| Step | Reasoning |
|---|---|
| Identify the relationship between the speed-time graph and the physical scenario. \[ a = \dfrac{\Delta v}{\Delta t} \] |
The question asks for a speed-time graph, and the slope of a speed-time graph represents the magnitude of the translational acceleration. |
| Determine the acceleration of the sliding block. \[ a_{\text{block}} = g \sin\theta \] |
We need the acceleration to determine the slope of the block’s graph. For a block on a frictionless ramp, the only force along the ramp is the component of gravity. |
| Determine the acceleration of the rolling sphere. \[ a_{\text{sphere}} = \dfrac{g \sin\theta}{1 + \dfrac{I}{MR^2}} \] |
We need the acceleration to compare it to the block’s acceleration. For an object rolling without slipping, some of the gravitational potential energy is converted into rotational kinetic energy, or equivalently, friction provides a torque that reduces the net force for translational acceleration. |
| Compare the two accelerations. \[ a_{\text{block}} = g \sin\theta \quad \text{and} \quad a_{\text{sphere}} = \dfrac{g \sin\theta}{1 + 2/5} = \dfrac{5}{7} g \sin\theta \] |
The relative magnitudes of the accelerations will determine which graph has the steeper slope. Since \(I > 0\) for a solid sphere, the denominator for the sphere’s acceleration is greater than 1. |
| Select the graph matching these characteristics. \[ a_{\text{block}} > a_{\text{sphere}} \] |
Both accelerations are constant (determined by constant mass, geometry, and gravity), meaning both graphs must be linear. Since the block’s acceleration is greater, its slope must be steeper. |
Why each choice is correct or incorrect:
(A) This is the correct answer because both objects have constant acceleration, and the sliding block has a higher acceleration than the rolling sphere.
(B) Incorrectly suggests that rolling increases translational acceleration; however, rotational inertia resists changes in motion and decreases translational acceleration compared to sliding.
(C) Incorrectly suggests that the acceleration is independent of whether the object rotates; for a rolling object, the torque from friction reduces the net translational force.
(D) Incorrectly suggests the sphere’s acceleration is non-constant; because the ramp angle and the sphere’s rotational inertia are constant, the acceleration remains constant, resulting in a linear velocity graph.
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.

A uniform rigid rod of mass \(M\) and length \(L\) is suspended vertically from a frictionless pivot at its top end. The rotational inertia of the rod about the pivot is \(I = \dfrac{1}{3}ML^2\). A small sphere of mass \(m\) is launched horizontally with initial speed \(v_0\) toward the rod. The sphere strikes the very bottom of the rod and sticks to it. The collision occurs so quickly that the rod does not noticeably move from its vertical position until after the collision is over. The sphere can be treated as a point mass.

A uniform disk of mass \(M\) and radius \(R\) is mounted on a horizontal, frictionless axle that passes through a point \(P\) located a distance \(\dfrac{R}{2}\) from the center of the disk. The disk is held in a position such that the center of the disk and the axle are at the same horizontal level, as shown in the figure. At the moment the disk is released from rest, which of the following correctly describes the torque about the axle and the magnitude of the upward vertical force \(F_{axle}\) exerted by the axle on the disk?

A small ball of mass \(m\) slides across a horizontal, frictionless surface with a velocity \(v\) toward a uniform rod of mass \(M\) and length \(L\). The rod is initially at rest and is free to rotate about a frictionless pivot at one of its ends. The ball strikes the rod at its midpoint and sticks to it. Which of the following correctly states whether the angular momentum of the ball-rod system about the pivot is conserved during the collision and provides a correct justification?

A figure skater with an initial rotational inertia \(I_0\) is spinning on ice with an initial angular velocity \(\omega_0\). Friction between the skater’s blades and the ice is negligible. The skater pulls their arms and one leg inward toward their axis of rotation, which decreases their rotational inertia to a final value \(I_f\). Let \(K_0\) represent the initial rotational kinetic energy of the skater. Which of the following expressions correctly represents the skater’s final rotational kinetic energy \(K_f\)?

Two uniform solid spheres, Sphere 1 and Sphere 2, are released from rest at the top of an incline that makes an angle \(\theta\) with the horizontal. Sphere 1 has mass \(M_1\) and Sphere 2 has mass \(M_2\), where \(M_1 = 2M_2\). Both spheres have the same radius \(R\) and roll down the incline without slipping. What is the ratio \(a_1/a_2\) of the linear acceleration of the center of mass of Sphere 1 to that of Sphere 2?

An engineer is testing two different designs for a rotating flywheel. Design A consists of a uniform solid disk with mass \(M\) and radius \(R\) mounted on a low-friction axle. When a motor applies a constant net torque \(\tau\) to Design A, it experiences an angular acceleration \(\alpha_0\). Design B consists of a uniform solid disk with mass \(2M\) and radius \(2R\). If the motor is adjusted to apply a constant net torque of \(4\tau\) to Design B, what is the resulting angular acceleration of Design B in terms of \(\alpha_0\)?

A spring of ideal spring constant \(k\) hangs vertically from a ceiling. When the spring is unextended, its bottom end is at position \(y = 0\). The positive \(y\)-direction is defined as downward. A block of mass \(m\) is attached to the spring and gently lowered until it hangs at rest at its equilibrium position \(y_{eq}\). The block is then pulled down an additional distance \(A\) to a maximum position \(y_{max} = y_{eq} + A\) and released from rest.
A block of mass \(M\) is attached to an ideal spring and undergoes simple harmonic motion on a frictionless horizontal surface. The equilibrium position of the block is at \(x = 0\). Which of the following graphs best represents the acceleration \(a\) of the block as a function of its displacement \(x\)?

A block of mass \(M\) is attached to the lower end of a vertical spring with spring constant \(k\), while the upper end of the spring is fixed to a ceiling. A second block of mass \(m\) is placed on top of the first block. The two-block system is set into vertical simple harmonic motion with amplitude \(A\). At the instant the blocks pass through the equilibrium position while moving downward, which of the following correctly describes the magnitude of the force exerted by the spring on the bottom block, \(F_{spring}\), and the magnitude of the normal force exerted by the bottom block on the top block, \(F_{normal}\)?

A block of mass \(m\) is attached to an ideal horizontal spring with spring constant \(k\). The system oscillates on a frictionless surface with amplitude \(A\). Which of the following expressions represents the kinetic energy of the block when its displacement from the equilibrium position is \(x = \dfrac{1}{3}A\)?
A
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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 |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
NEW! PHY AI accurately solves all questions
🔥 Get up to 30% off Elite Physics Tutoring
🧠 NEW! Learn Physics From Scratch Self Paced Course
🎯 Need exam style practice questions?