| Step | Reasoning |
|---|---|
| Identify that the question asks for the maximum angle \(\theta\) reached by the system, which can be found by relating the kinetic energy immediately after the collision to the change in gravitational potential energy. \[ K_{rot} = \Delta U_g \] |
The maximum angle is reached when all rotational kinetic energy has been converted into gravitational potential energy. |
| Determine the angular velocity \(\omega\) immediately after the collision using conservation of angular momentum about the pivot point. \[ L_i = L_f \implies m v_0 L = I_{sys} \omega \] |
External forces act at the pivot during the collision, so linear momentum is not conserved, but torque about the pivot is zero, so angular momentum is conserved. |
| Calculate the rotational inertia \(I_{sys}\) of the rod-projectile system. \[ I_{sys} = I_{rod} + I_{mass} = \dfrac{1}{3}ML^2 + mL^2 = \left( \dfrac{M+3m}{3} \right)L^2 \] |
The system consists of a uniform rod pivoted at its end and a point mass at its far end. |
| Solve for the rotational kinetic energy \(K_{rot}\) immediately after the collision using the results from steps 2 and 3. \[ K_{rot} = \dfrac{L_f^2}{2I_{sys}} = \dfrac{(m v_0 L)^2}{2 \left( \frac{M+3m}{3} \right) L^2} = \dfrac{3m^2 v_0^2}{2(M+3m)} \] |
This is the total mechanical energy that will be converted into potential energy. |
| Calculate the change in gravitational potential energy \(\Delta U_g\) as a function of \(\theta\). \[ \Delta U_g = Mg \Delta y_{rod} + mg \Delta y_{mass} = Mg \frac{L}{2}(1 – \cos\theta) + mgL(1 – \cos\theta) \\ \Delta U_g = \left( \frac{M}{2} + m \right) gL (1 – \cos\theta) = \frac{(M+2m)gL}{2}(1 – \cos\theta) \] |
Both the rod and the projectile increase in height. The rod’s center of mass is at \(L/2\) and the projectile is at \(L\). |
| Equate the kinetic and potential energy and solve for \(\cos\theta\). \[ \dfrac{3m^2 v_0^2}{2(M+3m)} = \dfrac{(M+2m)gL}{2}(1 – \cos\theta) \\ 1 – \cos\theta = \dfrac{3m^2 v_0^2}{(M+3m)(M+2m)gL} \\ \cos\theta = 1 – \dfrac{3m^2 v_0^2}{(M+3m)(M+2m)gL} \] |
Conservation of energy between the state immediately after collision and the state at maximum height. |
Why each choice is correct or incorrect:
(A) This is the correct answer.
(B) This result occurs if one forgets to include the rod’s potential energy term \(Mg(L/2)(1-\cos\theta)\), only accounting for the point mass.
(C) This result occurs if one incorrectly uses linear momentum conservation (treating the system as if it were not pivoted) and then uses the resulting energy in the pendulum equation.
(D) This result occurs if the student uses the rotational inertia of the rod alone (\(1/3 ML^2\)) for the system’s kinetic energy, ignoring the mass of the embedded projectile in the inertia calculation.
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 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
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] | |
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] |
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.
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
