AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the relationship between water level and displaced volume.
\[ V_{\text{level}} \propto V_{\text{displaced, total}} \]
The question asks for the change in water level, which is directly proportional to the total volume of water displaced by the objects in the pool.
Determine the initial volume of water displaced when the anchor is in the boat.
\[ F_B = (m_B + m_A)g \implies \rho_W V_{D1} g = (m_B + m_A)g \implies V_{D1} = \dfrac{m_B}{\rho_W} + \dfrac{m_A}{\rho_W} \]
In the initial state, the boat and anchor are floating together as a system. According to Archimedes’ Principle, a floating object displaces a volume of water whose weight is equal to the object’s weight.
Determine the final volume of water displaced after the anchor is thrown overboard and sinks.
\[ V_{D2} = V_{\text{boat}} + V_{\text{anchor}} = \dfrac{m_B}{\rho_W} + V_A \]
In the final state, the boat is still floating (displacing water based on its weight), but the anchor has sunk. A sunken object displaces a volume of water exactly equal to its own physical volume.
Compare the two displacement volumes using the density of the anchor.
\[ \rho_A > \rho_W \implies m_A = \rho_A V_A \implies V_A = \dfrac{m_A}{\rho_A} \]
To find the change, we must compare the volume of water with a weight equal to the anchor to the anchor’s actual physical volume. Since the anchor sinks, its density is greater than water.
Final comparison of the displacements.
\[ V_{D1} = \dfrac{m_B}{\rho_W} + \dfrac{m_A}{\rho_W} \quad \text{vs.} \quad V_{D2} = \dfrac{m_B}{\rho_W} + \dfrac{m_A}{\rho_A} \]
Substitute the expression for physical volume back into the final displacement to see if it is larger or smaller than the initial displacement.
Conclude the direction of change.
\[ \text{Because } \rho_A > \rho_W, \dfrac{m_A}{\rho_A} < \dfrac{m_A}{\rho_W} \implies V_{D2} < V_{D1} \]
Since the denominator in the second term of the second state is larger (\(\rho_A > \rho_W\)), the second volume is smaller.

Why each choice is correct or incorrect:

(A) Incorrectly identifies the change as an increase; while the anchor does occupy space at the bottom, it was already displacing significantly more water (equivalent to its weight) while in the boat.

(B) Incorrectly assumes the buoyant force is always equal to the weight; when the anchor is at the bottom, the normal force from the pool floor supports part of its weight, so the water displaces less volume.

(C) This is the correct answer; it correctly identifies that a sinking object displaces less water than a floating object of the same mass.

(D) While the level does decrease, the reasoning that the boat’s reduction in displacement is ‘greater’ than the anchor’s volume is the key comparison needed, but the justification in C is more precise regarding the anchor’s specific displacement transition.

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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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