AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the conserved quantities to determine the final state of the system.
\[ L_i = L_f \]
The question asks for a ratio of kinetic energies, which requires finding the final angular velocity of the system. In this collision, the pivot exerts an external force on the rod, so linear momentum is not conserved. However, the torque produced by the pivot force relative to the pivot is zero, meaning angular momentum about the pivot is conserved.
Calculate the initial and final angular momentum relative to the pivot.
\[ L_i = m v_0 L \]
\[ L_f = I_{sys} ̉\omega_f = \left( I_{rod} + I_{clay} \right) ̉\omega_f \]
\[ L_f = \left( \dfrac{1}{3}ML^2 + mL^2 \right) ̉\omega_f = L^2 \left( \dfrac{M}{3} + m \right) ̉\omega_f \]
To solve for the final angular velocity, we need expressions for the system’s angular momentum before and after the collision.
Solve for the final angular velocity.
\[ m v_0 L = L^2 \left( \dfrac{M + 3m}{3} \right) ̉\omega_f \]
\[ ̉\omega_f = \dfrac{3 m v_0}{L(M + 3m)} \]
The final angular velocity is required to calculate the final rotational kinetic energy.
Calculate the ratio of the final kinetic energy to the initial kinetic energy.
\[ K_i = \dfrac{1}{2}m v_0^2 \]
\[ K_f = \dfrac{1}{2} I_{sys} ̉\omega_f^2 = \dfrac{1}{2} \left[ \dfrac{L^2(M + 3m)}{3} \right] \left[ \dfrac{3 m v_0}{L(M + 3m)} \right]^2 \]
\[ K_f = \dfrac{1}{2} \left[ \dfrac{L^2(M + 3m)}{3} \right] \left[ \dfrac{9 m^2 v_0^2}{L^2(M + 3m)^2} \right] = \dfrac{3 m^2 v_0^2}{2(M + 3m)} \]
\[ \dfrac{K_f}{K_i} = \dfrac{\dfrac{3 m^2 v_0^2}{2(M + 3m)}}{\dfrac{1}{2}m v_0^2} = \dfrac{3m}{M + 3m} \]
The final kinetic energy of the system is purely rotational, while the initial kinetic energy is the translational energy of the clay.

Why each choice is correct or incorrect:

(A) Treats the collision as a 1D linear inelastic collision using linear momentum conservation, ignoring the rotational constraints of the pivot.

(B) This is the correct answer.

(C) Matches the denominator from the correct derivation but omits the factor of 3 in the numerator resulting from the kinetic energy of a rotating body.

(D) Uses the moment of inertia for a rod rotating about its center (1/12 ML^2) rather than its end (1/3 ML^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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