| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\sum F_{\text{ext},x}=0\] | With no friction between the boat and the water, there is no external horizontal force on the combined system (boat + fisherman). Internal forces (foot-on-boat, boat-on-foot) cancel within the system. |
| 2 | \[a_{\text{CM}}=\frac{\sum F_{\text{ext},x}}{M_{\text{tot}}}=0\] | Zero net external horizontal force implies the center of mass has zero horizontal acceleration. |
| 3 | \[v_{\text{CM}}=\text{constant}\] | If the system starts at rest relative to the water/shore, then \(v_{\text{CM}}=0\) at all times. |
| 4 | \[\Delta x_{\text{CM}}=0\] | With \(v_{\text{CM}}=0\), the center-of-mass position does not change relative to the shore/water while he walks. |
| 5 | \[m_f=m_b\] | The problem states the fisherman’s mass equals the boat’s mass. |
| 6 | \[x_{\text{CM}}=\frac{m_f x_f+m_b x_b}{m_f+m_b}=\frac{x_f+x_b}{2}\] | For two equal masses, the center of mass is the average of their positions (measured relative to the shore/water). |
| 7 | \[\Delta x_{\text{CM}}=\frac{\Delta x_f+\Delta x_b}{2}=0\Rightarrow \Delta x_f=-\Delta x_b\] | Because \(\Delta x_{\text{CM}}=0\), the fisherman’s displacement relative to the water is equal in magnitude and opposite in direction to the boat’s displacement (both displacements measured in the same ground/water frame). |
| 8 | \[\Delta x_{f/b}=\Delta x_f-\Delta x_b=L\] | Let \(L>0\) be how far the fisherman walks toward the shore relative to the boat (from the back toward the front). Relative displacement equals the difference of their ground-frame displacements. |
| 9 | \[\Delta x_f=-\Delta x_b\quad\text{and}\quad\Delta x_f-\Delta x_b=L\Rightarrow 2\Delta x_f=L\Rightarrow \Delta x_f=\frac{L}{2}\] | Combine Step 7 (CM condition) with Step 8 (how far he walked relative to the boat). This shows the fisherman’s displacement relative to the shore/water is \(+L/2\), i.e., he definitely moves closer to the shore. |
| 10 | \[\Delta x_b=-\frac{L}{2}\] | From \(\Delta x_f=-\Delta x_b\), the boat moves away from shore by \(L/2\) while he walks toward shore. |
| 11 | \[\text{Evaluate choices}\] | (a) True: \(\Delta x_f=L/2>0\), so he gets closer to shore. (b) False: he does not get farther. (c) False: \(\Delta x_{\text{CM}}=0\), so CM does not move toward shore. (d) True: CM does not move. (e) True: \(\Delta x_f>0\) means he moves forward relative to the water. |
| 12 | \[\boxed{\text{Correct: (a), (d), (e)}}\] | The fisherman moves shoreward relative to the water (and thus closer to shore), the boat recoils away, and the system’s center of mass stays fixed. |
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If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
Two identical blocks are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity \( v \) to the right. The spring is then released by remote control. At some later instant, the left block is moving at \( \frac{v}{2} \) to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant?
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A boy of mass \( m \) and a girl of mass \( 2m \) are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed \( v \) across the frictionless ice surface and he slides to the right. What is the total work done by the children?
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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] | |
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] |
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