| Step | Reasoning |
|---|---|
| Identify the relationship between the force on Cart A and the resulting change in its momentum. \[ \Delta p_A = F_{on A} \Delta t \] |
The question asks for the change in momentum of Cart A, which is determined by the impulse-momentum theorem \(\vec{J} = \vec{F}_{net}\Delta t = \Delta \vec{p}\). |
| Determine the magnitude and direction of the force acting on Cart A using Newton’s Third Law. \[ F_{on A} = -F_{on B} = -F \] |
The stem only provides the force Cart A exerts on Cart B (\(F\) to the right). Newton’s Third Law states that Cart B must exert an equal and opposite force on Cart A. |
| Calculate the change in momentum of Cart A. \[ \Delta p_A = (-F)\Delta t = -F\Delta t \] |
Applying the identified force to the impulse-momentum theorem for Cart A allows us to solve for its specific change. |
| Analyze the net external force on the two-cart system to determine the change in system momentum. \[ \Sigma F_{ext} = 0 \implies \Delta p_{sys} = 0 \] |
The change in momentum of a system is determined by the net external force: \(\Sigma \vec{F}_{ext} = \dfrac{\Delta \vec{p}_{sys}}{\Delta t}\). Since the track is frictionless and the collision forces are internal to the Cart A + Cart B system, the net external force is zero. |
Why each choice is correct or incorrect:
(A) This is the correct answer because the impulse on A is opposite its motion and internal forces do not change system momentum.
(B) Incorrect because the force on Cart A from Cart B acts to the left (opposite its motion), so its momentum must decrease.
(C) Incorrect because the force of the collision is internal to the two-cart system and cannot change the system’s total momentum.
(D) Incorrect because it identifies the change in momentum of Cart B (\(+F\Delta t\)) as the change in momentum of the total system.
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A stepped pulley consists of two rigidly connected concentric solid disks that rotate together on a frictionless horizontal axle. The outer disk has radius \(R\) and the inner disk has radius \(r\). The total rotational inertia of the pulley about the axle is \(I\).
A block of mass \(m_1\) hangs from a light string wrapped around the outer disk. A second block of mass \(m_2\) hangs from a light string wrapped around the inner disk. The strings do not slip. They are wrapped such that Block 1 exerts a counterclockwise torque and Block 2 exerts a clockwise torque on the pulley.
The system is released from rest, and Block 1 is observed to accelerate downward.
A student is tasked with designing an experiment to determine the local acceleration due to gravity, \(g\), using a simple pendulum. The student has access to the following equipment:
– A stable ring stand with a clamp
– A spool of lightweight string
– A set of small hooked spheres of various masses
– A meterstick
– A stopwatch
– A protractor
A projectile is launched from level ground with an initial velocity \(v_0\) at an angle \(\theta\) above the horizontal, where \(0^{\circ} < \theta < 90^{\circ}\). The projectile follows a trajectory through the air and eventually returns to the same level ground. Air resistance is negligible. Which of the following graphs best represents the speed \(v\) of the projectile as a function of time \(t\) from the moment of launch until it hits the ground?

An astronaut on a space station uses a Body Mass Measurement Device (BMMD) to determine their mass in microgravity. The device consists of a seat of mass \(m_s\) attached to a spring with an unknown spring constant \(k\). To calibrate the device, the astronaut attaches several calibration blocks of known mass \(m\) to the seat and measures the period of oscillation \(T\) for each mass. To determine the spring constant \(k\) from a linear graph, which of the following quantities should be plotted on the vertical and horizontal axes, and what is the physical significance of the slope of the resulting best-fit line?

A block of mass \(m\) is placed on a frictionless horizontal surface and attached to one end of an ideal spring with spring constant \(k\). The other end of the spring is fixed to a wall. The block is pulled to a displacement of magnitude \(A\) from its equilibrium position and released from rest, causing it to oscillate in simple harmonic motion. Which of the following is a correct expression for the speed of the block when it is at a position \(x = \dfrac{A}{2}\)?

Three blocks, labeled \(X\), \(Y\), and \(Z\), have masses \(m\), \(2m\), and \(3m\), respectively. The blocks are placed on a frictionless horizontal surface in the order \(X\)-\(Y\)-\(Z\) and are pushed to the right by a constant horizontal force of magnitude \(F\) applied to block \(X\). In this configuration, the magnitude of the contact force exerted by block \(Y\) on block \(Z\) is \(F_1\). The blocks are then rearranged on the same surface in the order \(X\)-\(Z\)-\(Y\). The same force \(F\) is applied to block \(X\), and the magnitude of the contact force exerted by block \(Z\) on block \(Y\) is \(F_2\). What is the ratio \(\dfrac{F_2}{F_1}\)?

A large, sealed cylindrical tank is filled with an ideal, incompressible fluid of density \(\rho\) to a total height \(H\). The air in the space above the fluid is maintained at a constant gauge pressure of \(P_G = 4\rho g H\). A small hole is opened in the side of the tank at a depth \(d = 0.5H\) below the fluid’s surface, and the fluid exits with speed \(v_1\). If the top of the tank were instead open to the atmosphere and the hole remained at the same depth, the fluid would exit with speed \(v_2\).
What is the ratio \(v_1 / v_2\)?

An incompressible, nonviscous fluid of density \(\rho\) flows through a horizontal pipe. At a wide section of the pipe, denoted as Point 1, the cross-sectional area is \(A_1\), the fluid speed is \(v_1\), and the pressure is \(P_1\). Further downstream, the pipe narrows to a constriction at Point 2 with a cross-sectional area \(A_2\). Which of the following is a correct expression for the fluid pressure \(P_2\) at the constriction?
A rigid, uniform disk is mounted on a low-friction axle and is initially at rest. A motor applies a net torque to the disk that increases at a constant rate, starting from zero at time \(t = 0\). Which of the following graphs best represents the angular acceleration \(\alpha\) of the disk as a function of time \(t\)?

A student measures the absolute pressure \(P\) as a function of depth \(h\) for three different incompressible liquids (Liquid 1, Liquid 2, and Liquid 3) held in open containers. The data for each liquid are shown in the graph. All three containers are located in the same laboratory.
Based on the graph, what is the ratio of the density of Liquid 3 to the density of Liquid 1, \(\dfrac{\rho_3}{\rho_1}\)?
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] | 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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