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A uniform rod of length \(L\) and mass \(M\) is launched into the air. At time \(t = 0\), the rod’s center of mass is located at the origin \((0,0)\) of an \(xy\)-coordinate system. The rod is perfectly vertical, with a labeled point \(P\) at the top end of the rod. At this instant, the center of mass has an initial horizontal velocity \(v_{0x}\) to the right and an initial vertical velocity \(v_{0y}\) upward. The rod is rotating counterclockwise in the \(xy\)-plane with a constant angular velocity \(\omega_0\). Air resistance is negligible.

A rigid baton consists of a thin, uniform rod of length \(L\) and mass \(M\). A small, solid sphere, also of mass \(M\), is securely attached to the right end of the rod, labeled End A. The left end of the rod is labeled End B. The center of mass of the baton is located at a distance of \(L/4\) from End A, and is marked with an ‘X’, as shown in Figure 1.
In Experiment 1, the baton is tossed into the air such that it moves in a vertical plane, spinning clockwise as it travels in a parabolic trajectory. Air resistance is negligible.

A baton consists of a rigid rod of length \(L\) and negligible mass. A small sphere labeled A of mass \(m\) is attached to the left end, and a small sphere labeled B of mass \(3m\) is attached to the right end. The center of mass of the baton is marked with a red dot.
A student tosses the baton into the air. Immediately after leaving the student’s hand at time \(t = 0\), the center of mass of the baton has an initial horizontal velocity \(v_x\) and an initial upward vertical velocity \(v_{y0}\), and the baton is rotating clockwise with an initial angular velocity \(\omega_0\). Air resistance is negligible. The baton remains in the air for a time \(t_f\).

A wooden crate of mass \(m\) is projected up a rough loading ramp with an initial speed \(v_0\). The ramp is inclined at an angle \(\theta\) above the horizontal, and the coefficient of kinetic friction between the crate and the ramp is \(\mu_k\). The crate slides a distance \(d\) along the ramp before momentarily coming to rest. Which of the following correctly describes the mechanical energy of the crate-Earth system as the crate slides up the ramp?

A solid sphere of mass \(M\) and radius \(R\) is placed at rest at a vertical height \(H\) on an incline that makes an angle \(\theta\) with the horizontal. The surface of the incline has non-negligible friction, and the sphere rolls without slipping down the incline. The rotational inertia of a solid sphere about its center of mass is \(I = \dfrac{2}{5}MR^2\).

A planet of mass \(M\) has two identical satellites, Satellite A and Satellite B, each of mass \(m\). Satellite A is in a stable circular orbit of radius \(R_A\) around the planet. Satellite B is in a stable elliptical orbit around the same planet. The closest approach of Satellite B to the planet’s center (periapsis) is \(R_A\), and its farthest point (apoapsis) is \(R_B\), where \(R_B > R_A\). Figure 1 shows both satellites at the instant they are at the closest approach distance \(R_A\).

A block of mass \(m\) is pushed up a rough incline at a constant speed by an external force of magnitude \(F_{ext}\) acting parallel to the incline. The block is moved from the bottom of the incline to a final height \(h\). Which of the following correctly compares the work done by the external force, \(W_{ext}\), to the change in the gravitational potential energy of the block-Earth system, \(\Delta U_g\), and provides a correct justification?
A ball of mass \(m\) is launched vertically upward from the ground with an initial speed \(v_0\). While the ball is in motion, the air exerts a non-negligible resistive force of constant average magnitude \(F_D\). The ball reaches a maximum height \(H\) above the ground. Which of the following is a correct expression for \(H\)?

A satellite of mass \(m\) is in a stable circular orbit of radius \(R\) around a planet of mass \(M\). The satellite has an orbital angular momentum \(L\) relative to the center of the planet. The satellite’s engines are fired to transition the satellite into a new stable circular orbit of radius \(9R\). Which of the following is a correct expression for the new orbital angular momentum \(L_{new}\) and a correct justification for that expression?
Two satellites, P and Q, of equal mass \(m\) move in stable circular orbits around the same planet. Satellite P orbits at a radius \(R_{P}\) and has a translational kinetic energy \(K_{P}\). Satellite Q orbits at a radius \(R_{Q}\). Which of the following is a correct expression for the magnitude of the angular momentum \(L_{Q}\) of satellite Q?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque 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 Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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