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UBQ Credits
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_{c}=32\;\text{kg},\;m_{b}=71\;\text{kg},\;m_{p}=7.1\;\text{kg}\] | Identify the masses of the child, boat, and package. |
| 2 | \[v_{i}=0\] | The child–boat–package system is initially at rest. |
| 3 | \[p_{i}=0\] | Initial linear momentum of the entire system is zero because \(v_{i}=0\). |
| 4 | \[p_{f}=m_{p}v_{p}+\left(m_{c}+m_{b}\right)v_{x}\] | Final momentum equals the vector sum of momentum of the package and momentum of the combined child+boat, where \(v_{p}=12.2\,\text{m/s}\) is the package speed and \(v_{x}\) is the recoil speed of the child+boat. |
| 5 | \[p_{i}=p_{f}\] | Apply conservation of linear momentum; no external horizontal forces act. |
| 6 | \[0=m_{p}v_{p}+\left(m_{c}+m_{b}\right)v_{x}\] | Set initial momentum equal to final momentum. |
| 7 | \[v_{x}=-\frac{m_{p}v_{p}}{m_{c}+m_{b}}\] | Algebraically solve for the unknown recoil velocity \(v_{x}\). |
| 8 | \[v_{x}=-\frac{\left(7.1\,\text{kg}\right)\left(12.2\,\text{m/s}\right)}{32\,\text{kg}+71\,\text{kg}}\] | Insert the numerical values for the masses and package speed. |
| 9 | \[v_{x}\approx-0.84\,\text{m/s}\] | Compute the magnitude; the negative sign indicates motion opposite the thrown package. |
| 10 | \[\boxed{\,v_{x}\approx0.84\,\text{m/s (opposite direction)}\,}\] | State the final speed of the child–boat system, specifying it moves opposite to the package’s motion. |
Just ask: "Help me solve this problem."
If you want to double the momentum of a gas molecule, by what factor must you increase its kinetic energy?
A \(2 \, \text{kg}\) object slides east at \(4 \, \text{m/s}\) and collides with a stationary \(3 \, \text{kg}\) object. After the collision, the \(2 \, \text{kg}\) object is traveling at an unknown velocity at \(15^\circ\) north of east and the \(3 \, \text{kg}\) object is traveling at \(38^\circ\) south of east. What is each object’s final velocity?
Two ice skaters suddenly push off against one another starting from a stationary position. The 45 kg skater acquires a speed of 0.375 m/s relative to the ice. What speed does the 60 kg skater acquire relative to the ice?
A bowling ball moving with speed v collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
| 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?
\(0.84\,\text{m/s}\)
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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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