AP Physics

Unit 1 - Vectors and Kinematics

MCQ
Mathematical
Intermediate

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Step Reasoning
Identify the goal and determine the equivalent capacitance of the parallel sub-combination (C_2 and C_3).
\[ C_{23} = C_2 + C_3 = C + C = 2C \]
To find the potential difference V_1 across C_1, we first need to simplify the branch containing C_1, C_2, and C_3 into two series components.
Determine the potential difference V_1 across capacitor C_1.
\[ V_1 = \left( \dfrac{C_{23}}{C_1 + C_{23}} \right) V_0 = \left( \dfrac{2C}{2C + 2C} \right) V_0 = \dfrac{1}{2} V_0 \]
Capacitor C_1 (capacitance 2C) is in series with the combination C_23 (equivalent capacitance 2C) across the battery potential difference V_0.
Determine the potential difference V_2 across capacitor C_2.
\[ V_2 = V_0 – V_1 = V_0 – \dfrac{1}{2} V_0 = \dfrac{1}{2} V_0 \]
The remaining potential difference across the parallel combination C_23 is V_0 – V_1.
Calculate the energy U_2 stored in capacitor C_2 and the total energy U_total stored in all capacitors.
\[ U_2 = \dfrac{1}{2} C_2 V_2^2 = \dfrac{1}{2} C \left(\dfrac{1}{2} V_0\right)^2 = \dfrac{1}{8} C V_0^2 \]
\[ C_{\text{upper}} = \dfrac{C_1 C_{23}}{C_1 + C_{23}} = \dfrac{(2C)(2C)}{2C + 2C} = C \]
\[ C_{\text{total}} = C_{\text{upper}} + C_4 = C + C = 2C \]
\[ U_{\text{total}} = \dfrac{1}{2} C_{\text{total}} V_0^2 = \dfrac{1}{2} (2C) V_0^2 = C V_0^2 \]
The energy ratio requires expressing U_2 and U_total in terms of C and V_0.
Compute the ratio of stored energies U_2 / U_total.
\[ \dfrac{U_2}{U_{\text{total}}} = \dfrac{\dfrac{1}{8} C V_0^2}{C V_0^2} = \dfrac{1}{8} \]
Dividing U_2 by U_total yields the required ratio.

Why each choice is correct or incorrect:

(A) This is the correct answer.

(B) Calculates the ratio relative to the upper branch energy U_upper = (1/2) C V_0^2, getting (1/8)/(1/2) = 1/4, rather than relative to total energy in all four capacitors.

(C) Incorrectly treats C_1 as being in series with a single capacitor C rather than the parallel combination C_23 = 2C, leading to V_1 = (2/3)V_0.

(D) Combines the voltage division error with an incorrect energy accounting.

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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