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UBQ Credits
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]v^2 = v’^2 + 2ad[/katex] | Kinematic equation for motion under constant acceleration, where [katex]v[/katex] is the final velocity, [katex]v'[/katex] is the initial velocity, [katex]a[/katex] is acceleration, and [katex]d[/katex] is the distance. |
| 2 | [katex]a = -\mu_k g[/katex] | Acceleration due to kinetic friction, where [katex]\mu_k[/katex] is the coefficient of kinetic friction and [katex]g[/katex] is the acceleration due to gravity (approximately [katex]9.81 , \text{m/s}^2[/katex]). Given: [katex]\mu_k = 0.40[/katex]. |
| 3 | [katex]v = \sqrt{v’^2 + 2ad}[/katex] | Solve for [katex]v[/katex]. Given: [katex]v’ = 25.43 , \text{m/s}, d = 10.0 , \text{m}[/katex]. |
| 4 | [katex]v = 23.84 , \text{m/s}[/katex] | Velocity after sliding 10.0 meters |
Distance traveled by the combined system (2 blocks and the bullet)
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]m_1v_1 + m_2v_2 = (m_1 + m_2)v'[/katex] | Conservation of momentum for the collision between the bullet-block system and the second block, where [katex]m_1[/katex] and [katex]v_1[/katex] are the mass and velocity of the bullet-block system, [katex]m_2[/katex] and [katex]v_2[/katex] are the mass and velocity of the second block, and [katex]v'[/katex] is the final velocity of the combined system. |
| 2 | [katex]v’ = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}[/katex] | Solve for [katex]v'[/katex]. Given: [katex]m_1 = 0.585 , \text{kg}, v_1 = 23.84 , \text{m/s}, m_2 = 2.50 , \text{kg}, v_2 = 0 , \text{m/s}[/katex]. |
| 3 | [katex]0 = v’^2 + 2ad[/katex] | Kinematic equation for motion under constant acceleration when the final velocity is 0. |
| 4 | [katex]d = \frac{-v’^2}{2a}[/katex] | Solve for [katex]d[/katex]. The acceleration [katex]a[/katex] remains [katex]-\mu_k g[/katex] as before. |
| 5 | [katex]v’ = 4.52 , \text{m/s}[/katex] | Final velocity of the combined system after the second collision |
| 6 | [katex]d = 2.60 , \text{m}[/katex] | Distance traveled by the combined system before stopping. |
Just ask: "Help me solve this problem."
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the magnitude of the change in the momentum of the ball during the rebound?
A big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s2. How far up does it move?
In which of the following is the rate of change of the particle’s momentum zero?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
A 3800 kg open railroad car coasts along with a constant speed of 8.60 m/s along a level track. Snow begins to fall vertically and fills the car at rate of 3.50 kg/min. Ignoring friction with the tracks, what is the speed of the car after 90 min?
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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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