AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Relate the flux through the paraboloid surface to the flux through a flat circular cap using Gauss’s law.
\[ \oint \vec{E} \cdot d\vec{A} = \Phi_{\text{paraboloid}} – \Phi_{\text{disk}} = 0 \implies \Phi_{\text{paraboloid}} = \Phi_{\text{disk}} \]
Integrating electric field vectors directly over a curved paraboloid surface is complex. By considering a closed Gaussian surface formed by the open paraboloid and a flat circular disk cap across the top opening of radius \(R\), no net charge is enclosed inside this region. Therefore, the flux exiting through the paraboloid must equal the flux entering through the circular cap.
Set up the integral for the electric flux through the flat circular disk.
\[ d\Phi_{\text{disk}} = E \cos\theta \, dA = \left( \dfrac{q}{4\pi\varepsilon_0 (r’^2 + d^2)} \right) \left( \dfrac{d}{\sqrt{r’^2 + d^2}} \right) (2\pi r’ dr’) \]
A ring element of radius \(r’\) and width \(dr’\) on the disk has an area \(dA = 2\pi r’ dr’\). The distance from charge \(q\) to any point on this ring is \(r = \sqrt{r’^2 + d^2}\), and the angle \(\theta\) between the electric field and the normal to the disk satisfies \(\cos\theta = \dfrac{d}{\sqrt{r’^2 + d^2}}\).
Integrate from \(r’ = 0\) to \(r’ = R\) to determine the total flux.
\[ \Phi_{\text{disk}} = \dfrac{q d}{2\varepsilon_0} \int_{0}^{R} \dfrac{r’ \, dr’}{(r’^2 + d^2)^{3/2}} = \dfrac{q d}{2\varepsilon_0} \left[ -\dfrac{1}{\sqrt{r’^2 + d^2}} \right]_0^R = \dfrac{q}{2\varepsilon_0} \left( 1 – \dfrac{d}{\sqrt{R^2 + d^2}} \right) \]
Summing the contributions from all concentric ring elements from the center of the disk to its outer boundary \(R\) yields the total flux passing through the cap (and thus through the paraboloid).

Why each choice is correct or incorrect:

(A) Omitted the factor of \(\dfrac{1}{2}\) that arises from integrating over the solid angle subtended by the cap or evaluating the \(2\pi\) factor from the area integral against the \(4\pi\varepsilon_0\) constant.

(B) Incorrectly used only the upper-limit term \(\dfrac{d}{\sqrt{R^2+d^2}}\), forgetting to subtract it from the lower-limit evaluation at \(r’=0\) which gives \(1\).

(C) This is the correct answer.

(D) Inverted the geometric ratio for \(\cos\theta\), replacing \(\cos\theta = \dfrac{d}{\sqrt{R^2+d^2}}\) with \(\dfrac{R}{\sqrt{R^2+d^2}}\).

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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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