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1. Maximum Height Reached by the Coin
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]v^2 = u^2 + 2as[/katex] | Kinematic equation for motion, with [katex]v[/katex] as final velocity, [katex]u[/katex] as initial velocity, [katex]a[/katex] as acceleration, and [katex]s[/katex] as displacement. |
| 2 | [katex]0 = (11, \text{m/s})^2 – 2 \times 9.81, \text{m/s}^2 \times s[/katex] | At maximum height, final velocity [katex]v = 0[/katex], initial velocity [katex]u = 11, \text{m/s}[/katex] upward, acceleration [katex]a = -9.81, \text{m/s}^2[/katex] (gravity acts downward). |
| 3 | Solve for [katex]s[/katex] | Calculate the displacement [katex]s[/katex]. |
2. Position 4.20 Seconds After Being Released
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]s = ut + \frac{1}{2}at^2[/katex] | Kinematic equation for displacement. |
| 2 | [katex]s = 11, \text{m/s} \times 4.20, \text{s} – \frac{1}{2} \times 9.81, \text{m/s}^2 \times (4.20, \text{s})^2[/katex] | Substitute values for [katex]u[/katex], [katex]a[/katex], and [katex]t[/katex] (time after release). |
3. Velocity 4.20 Seconds After Being Released
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]v = u + at[/katex] | Kinematic equation for velocity. |
| 2 | [katex]v = 11, \text{m/s} – 9.81, \text{m/s}^2 \times 4.20, \text{s}[/katex] | Substitute values for [katex]u[/katex], [katex]a[/katex], and [katex]t[/katex]. |
4. Time Before Coin Hits the Ground
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex]s = ut + \frac{1}{2}at^2[/katex] | Kinematic equation for displacement. |
| 2 | [katex]-250, \text{m} = 11, \text{m/s} \times t – \frac{1}{2} \times 9.81, \text{m/s}^2 \times t^2[/katex] | Substitute values for [katex]s[/katex] (downward displacement), [katex]u[/katex], and [katex]a[/katex]. Solve for [katex]t[/katex]. |
Let’s perform the calculations for each part.
The results for each part are as follows:
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What does displacement mean in the context of motion?
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball \( B \) passes ball \( A \):
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
A helicopter is ascending vertically with a speed of \( 5.40 \) \( \text{m/s} \). At a height of \( 105 \) \( \text{m} \) above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance and using ONLY simple kinematics (DO NOT use energy to explain this). Explain it like you would to a 5th grader and select the correct choice:
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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