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| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]F_f = \mu N[/katex] | The force of friction, [katex]F_f[/katex], is the product of the coefficient of friction, [katex]\mu[/katex], and the normal force, [katex]N[/katex]. |
| 2 | [katex]N = (m_1 + m_2)g[/katex] | Normal force is the combined weight of the two cars post-collision. |
| 3 | [katex]F_f = \mu (m_1 + m_2)g[/katex] | Combining formulas for friction force. |
| 4 | [katex]W = F_f d[/katex] | Work done by friction, [katex]W[/katex], equals the force of friction times the distance slid, [katex]d[/katex]. |
| 5 | [katex]\frac{1}{2}(m_1 + m_2)v^2 = \mu (m_1 + m_2)g d[/katex] | The work done by friction equals the initial kinetic energy of the system post-collision. |
| 6 | [katex]v^2 = 2 \mu g d[/katex] | Simplifying the equation for [katex]v[/katex], the combined speed post-collision. |
| 7 | [katex]v = \sqrt{2 \mu g d}[/katex] | Solving for [katex]v[/katex]. |
| 8 | [katex]m_1 v_1 = (m_1 + m_2)v[/katex] | Conservation of momentum to relate the initial velocity of the 2,000 kg car to the combined velocity post-collision. |
| 9 | [katex]v_1 = \frac{(m_1 + m_2)v}{m_1}[/katex] | Isolating [katex]v_1[/katex] for the initial velocity of the 2,000 kg car. |
| 10 | [katex]v = \sqrt{2 \cdot 0.7 \cdot 9.81 \cdot 6}[/katex] | Plugging in values for [katex]\mu = 0.7[/katex], [katex]g = 9.81,m/s^2[/katex], and [katex]d = 6,m[/katex]. |
| 11 | [katex]v \approx 9.077,m/s[/katex] | Calculating the combined speed post-collision. |
| 12 | [katex]v_1 = \frac{2000,kg}{(2000,kg + 1000,kg)} \cdot 9.077,m/s[/katex] | Use conservation of momentum. Plug in masses and the combined speed to find the initial velocity of the 2,000 kg car. |
| 13 | [katex]v_1 \approx 13.616,m/s[/katex] | Calculating the initial velocity of the 2,000 kg car. |
Just ask: "Help me solve this problem."
Which of the following graphs represent an object having zero acceleration? (There could be more than one answer)
A 0.025 kg golf ball moving at 18.0 m/s crashes through the window of a house in 5.0 × 10-4 s. After the crash, the ball continues in the same direction with a speed of 10.0 m/s.
A constant force of 8.0 N is exerted on a 16 kg object initially at rest. How much speed will the object gain after 4 seconds?
In a 4.0-kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6.0 minutes. What is the average speed of the runner? Use standard units: m/s.
An object is thrown straight upward at 64 m/s.
13.616 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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