 1.67 \times 10^{-27}\text{ kg}\)) is directed at the same foil.
(a) In a coherent, paragraph-length response, explain why increasing the accelerating potential difference \(\Delta V\) of the electron beam causes the diameter \(D\) of the diffraction ring to decrease. Justify your answer using physics principles, connecting the potential difference to the wave behavior of the electrons.
(b) Calculate the de Broglie wavelength of an electron entering the crystalline foil in Trial 1. Assume that relativistic effects are negligible.
(c) In Trial 3, the beam of neutrons produces a first-order ring of diameter \(D = 2.4\text{ cm}\), identical to the diameter in Trial 1. Indicate whether the kinetic energy of a neutron in Trial 3 is greater than, less than, or equal to the kinetic energy of an electron in Trial 1. Justify your answer.
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解題
(a) When an electron of charge \(e\) is accelerated through a potential difference \(\Delta V\), the electric field does work \(W = e\Delta V\) on the electron, converting electric potential energy into kinetic energy (\(K = \frac{1}{2}m_e v^2 = e\Delta V\)). An increase in \(\Delta V\) results in a greater kinetic energy and therefore a greater speed \(v\) and greater linear momentum \(p = m_e v = \sqrt{2m_e e\Delta V}\). According to the de Broglie relation, the wavelength associated with a moving particle is inversely proportional to its momentum (\(\lambda_e = \frac{h}{p}\)). Thus, a higher potential difference produces a shorter de Broglie wavelength. From the wave theory of diffraction (or \(D \approx \frac{2L\lambda_e}{d}\)), waves of shorter wavelength undergo less diffraction, resulting in a smaller angle of diffraction and a smaller diameter \(D\) of the diffraction ring.
(b) For an electron accelerated through \(\Delta V_1 = 150\text{ V}\): \[ K_1 = e\Delta V_1 = (1.60 \times 10^{-19}\text{ C})(150\text{ V}) = 2.40 \times 10^{-17}\text{ J} \]
The momentum of the electron is: \[ p = \sqrt{2m_e K_1} = \sqrt{2(9.11 \times 10^{-31}\text{ kg})(2.40 \times 10^{-17}\text{ J})} = 6.613 \times 10^{-24}\text{ kg}\cdot\text{m/s} \]
The de Broglie wavelength is: \[ \lambda_e = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}{6.613 \times 10^{-24}\text{ kg}\cdot\text{m/s}} = 1.00 \times 10^{-10}\text{ m} \]
(c) Less than. Justification: Since the diffraction ring diameter for the neutrons in Trial 3 is equal to that of the electrons in Trial 1 (\(D_3 = D_1 = 2.4\text{ cm}\)), the de Broglie wavelength of the neutrons must be equal to the de Broglie wavelength of the electrons (\(\lambda_n = \lambda_e\)). Because \(\lambda = \frac{h}{p}\), both particles must have the same momentum \(p\). Kinetic energy in terms of momentum is given by \(K = \frac{p^2}{2m}\). Because the mass of a neutron (\(m_n \approx 1.67 \times 10^{-27}\text{ kg}\)) is much greater than the mass of an electron (\(m_e \approx 9.11 \times 10^{-31}\text{ kg}\)), a neutron with the same momentum has significantly less kinetic energy than the electron.
評分準則
Part (a): 5 points total - 1 point: For correctly relating the accelerating potential difference \(\Delta V\) to the work done on the electron or its kinetic energy (\(K = q\Delta V\)). - 1 point: For indicating that an increase in kinetic energy or speed results in an increase in linear momentum (\(p = mv\) or \(p = \sqrt{2mK}\)). - 1 point: For using the de Broglie relation (\(\lambda = h/p\)) to show that greater momentum leads to a shorter wavelength. - 1 point: For connecting shorter de Broglie wavelength to less diffraction / smaller ring diameter (e.g., \(D \propto \lambda\)). - 1 point: For a logical, relevant, and internally consistent argument that addresses all parts of the question in paragraph form.
Part (b): 3 points total - 1 point: For writing a correct expression for the de Broglie wavelength in terms of momentum or kinetic energy (e.g., \(\lambda = h/p\) or \(\lambda = h/\sqrt{2mK}\)). - 1 point: For substituting the correct values of electron mass, charge/potential difference (or kinetic energy), and Planck's constant. - 1 point: For the correct numerical answer with units (\(1.00 \times 10^{-10}\text{ m}\) or \(0.100\text{ nm}\); accepts values within \(0.99\times 10^{-10}\) to \(1.01\times 10^{-10}\text{ m}\)).
Part (c): 2 points total - 1 point: For selecting 'Less than' with an attempt at a relevant justification. - 1 point: For a valid justification that connects equal ring diameters to equal de Broglie wavelengths (equal momenta \(p\)), and uses \(K = p^2/(2m)\) along with \(m_n > m_e\) to conclude \(K_n < K_e\).
題目 2 · Long Free-Response
12 分
A group of students is tasked with designing an experiment to investigate the properties of a parallel-plate capacitor and determine an experimental value for the vacuum permittivity \(\varepsilon_0\).
In Experiment 1, two circular parallel conducting plates, each of radius \(R = 0.10\text{ m}\), are mounted on an insulated track with an adjustable separation distance \(d\). The students have access to a capacitance meter, connecting wires, and a metric ruler/caliper.
(a) Describe an experimental procedure for collecting data that would allow the students to determine an experimental value for \(\varepsilon_0\). Provide enough detail so that another student could replicate the experiment, including any steps necessary to reduce experimental uncertainty.
(b) i. The capacitor plates are connected to a DC power supply of constant potential difference \(\Delta V\). On a set of axes with electric field magnitude \(E\) on the vertical axis and plate separation \(d\) on the horizontal axis, describe or sketch the expected relationship between \(E\) and \(d\) as the separation distance \(d\) increases.
ii. With the capacitor still connected to the constant voltage power supply, describe or sketch the expected relationship between the stored electrical energy \(U_C\) on the vertical axis and the plate separation \(d\) on the horizontal axis as \(d\) increases.
iii. Briefly justify why the curve or line described in part (b)(ii) has the shape you indicated.
In Experiment 2, the students adjust the separation \(d\) between the plates in air (assume \(\kappa \approx 1.0\)) and measure the capacitance \(C\) using the capacitance meter. The table below displays their collected data.
(c) i. Indicate what measured and/or calculated quantities could be graphed to yield a straight line that can be used to determine the experimental value of \(\varepsilon_0\). Use the blank column in the table to list any calculated quantities.
ii. Briefly explain how the graph of the quantities identified in part (c)(i) should be plotted and how a line of best fit should be drawn.
(d) Using the slope of the linearized graph, calculate an experimental value for the vacuum permittivity \(\varepsilon_0\).
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解題
### Part (a) Procedure: 1. Measure the diameter of each circular conducting plate using a metric ruler or calipers to verify the radius \(R = 0.10\text{ m}\) and calculate the plate area \(A = \pi R^2\). 2. Set the two parallel plates at an initial separation distance \(d\) using the metric ruler or caliper to ensure the plates are parallel and uniform across their surface. 3. Connect the capacitance meter leads to the two plates and record the capacitance \(C\) and the plate separation \(d\). 4. Repeat the capacitance measurement at least five different plate separations \(d\) over a wide range (e.g., from \(2.0\text{ mm}\) to \(8.0\text{ mm}\)). 5. To reduce experimental uncertainty, perform multiple trials at each separation distance, keep connecting leads stationary to minimize stray capacitance, and average the measurements.
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### Part (b) i. Electric field vs. separation: For a constant potential difference \(\Delta V\), \(E = \frac{\Delta V}{d}\). The curve is inversely proportional to \(d\) (a decreasing hyperbolic curve, concave upward, asymptotic to both axes).
ii. Stored energy vs. separation: For a constant potential difference \(\Delta V\), stored energy is \(U_C = \frac{1}{2} C (\Delta V)^2 = \frac{\varepsilon_0 A (\Delta V)^2}{2d}\). The curve is inversely proportional to \(d\) (a decreasing curve, concave upward, approaching zero as \(d\) becomes large).
iii. Justification: Because the capacitor remains connected to a constant voltage supply, the potential difference \(\Delta V\) is fixed. Capacitance is inversely proportional to separation distance (\(C \propto 1/d\)). Since stored energy is \(U_C = \frac{1}{2} C (\Delta V)^2\), energy is directly proportional to capacitance, and therefore \(U_C\) decreases inversely with \(d\).
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### Part (c) i. Linearization: From \(C = \frac{\varepsilon_0 A}{d}\): - Vertical Axis: \(C\) (in \(\text{pF}\) or \(\text{F}\)) - Horizontal Axis: \(\frac{1}{d}\) (in \(\text{mm}^{-1}\) or \(\text{m}^{-1}\))
ii. Graphing description: Plot \(C\) on the vertical axis against \(1/d\) on the horizontal axis with a linear scale occupying more than half the grid. Draw a single straight line of best fit that evenly balances points above and below the line, passing close to the origin.
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### Part (d) Slope calculation: Using two points on the best-fit line: $$\text{Slope} = \frac{\Delta C}{\Delta (1/d)} = \frac{(139 - 35) \times 10^{-12}\text{ F}}{(500 - 125)\text{ m}^{-1}} = \frac{104 \times 10^{-12}\text{ F}}{375\text{ m}^{-1}} \approx 2.77 \times 10^{-13}\text{ F}\cdot\text{m}$$
#### Part (a) [3 points] - 1 point: For specifying measurements of plate dimensions/area and separation distance \(d\) using appropriate tools (ruler/caliper). - 1 point: For indicating that capacitance \(C\) should be measured across a range of multiple distinct separation distances \(d\). - 1 point: For detailing steps to minimize experimental uncertainty (e.g., repeating trials at each separation, keeping plates parallel, minimizing stray lead effects).
#### Part (b) [3 points] - 1 point: For correctly indicating that electric field \(E\) decreases as \(1/d\) (a decreasing curve, concave up). - 1 point: For correctly indicating that internal energy \(U_C\) decreases as \(1/d\) (a decreasing curve, concave up). - 1 point: For a valid justification linking constant potential difference \(\Delta V\) to \(U_C = \frac{1}{2}C(\Delta V)^2\) and \(C \propto 1/d\).
#### Part (c) [4 points] - 1 point: For identifying appropriate variables to yield a linear relationship (e.g., \(C\) vs. \(1/d\) or \(1/C\) vs. \(d\)). - 1 point: For correctly calculating the values for the linearized quantity in the table with consistent units. - 1 point: For correctly describing linear axes scales that span more than half of the graph area and accurate data plotting. - 1 point: For describing a proper best-fit line reflecting the trend of the data points.
#### Part (d) [2 points] - 1 point: For correctly relating the slope of the line to the physical constant via the equation \(\text{Slope} = \varepsilon_0 A\) (or \(\text{Slope} = 1/(\varepsilon_0 A)\)). - 1 point: For calculating an experimental value for \(\varepsilon_0\) with appropriate units that is approximately equal to \(8.8 \times 10^{-12}\text{ F/m}\) (acceptable range: \(8.0 \times 10^{-12}\) to \(9.5 \times 10^{-12}\text{ F/m}\) depending on best-fit line choice).
題目 3 · free-response
12 分
A circuit consists of an ideal battery of electromotive force (emf) \(\mathcal{E}\) and four resistors labeled \(A\), \(B\), \(C\), and \(D\). Resistors \(A\) and \(B\) each have resistance \(R\), while Resistors \(C\) and \(D\) each have resistance \(2R\).
Resistor \(A\) is connected in series with a parallel combination consisting of two branches: • Branch 1 contains Resistor \(B\). • Branch 2 contains Resistor \(C\) and Resistor \(D\) connected in series with each other.
(a) For parts (a)(i) and (a)(ii), express your answers in terms of numerical values, \(\mathcal{E}\), and \(R\) only.
i. Derive an expression for the equivalent resistance \(R_{\text{eq}}\) of the entire circuit.
ii. Derive an expression for the electric potential difference \(\Delta V_B\) across Resistor \(B\).
(b) The vertical axis of a bar chart represents the magnitude of electric potential difference \(|\Delta V|\), with grid divisions marked from \(0\) to \(\mathcal{E}\).
Describe the relative heights of the bars representing the potential differences across Resistor \(A\), Resistor \(B\), Resistor \(C\), and Resistor \(D\) relative to the battery emf \(\mathcal{E}\).
(c) A student claims: "Because Resistors \(C\) and \(D\) each have a resistance of \(2R\), the total resistance of Branch 2 is larger than that of Branch 1 (which has resistance \(R\)). Therefore, more electrical power will be dissipated in Branch 2 than in Branch 1."
State whether the student's claim is correct or incorrect. Justify your answer by referring to the derivations from part (a) or the relationships between potential differences across the branches.
(d) Resistor \(D\) is now removed from Branch 2 and replaced with an ideal connecting wire of zero resistance, while all other components and connections remain unchanged.
Indicate whether the rate of energy dissipation (power) in Resistor \(A\) increases, decreases, or remains the same.
_____ Increases _____ Decreases _____ Remains the same
Briefly justify your answer.
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解題
**(a)(i) Derivation of Equivalent Resistance \(R_{\text{eq}}\):** 1. Find the resistance of Branch 2 containing resistors \(C\) and \(D\) in series: \[ R_{\text{branch 2}} = R_C + R_D = 2R + 2R = 4R \] 2. Find the equivalent resistance of the parallel combination of Branch 1 (\(R_B = R\)) and Branch 2 (\(4R\)): \[ \frac{1}{R_p} = \frac{1}{R_B} + \frac{1}{R_{\text{branch 2}}} = \frac{1}{R} + \frac{1}{4R} = \frac{5}{4R} \implies R_p = \frac{4}{5}R \] 3. Add the series resistor \(R_A = R\): \[ R_{\text{eq}} = R_A + R_p = R + \frac{4}{5}R = \frac{9}{5}R \]
**(a)(ii) Derivation of Potential Difference \(\Delta V_B\):** 1. The total current leaving the battery is: \[ I_{\text{total}} = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{\mathcal{E}}{\frac{9}{5}R} = \frac{5\mathcal{E}}{9R} \] 2. All total current flows through Resistor \(A\), so the potential drop across Resistor \(A\) is: \[ \Delta V_A = I_{\text{total}} R_A = \left(\frac{5\mathcal{E}}{9R}\right) R = \frac{5}{9}\mathcal{E} \] 3. By Kirchhoff's loop rule, the potential difference across the parallel network (which equals \(\Delta V_B\)) is: \[ \Delta V_B = \mathcal{E} - \Delta V_A = \mathcal{E} - \frac{5}{9}\mathcal{E} = \frac{4}{9}\mathcal{E} \]
(b) Relative Bar Heights: • Battery bar height: \(1.0\,\mathcal{E}\) (9 grid units if scaled out of 9) • Resistor \(A\) bar height: \(\frac{5}{9}\mathcal{E}\) • Resistor \(B\) bar height: \(\frac{4}{9}\mathcal{E}\) • Resistor \(C\) bar height: \(\frac{2}{9}\mathcal{E}\) (half of \(\Delta V_B\) since \(R_C = R_D = 2R\)) • Resistor \(D\) bar height: \(\frac{2}{9}\mathcal{E}\)
(c) Analysis of Student's Claim: • The claim is incorrect. • Justification: Branch 1 and Branch 2 are connected in parallel, meaning they have the identical potential difference across them: \(\Delta V_{\text{branch 1}} = \Delta V_{\text{branch 2}} = \Delta V_B = \frac{4}{9}\mathcal{E}\). • The electric power dissipated in a branch is given by \(P = \frac{(\Delta V)^2}{R_{\text{branch}}}\). • For Branch 1: \(P_1 = \frac{(\Delta V_B)^2}{R}\). • For Branch 2: \(P_2 = \frac{(\Delta V_B)^2}{4R} = \frac{1}{4}P_1\). • Therefore, Branch 2 dissipates one-fourth the power of Branch 1 because its larger resistance draws less current at the same potential difference.
**(d) Effect of Replacing Resistor \(D\) with a Wire:** • Increases • Justification: Replacing Resistor \(D\) with an ideal wire reduces the resistance of Branch 2 from \(4R\) to \(2R\). This decreases the equivalent resistance of the parallel combination from \(\frac{4}{5}R\) to \(R_p' = \frac{R(2R)}{R + 2R} = \frac{2}{3}R\), which reduces the total circuit resistance from \(\frac{9}{5}R\) to \(R_{\text{eq}}' = R + \frac{2}{3}R = \frac{5}{3}R\). • Since the battery emf \(\mathcal{E}\) is unchanged, the total current delivered to the circuit \(I_A = \frac{\mathcal{E}}{R_{\text{eq}}}\) increases. • Because the power dissipated in Resistor \(A\) is \(P_A = I_A^2 R\), the increase in current results in an increased rate of energy dissipation in Resistor \(A\).
評分準則
Part (a)(i) — 2 points • 1 point: For correctly combining resistors in series and parallel to obtain the parallel combination resistance \(R_p = \frac{4}{5}R\). • 1 point: For the correct final expression for equivalent resistance \(R_{\text{eq}} = \frac{9}{5}R\).
Part (a)(ii) — 2 points • 1 point: For calculating the total circuit current \(I = \frac{5\mathcal{E}}{9R}\) or applying an equivalent voltage-divider expression. • 1 point: For the correct expression for the potential difference \(\Delta V_B = \frac{4}{9}\mathcal{E}\).
Part (b) — 3 points • 1 point: For drawing/describing non-zero bars for \(C\) and \(D\) that are equal in height (\(\Delta V_C = \Delta V_D\)). • 1 point: For showing that the bar height for \(B\) is equal to the sum of the bar heights of \(C\) and \(D\) (\(\Delta V_B = \Delta V_C + \Delta V_D\)). • 1 point: For drawing all four bars correctly scaled relative to the battery emf (\(\Delta V_A = \frac{5}{9}\mathcal{E}\), \(\Delta V_B = \frac{4}{9}\mathcal{E}\), \(\Delta V_C = \Delta V_D = \frac{2}{9}\mathcal{E}\), such that \(\Delta V_A + \Delta V_B = \mathcal{E}\)).
Part (c) — 3 points • 1 point: For indicating that the statement is incorrect, with an attempt at a physics-based justification. • 1 point: For stating that the potential difference across Branch 1 and Branch 2 is the same because they are in parallel. • 1 point: For using an appropriate power relationship (e.g., \(P = \frac{(\Delta V)^2}{R}\) or \(P = I\Delta V\) combined with smaller current in Branch 2) to correctly explain that the branch with greater resistance dissipates less power.
Part (d) — 2 points • 1 point: For selecting "Increases" with an attempt at a relevant justification. • 1 point: For correctly explaining that replacing resistor \(D\) lowers the total equivalent circuit resistance, thereby increasing the total current flowing through Resistor \(A\) (or increasing the potential difference across Resistor \(A\)), which increases the power dissipated in \(A\).
題目 4 · Short Answer / Particle Dynamics
10 分
Two positively charged ions, Ion A and Ion B, are accelerated from rest across an electric potential difference of magnitude \(\Delta V_0\).
• Ion A has mass \(m_A\) and charge \(+q_0\). • Ion B has mass \(3m_A\) and charge \(+2q_0\).
(a) Calculate the ratio of the magnitude of the momentum of Ion B to the magnitude of the momentum of Ion A, \(\frac{p_B}{p_A}\), immediately after exiting the potential difference.
After exiting the accelerating region, both ions enter a uniform magnetic field of magnitude \(B_0\) that is directed perpendicular to their velocity vectors.
(b) i. Derive an expression for the radius of curvature \(R_A\) of the circular trajectory of Ion A in the magnetic field in terms of \(m_A, q_0, \Delta V_0, B_0\), and physical constants, as appropriate. ii. Calculate the numerical value of the ratio of the radius of curvature of Ion B to that of Ion A, \(\frac{R_B}{R_A}\).
(c) Suppose the ions enter the magnetic field moving in the \(+x\)-direction, and the magnetic field is directed into the page (in the \(-z\)-direction). i. Indicate whether the initial magnetic deflection of Ion A and Ion B is in the \(+y\)-direction or the \(-y\)-direction. Justify your answer using physics principles. ii. A uniform electric field is applied in the region containing the magnetic field so that Ion A travels in a straight line along the \(+x\)-axis at constant speed. Determine the direction of the required electric field and derive an expression for its magnitude \(E\) in terms of \(m_A, q_0, \Delta V_0, B_0\), and physical constants.
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解題
(a) Work done by the electric field equals the kinetic energy gained from rest: \(K = q\Delta V_0\) For Ion A: \(K_A = q_0 \Delta V_0\) For Ion B: \(K_B = (2q_0)\Delta V_0 = 2q_0 \Delta V_0\) Using the relationship between kinetic energy and momentum \(p = \sqrt{2mK}\): \(p_A = \sqrt{2m_A(q_0 \Delta V_0)} = \sqrt{2m_A q_0 \Delta V_0}\) \(p_B = \sqrt{2(3m_A)(2q_0 \Delta V_0)} = \sqrt{12m_A q_0 \Delta V_0} = \sqrt{6}\sqrt{2m_A q_0 \Delta V_0}\) Therefore, the ratio is: \(\frac{p_B}{p_A} = \frac{\sqrt{12m_A q_0 \Delta V_0}}{\sqrt{2m_A q_0 \Delta V_0}} = \sqrt{6}\approx 2.45\)
(b) i. Inside the magnetic field, the magnetic force acts as the centripetal force: \(F_{\text{net}} = F_M \implies \frac{m_A v_A^2}{R_A} = q_0 v_A B_0\) \(R_A = \frac{m_A v_A}{q_0 B_0} = \frac{p_A}{q_0 B_0}\) Substituting \(v_A = \sqrt{\frac{2q_0 \Delta V_0}{m_A}}\) or \(p_A = \sqrt{2m_A q_0 \Delta V_0}\): \(R_A = \frac{\sqrt{2m_A q_0 \Delta V_0}}{q_0 B_0} = \frac{1}{B_0}\sqrt{\frac{2m_A \Delta V_0}{q_0}}\)
ii. For any charged particle in a magnetic field, \(R = \frac{p}{qB_0}\). \(\frac{R_B}{R_A} = \frac{p_B / (q_B B_0)}{p_A / (q_A B_0)} = \left(\frac{p_B}{p_A}\right)\left(\frac{q_A}{q_B}\right)\) Substituting the known ratios \(\frac{p_B}{p_A} = \sqrt{6}\) and \(\frac{q_A}{q_B} = \frac{q_0}{2q_0} = \frac{1}{2}\): \(\frac{R_B}{R_A} = \frac{\sqrt{6}}{2} = \sqrt{\frac{3}{2}} \approx 1.22\)
(c) i. The magnetic force on a moving charge is \(\vec{F}_M = q(\vec{v}\times\vec{B})\). With \(\vec{v}\) in the \(+x\)-direction and \(\vec{B}\) into the page (\(-z\)-direction), the right-hand rule gives a cross product \(\hat{i}\times(-\hat{k}) = +\hat{j}\) (toward the \(+y\)-direction). Since both ions carry positive charges (\(q > 0\)), the magnetic force points in the \(+y\)-direction, deflecting both particles toward the \(+y\)-direction.
ii. For Ion A to move in a straight line at constant velocity, the net force must be zero: \(\vec{F}_{\text{net}} = \vec{F}_E + \vec{F}_M = 0 \implies \vec{F}_E = -\vec{F}_M\) Since \(\vec{F}_M\) points in the \(+y\)-direction, the electric force \(\vec{F}_E = q_0\vec{E}\) must point in the \(-y\)-direction. Because \(q_0 > 0\), the electric field \(\vec{E}\) must be directed in the \(-y\)-direction. Balancing magnitudes: \(q_0 E = q_0 v_A B_0 \implies E = v_A B_0\). Substituting \(v_A = \sqrt{\frac{2q_0 \Delta V_0}{m_A}}\): \(E = B_0\sqrt{\frac{2q_0 \Delta V_0}{m_A}}\)
評分準則
Part (a): 2 points • 1 point: For correctly relating the kinetic energy acquired by a particle to the potential difference (\(K = q\Delta V\)) and expressing momentum in terms of kinetic energy or potential difference (e.g., \(p = \sqrt{2mq\Delta V}\)). • 1 point: For a correct calculation of the ratio \(\frac{p_B}{p_A} = \sqrt{6}\) (or \(\approx 2.45\)).
Part (b)(i): 3 points • 1 point: For setting the magnetic force equal to the centripetal force using Newton's second law (\(q_0 v_A B_0 = \frac{m_A v_A^2}{R_A}\) or \(R_A = \frac{p_A}{q_0 B_0}\)). • 1 point: For correctly determining the speed \(v_A = \sqrt{\frac{2q_0 \Delta V_0}{m_A}}\) or momentum \(p_A = \sqrt{2m_A q_0 \Delta V_0}\). • 1 point: For a correct final expression for \(R_A\) algebraically simplified in terms of given variables.
Part (b)(ii): 2 points • 1 point: For correctly applying the functional dependence \(R \propto \frac{p}{q}\) or substituting the mass, charge, and velocity of Ion B into the radius expression. • 1 point: For a correct numerical or simplified radical ratio \(\frac{R_B}{R_A} = \frac{\sqrt{6}}{2} = \sqrt{1.5} \approx 1.22\).
Part (c)(i): 2 points • 1 point: For stating that the deflection is in the \(+y\)-direction. • 1 point: For a valid justification referencing the right-hand rule for the cross product \(\vec{v}\times\vec{B}\) and the positive sign of the charge.
Part (c)(ii): 1 point • 1 point: For indicating the electric field is in the \(-y\)-direction AND providing the correct expression \(E = B_0\sqrt{\frac{2q_0 \Delta V_0}{m_A}}\) derived from balancing electric and magnetic forces.