AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the final goal and the governing principle.
\[M g H = \dfrac{1}{2} M v_f^2 + \dfrac{1}{2} I \omega_f^2\]
The question asks for the final speed \(v_f\) of a disk undergoing a two-part descent. Since no non-conservative work is done (static friction does no work during rolling without slipping), mechanical energy is conserved.
Determine the angular velocity at the bottom of the incline.
\[\omega_f = \omega_{mid}\]
On the second (frictionless) half of the incline, there is no torque relative to the center of mass because the normal force and gravity act through the center and there is no friction. Therefore, the angular velocity remains constant from the midpoint to the bottom.
Calculate the angular velocity and translational speed at the midpoint.
\[\begin{align*} M g \left( \dfrac{H}{2} \right) &= \dfrac{1}{2} M v_{mid}^2 + \dfrac{1}{2} I \omega_{mid}^2 \\ M g \left( \dfrac{H}{2} \right) &= \dfrac{1}{2} M v_{mid}^2 + \dfrac{1}{2} \left( \dfrac{1}{2} M R^2 \right) \left( \dfrac{v_{mid}}{R} \right)^2 \\ M g \dfrac{H}{2} &= \dfrac{3}{4} M v_{mid}^2 \implies v_{mid}^2 = \dfrac{2}{3} g H \end{align*}\]
During the first half, the disk rolls without slipping, so \(v_{mid} = \omega_{mid} R\). We use energy conservation for the first half of the descent (height change of \(H/2\)).
Calculate the constant rotational kinetic energy for the final state.
\[K_{rot, f} = \dfrac{1}{2} I \omega_{mid}^2 = \dfrac{1}{4} M v_{mid}^2 = \dfrac{1}{4} M \left( \dfrac{2}{3} g H \right) = \dfrac{1}{6} M g H\]
Since \(\omega\) is constant on the second half, the rotational energy at the bottom is the same as at the midpoint.
Solve for the final translational speed.
\[\begin{align*} M g H &= \dfrac{1}{2} M v_f^2 + \dfrac{1}{6} M g H \\ \dfrac{5}{6} M g H &= \dfrac{1}{2} M v_f^2 \\ v_f^2 &= \dfrac{5}{3} g H \implies v_f = \sqrt{\dfrac{5}{3} g H} \end{align*}\]
Substitute the constant rotational energy back into the total energy conservation equation for the full descent.

Why each choice is correct or incorrect:

(A) Incorrect because it treats the object as a point mass and ignores the energy stored in rotation during the first half of the descent.

(B) Incorrect because it assumes the rolling-without-slipping constraint (v=wR) holds for the entire descent, but there is no friction to increase the angular velocity on the second half.

(C) This is the correct answer.

(D) Incorrect because it uses the moment of inertia for a thin hoop (I = MR^2) rather than a solid disk (I = 1/2 MR^2) in the energy calculations.

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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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