AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the target equation using conservation of energy.
\[ \frac{1}{2}m v^2 = q\left(V(2L) – V(3L)\right) \]
The particle is acted on only by the conservative electric force, so the gain in kinetic energy equals the loss in electric potential energy.
Derive an expression for the electric potential \(V(x)\) at an arbitrary position \(x > L\) along the axis.
\[ V(x) = \int_0^L \frac{1}{4\pi\varepsilon_0} \frac{\frac{Q}{L} dx’}{x – x’} = \frac{Q}{4\pi\varepsilon_0 L} \left[ -\ln(x – x’) \right]_0^L = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{x}{x – L}\right) \]
To calculate the potential difference, we must integrate the potential contributions \(dV\) from charge elements \(dq = \frac{Q}{L} dx’\) across the length of the rod from \(x’ = 0\) to \(x’ = L\).
Evaluate the electric potential at the initial position \(x = 2L\) and final position \(x = 3L\).
\[ V(2L) = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{2L}{2L – L}\right) = \frac{Q}{4\pi\varepsilon_0 L} \ln(2) \]
\[ V(3L) = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{3L}{3L – L}\right) = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{3}{2}\right) \]
These specific values are required to determine the potential difference experienced by the particle during its motion.
Calculate the potential difference and solve for the speed \(v\).
\[ V(2L) – V(3L) = \frac{Q}{4\pi\varepsilon_0 L} \left[ \ln(2) – \ln\left(\frac{3}{2}\right) \right] = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{2}{3/2}\right) = \frac{Q}{4\pi\varepsilon_0 L} \ln\left(\frac{4}{3}\right) \]
\[ \frac{1}{2}m v^2 = \frac{q Q}{4\pi\varepsilon_0 L} \ln\left(\frac{4}{3}\right) \implies v = \sqrt{ \frac{q Q}{2\pi\varepsilon_0 m L} \ln\left(\frac{4}{3}\right) } \]
Subtracting \(V(3L)\) from \(V(2L)\) gives the potential drop, which converts directly into kinetic energy.

Why each choice is correct or incorrect:

(A) Treats the rod as a point charge \(Q\) located at the origin, yielding \(V(x) = \frac{Q}{4\pi\varepsilon_0 x}\) and \(V(2L) – V(3L) = \frac{Q}{24\pi\varepsilon_0 L}\), which misses the continuous logarithmic potential structure of a line charge.

(B) Correctly integrates the potential and computes \(\Delta V\), but forgets the factor of 2 when isolating \(v\) from \(\frac{1}{2}mv^2\).

(C) Uses only the potential at the final position \(V(3L) = \frac{Q}{4\pi\varepsilon_0 L}\ln\left(\frac{3}{2}\right)\) instead of the potential difference \(V(2L) – V(3L)\).

(D) This is the correct answer.

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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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