| Step | Reasoning |
|---|---|
| Identify the relationship between the motion of the center of mass and the forces acting on the system. \[ \sum \vec{F}_{ext} = M_{sys} \vec{a}_{cm} \] |
The question asks for the velocity of the center of mass of the system after an internal interaction. |
| Determine the net external force acting on the space probe system. | To find the acceleration of the center of mass, we must identify all forces that originate from outside the system boundary. |
| Analyze the nature of the separation force. \[ \vec{F}_{separation} = \vec{F}_{int} \] |
The separation force is described as being generated by an internal mechanism within the probe. |
| Conclude that the center of mass acceleration is zero. \[ \sum \vec{F}_{ext} = 0 \implies \vec{a}_{cm} = 0 \] |
In deep space, external gravitational and resistive forces are negligible. Since the separation forces are internal, they cancel out in the summation of forces for the system. |
| Relate the zero acceleration to the velocity of the center of mass. \[ \vec{v}_{cm, f} = \vec{v}_{cm, i} = v_0 \] |
If the acceleration of the center of mass is zero, its velocity must remain constant regardless of any internal changes in the system’s configuration. |
Why each choice is correct or incorrect:
(A) The student correctly identifies the definition of center of mass velocity but fails to realize that the propulsion unit’s velocity must increase to a value that keeps the total momentum (and thus the weighted average) constant.
(B) Work is a scalar and the work done by the spring/mechanism on the two blocks does not sum to zero; more importantly, the conservation of center of mass velocity is a consequence of Newton’s Second Law and momentum, not work-energy balance.
(C) This is the correct answer.
(D) The value \(1.25 v_0\) is the new velocity of the propulsion unit module individually, found by solving \(M v_0 = (0.2M)(0) + (0.8M)v_{propulsion}\). The student has mistaken the velocity of a component for the velocity of the system’s center of mass.
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 projectile of mass \(m\) is fired horizontally with speed \(v_0\) toward a stationary block of mass \(M\) that is suspended from a rigid support by a light string of length \(L\). The projectile passes through the block and emerges from the other side traveling in the same direction with a speed of \(\alpha v_0\), where \(0 < \alpha < 1\). If the block subsequently swings upward, which of the following expressions represents the maximum height \(h\) reached by the block in terms of the given quantities and fundamental constants?

A block of mass \(m\) moves with speed \(v\) on a horizontal frictionless surface toward a stationary block of mass \(2m\). The stationary block is located at the lowest point of a vertical circular track of radius \(R\). The blocks collide and stick together, then immediately begin to move along the circular track. What is the magnitude of the normal force exerted by the track on the combined blocks immediately after the collision?

Two skaters, Skater X of mass \(M\) and Skater Y of mass \(3M\), are initially at rest on a horizontal, frictionless ice surface. The skaters push off from each other and move in opposite directions. Which of the following correctly compares the final kinetic energy \(K_X\) of Skater X and the final kinetic energy \(K_Y\) of Skater Y after they have separated?

Puck A with mass \(M\) is moving with speed \(v_0\) in the positive \(x\)-direction across a horizontal, frictionless surface. Puck B with mass \(2M\) is moving with speed \(v_0\) in the positive \(y\)-direction. The pucks collide at the origin and stick together, moving as a single object after the collision. What is the ratio of the total kinetic energy of the combined system after the collision to the total kinetic energy of the two-puck system before the collision?

An object of mass \(4M\) is at rest on a horizontal, frictionless surface. An internal explosion splits the object into two fragments of masses \(M\) and \(3M\). Immediately after the explosion, the fragment of mass \(M\) is observed to move to the left with speed \(v\). Which of the following expressions represents the total mechanical energy released during the explosion?

A launcher of total mass M, which includes a projectile of mass m, is initially at rest on a frictionless horizontal table at a height H above the floor. The launcher is positioned at the left edge of the table. The launcher fires the projectile horizontally to the right with a speed v_0 relative to the floor. The launcher recoils to the left, immediately leaves the table, and hits the floor. Which of the following is a correct expression for the horizontal distance x from the left edge of the table to the point where the launcher lands?

Two blocks of mass \(m\) and \(3m\) are held at rest on a horizontal, frictionless surface with a compressed spring of negligible mass between them. The blocks are released and move in opposite directions. Which of the following statements correctly describes the total momentum \(\vec{p}_{sys}\) and total kinetic energy \(K_{sys}\) of the two-block system after the blocks have lost contact with the spring?

A block of mass \(M\) is initially at rest on a horizontal, frictionless surface. A constant horizontal force of magnitude \(F\) is applied to the block for a time interval \(\Delta t\). Immediately after the force is removed, the block slides for an additional time interval \(t\). Which of the following expressions represents the distance \(d\) traveled by the block during the second time interval \(t\)?

Two pucks, each of mass \(M\), slide on a horizontal frictionless surface. Puck 1 moves with speed \(v_1\) in the \(+x\)-direction, and Puck 2 moves with speed \(v_2\) in the \(+y\)-direction. The pucks collide and stick together, moving with a final speed \(v_f\) at an angle \(\theta\) relative to the \(x\)-axis. If \(\tan \theta = \dfrac{3}{4}\), what is the ratio \(\dfrac{v_f}{v_1}\)?

A block of mass \(M\) is sliding in the \(+x\)-direction across a rough horizontal surface where the coefficient of kinetic friction between the block and the surface is \(\mu\). At time \(t_0\), an internal explosion splits the block into two fragments of masses \(m_1\) and \(m_2\). Immediately after the explosion, both fragments are still moving in the \(+x\)-direction. How does the magnitude of the acceleration of the center of mass of the two-fragment system, \(a_{cm}\), immediately after the explosion compare to the magnitude of the block’s acceleration, \(a_{block}\), before the explosion?
C
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?