| Step | Reasoning |
|---|---|
| Identify the target expression for angular momentum and relate it to kinetic energy. \[L_{Q} = m v_{Q} R_{Q}\] |
The question asks for the angular momentum of satellite Q (\(L_{Q}\)) in terms of variables including the kinetic energy of satellite P (\(K_{P}\)). |
| Express the velocity of satellite Q in terms of its kinetic energy. \[K_{Q} = \dfrac{1}{2} m v_{Q}^{2} \implies v_{Q} = \sqrt{\dfrac{2K_{Q}}{m}}\] |
Angular momentum depends on velocity, and the choices involve kinetic energy terms. |
| Relate the kinetic energy of the two satellites using orbital mechanics principles. \[\dfrac{G M m}{r^{2}} = \dfrac{m v^{2}}{r} \implies m v^{2} = \dfrac{G M m}{r} \implies K = \dfrac{G M m}{2r}\] |
The expression must be in terms of \(K_{P}\), so we must find the functional dependence of orbital kinetic energy on the radius. |
| Determine the ratio of kinetic energies based on the inverse relationship with radius. \[K_{Q} R_{Q} = K_{P} R_{P} \implies K_{Q} = K_{P} \dfrac{R_{P}}{R_{Q}}\] |
Knowing \(K \propto 1/r\) allows us to substitute \(K_{Q}\) with an expression containing \(K_{P}\). |
| Combine the expressions to solve for \(L_{Q}\). \[\begin{align*} L_{Q} &= m R_{Q} \sqrt{\dfrac{2 K_{Q}}{m}} \\ L_{Q} &= m R_{Q} \sqrt{\dfrac{2 K_{P} R_{P}}{m R_{Q}}} \\ L_{Q} &= \sqrt{m^{2} R_{Q}^{2} \dfrac{2 K_{P} R_{P}}{m R_{Q}}} \\ L_{Q} &= \sqrt{2 m K_{P} R_{P} R_{Q}} \end{align*}\] |
Substituting the kinetic energy relationship into the angular momentum definition yields the final symbolic form. |
Why each choice is correct or incorrect:
(A) This is the correct answer.
(B) This expression simplifies to \(L_{P} = m v_{P} R_{P}\), representing the angular momentum of the first satellite, not the second.
(C) This assumes the satellites have the same kinetic energy regardless of their orbital radius, which violates the requirement that gravitational force provides the centripetal acceleration.
(D) This result stems from an algebraic error or an incorrect proportional assumption, such as thinking that kinetic energy decreases with the square of the radius.
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A car travels along a straight road between two cities separated by a total distance of \(2D\). The car travels the first distance \(D\) at a constant speed \(v_0\) and the remaining distance \(D\) at a constant speed \(3v_0\). Which of the following correctly identifies the average speed \(v_{avg}\) of the car for the entire trip and provides a valid justification?
A student conducts a laboratory experiment where a cart is moved \(2.0 \text{ m}\) to the right and then \(1.0 \text{ m}\) to the left along a straight, horizontal track. The student calculates the total distance traveled and the final displacement of the cart. Which of the following correctly classifies these quantities and provides a valid justification?
An experimental automated cart is tested on a linear track. A computer-controlled sensor measures the cart’s velocity \(v\) as a function of time \(t\), as shown in the graph. What is the displacement of the cart during the time interval from \(t = 0 \text{ s}\) to \(t = 10 \text{ s}\)?
A hiker starts at a trailhead and walks \(3.0 \text{ km}\) due North. The hiker then turns and walks \(5.0 \text{ km}\) in a direction \(37^\circ\) South of East to reach a campsite. (Note: \(\sin 37^\circ \approx 0.60\); \(\cos 37^\circ \approx 0.80\)). What is the magnitude of the hiker’s total displacement from the trailhead to the campsite?
A laboratory cart is restricted to motion along a horizontal track. A motion sensor records the direction of the cart’s velocity and the direction of its acceleration at three different times, as shown in the table below.
| Time | Direction of Velocity | Direction of Acceleration |
| :— | :— | :— |
| \(t_1\) | Right | Left |
| \(t_2\) | Left | Left |
| \(t_3\) | Left | Right |
Which of the following correctly describes the motion of the cart at each time?
A test rocket moves along a straight, horizontal track. A sensor records the rocket’s acceleration as a function of time, as shown in the graph below.
What is the average acceleration of the rocket during the time interval from \(t = 0 \text{ s}\) to \(t = 5 \text{ s}\)?
A sprinter crosses the finish line of a race moving with a velocity of \(12 \text{ m/s}\). The sprinter continues to run at this constant velocity for a reaction time of \(1.0 \text{ s}\) before beginning to slow down with a constant acceleration. If the sprinter comes to a complete stop exactly \(4.0 \text{ s}\) after crossing the finish line, what is the magnitude of the sprinter’s acceleration during the braking phase?
A surveyor starts at point \(P\) and walks a distance \(d\) due north. The surveyor then turns and walks an equal distance \(d\) in a direction \(60^{\circ}\) west of north to reach point \(Q\). What is the magnitude of the surveyor’s total displacement from point \(P\) to point \(Q\)?
A train is traveling at a constant speed \( v_0 \) when the engineer applies the brakes, resulting in a constant deceleration of magnitude \( a_0 \) that brings the train to a stop in a distance \( d \). In a second trial, the train is traveling at a speed \( 2v_0 \) when the brakes are applied and is brought to a stop with a constant deceleration of magnitude \( 2a_0 \). Which of the following is the stopping distance for the train in the second trial?
A small experimental vehicle starts from rest and accelerates with a constant acceleration \(a\) over a horizontal distance \(d\). After this distance, the vehicle’s engine is adjusted such that it continues to accelerate at a constant rate of \(\dfrac{a}{2}\) for an additional horizontal distance \(d\). Which of the following expressions represents the speed of the vehicle after it has traveled the total distance \(2d\)?
A
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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] | |
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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