AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the relationship for the horizontal linear impulse exerted by the pivot.
\[ J_{\text{pivot}} = P_f – P_i \]
The impulse on the system equals its total change in linear momentum according to the impulse-momentum theorem.
Express the initial linear momentum \(P_i\) and formulate the final linear momentum \(P_f\) in terms of the post-collision angular speed \(\omega_f\).
\[ P_i = m v_0 \]
\[ P_f = M v_{\text{cm}} + m v_{\text{block}} = M (x_{\text{cm}} \omega_f) + m (L \omega_f) = (M x_{\text{cm}} + m L) \omega_f \]
Before the collision, only mass \(m\) is moving; after the collision, both the rod’s center of mass and mass \(m\) rotate with angular speed \(\omega_f\).
Use calculus to calculate the density constant \(C\), the rod’s rotational inertia \(I_{\text{rod}}\), and its center of mass position \(x_{\text{cm}}\).
\[ M = \int_0^L C x \, dx = \frac{1}{2} C L^2 \implies C = \frac{2M}{L^2} \]
\[ I_{\text{rod}} = \int_0^L x^2 \lambda(x) \, dx = \int_0^L x^2 \left(\frac{2M x}{L^2}\right) dx = \frac{2M}{L^2} \left[\frac{x^4}{4}\right]_0^L = \frac{1}{2} M L^2 \]
\[ x_{\text{cm}} = \frac{1}{M} \int_0^L x \lambda(x) \, dx = \frac{1}{M} \int_0^L x \left(\frac{2M x}{L^2}\right) dx = \frac{2}{L^2} \left[\frac{x^3}{3}\right]_0^L = \frac{2}{3} L \]
Because the density \(\lambda(x) = C x\) is non-uniform, \(I_{\text{rod}}\) and \(x_{\text{cm}}\) must be determined by integration.
Apply conservation of angular momentum about the pivot to solve for \(\omega_f\).
\[ L_i = L_f \]
\[ m v_0 L = (I_{\text{rod}} + m L^2) \omega_f = \left(\frac{1}{2} M L^2 + m L^2\right) \omega_f \]
\[ \omega_f = \frac{m v_0 L}{\left(\frac{1}{2} M + m\right) L^2} = \frac{2 m v_0}{(M + 2m) L} \]
The pivot exerts no torque about itself, so total angular momentum about the pivot is conserved during the collision.
Substitute \(x_{\text{cm}}\) and \(\omega_f\) into the expressions for \(P_f\) and \(J_{\text{pivot}}\).
\[ P_f = \left[ M \left(\frac{2}{3} L\right) + m L \right] \left[ \frac{2 m v_0}{(M + 2m) L} \right] = \left( \frac{2}{3} M + m \right) \frac{2 m v_0}{M + 2m} = \frac{\left(\frac{4}{3} M + 2m\right) m v_0}{M + 2m} \]
\[ J_{\text{pivot}} = P_f – P_i = \frac{\left(\frac{4}{3} M + 2m\right) m v_0}{M + 2m} – m v_0 = m v_0 \left( \frac{\frac{4}{3} M + 2m – (M + 2m)}{M + 2m} \right) = \frac{M m v_0}{3(M + 2m)} \]
This evaluates the final net momentum and yields the magnitude of the impulse exerted by the pivot.

Why each choice is correct or incorrect:

(A) This is the correct answer.

(B) This value results from assuming the rod has uniform mass density with \(I_{\text{rod}} = \frac{1}{3} M L^2\) and \(x_{\text{cm}} = \frac{1}{2} L\).

(C) This value results from accounting only for the rod’s linear momentum \(P_{\text{rod}} = M x_{\text{cm}} \omega_f\) and neglecting the block’s momentum \(m L \omega_f\) after collision.

(D) This value results from incorrectly combining the algebraic terms in \(P_f – P_i\), dropping a factor of 2 in the numerator difference.

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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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