| Step | Reasoning |
|---|---|
| Identify the relationship between the measured period and the added mass. \[ T = 2\pi\sqrt{\dfrac{m + m_s}{k}} \] |
The question asks for a linear graph to determine \(k\), so we must start with the governing physical law for a spring-mass oscillator. |
| Linearize the equation by squaring both sides to isolate the variables. \[ T^2 = \dfrac{4\pi^2}{k}(m + m_s) \] |
The period is proportional to the square root of the mass; squaring the equation transforms it into a linear form \(y = mx + b\). |
| Distribute the terms to identify the slope and vertical intercept. \[ T^2 = \left(\dfrac{4\pi^2}{k}\right)m + \left(\dfrac{4\pi^2 m_s}{k}\right) \] |
The student varies the calibration mass \(m\), which is the independent variable (horizontal axis), while \(T^2\) is the dependent variable (vertical axis). |
| Relate the algebraic components to the graph properties. \[ y = T^2, \quad x = m, \quad \text{slope} = \dfrac{4\pi^2}{k} \] |
By comparing the linearized equation to the form \(y = mx + b\), we can determine exactly what the slope represents. |
Why each choice is correct or incorrect:
(A) The relationship between \(T\) and \(m\) is a square-root function, not a linear one; therefore, a plot of \(T\) vs. \(m\) would be a curve.
(B) This is the correct answer; plotting the square of the period against the varied mass yields a linear slope of \(4\pi^2/k\).
(C) Uses the correct axes but incorrectly calculates the slope as the reciprocal of the actual value derived from the period equation.
(D) While plotting \(m\) vs. \(T^2\) is a valid linearization, the slope for those specific axes would be \(k / 4\pi^2\), not \(4\pi^2 / k\).
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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?

Cart A with mass \(m\) is moving with velocity \(v\) to the right on a horizontal, frictionless track toward Cart B with mass \(2m\), which is initially at rest. The carts collide, and during the collision, Cart A exerts a constant average force of magnitude \(F\) on Cart B for a time interval \(\Delta t\).
Which of the following correctly identifies the change in momentum of Cart A, \(\Delta p_A\), and the change in momentum of the two-cart system, \(\Delta p_{sys}\), during this time interval?
| | \(\Delta p_A\) | \(\Delta p_{sys}\) |
|—|—|—|
| (A) | \(-F\Delta t\) | \(0\) |
| (B) | \(+F\Delta t\) | \(0\) |
| (C) | \(-F\Delta t\) | \(-F\Delta t\) |
| (D) | \(-F\Delta t\) | \(+F\Delta t\) |

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, \(\dfrac{\rho_3}{\rho_1}\)?
B
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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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