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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[m_1 = 88\;\text{kg},\; m_2 = 55\;\text{kg},\; m_b = 70\;\text{kg},\; s = 3.1\;\text{m}\] | Define the masses of the two people, the boat, and the seat separation \(s\). |
| 2 | \[r_2 – r_1 = s\] | Let \(r_1\) be the seat position of the \(88\,\text{kg}\) person and \(r_2\) that of the \(55\,\text{kg}\) person, measured from the boat’s center of mass; their difference equals \(s\). |
| 3 | \[X_{\text{cm,i}} = \frac{m_1(B_i + r_1) + m_2(B_i + r_2) + m_b B_i}{m_1+m_2+m_b}\] | Write the initial horizontal center-of-mass position of the entire isolated system relative to the water; \(B_i\) is the boat’s center position initially. |
| 4 | \[X_{\text{cm,f}} = \frac{m_1(B_f + r_2) + m_2(B_f + r_1) + m_b B_f}{m_1+m_2+m_b}\] | After exchanging seats, each person occupies the other seat, so their positions swap inside the fraction. \(B_f\) is the boat’s final position. |
| 5 | \[X_{\text{cm,i}} = X_{\text{cm,f}}\] | With no external horizontal forces, the center of mass of the system remains fixed relative to the water. |
| 6 | \[M(B_i – B_f) = (m_1 – m_2)s\] | After equating Steps 3 and 4, collect terms; here \(M = m_1 + m_2 + m_b\). |
| 7 | \[B_f – B_i = -\frac{m_1 – m_2}{M}\,s\] | Solve for the boat’s displacement relative to the water; the minus sign shows direction. |
| 8 | \[|\Delta x_{\text{boat}}| = \frac{88-55}{88+55+70}\,(3.1) = 0.48\;\text{m}\] | Insert the numerical values: \(m_1 – m_2 = 33\,\text{kg}\) and \(M = 213\,\text{kg}\). |
| 9 | \[\boxed{\;\Delta x_{\text{boat}} = 0.48\,\text{m}\;}\] | Magnitude of the boat’s motion relative to the water. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[B_f – B_i < 0\] | The negative sign from Step 7 (part a) shows the boat moves opposite the positive axis defined from the \(88\,\text{kg}\) person toward the \(55\,\text{kg}\) person. |
| 2 | \[\text{Motion toward 88 kg seat}\] | Hence the boat slides in the direction of the heavier person’s original position. |
Just ask: "Help me solve this problem."
A super dart of mass \(20 \, \text{g}\), traveling at \(350 \, \text{m/s}\), strikes a steel plate at an angle of \(30^\circ\) with the plane of the plate, as shown in the figure. It bounces off the plate at the same angle but at a speed of \(320 \, \text{m/s}\). What is the magnitude of the impulse that the plate gives to the bullet?
| Experiment | Initial Velocity of Cart X \( (\text{m/s}) \) | Initial Velocity of Cart Y \( (\text{m/s}) \) | Final Velocity of Cart X \( (\text{m/s}) \) | Final Velocity of Cart Y \( (\text{m/s}) \) |
|---|---|---|---|---|
| \( 1 \) | \( 1 \) | \( 0 \) | \( 0 \) | \( 1 \) |
| \( 2 \) | \( 1 \) | \( -1 \) | \( -1 \) | \( 1 \) |
| \( 3 \) | \( 2 \) | \( 1 \) | \( 1 \) | \( 2 \) |
A student performs several experiments in which two carts collide as they travel along a horizontal surface. Cart X and Cart Y both have a mass of \( 1 \) \( \text{kg} \). Data collected from the three experiments are shown in the table above. During which experiment does the center of mass of the system of two carts have the greatest change in its momentum?
If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?
A child (\(m = 32 \, \text{kg}\)) in a boat (\(m = 71 \, \text{kg}\)) throws a \(7.1 \, \text{kg}\) package out horizontally with a speed of \(12.2 \, \text{m/s}\). Calculate the velocity of the boat immediately after, assuming it was initially at rest. Ignore water resistance.
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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] |
General Metric Conversion Chart
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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