| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( h \) | Height from which both objects are dropped. |
| 2 | \( m_1 = 100 \text{ pounds} \) | Mass of the rock. |
| 3 | \( m_2 = 1 \text{ pound} \) | Mass of the metal arrow. |
| 4 | \( g = 9.8 \, \text{m/s}^2 \) | Gravitational acceleration acting on both objects. |
| 5 | \( t = \sqrt{\frac{2h}{g}} \) | Formula for the time taken to fall from height \( h \) under gravity (no air resistance). This formula does not depend on mass. |
| 6 | Both the rock and the arrow hit the ground at the same time. | Since the time to fall \( t \) is independent of mass, both objects will take the same time to reach the ground when dropped from the same height \( h \), assuming no air resistance. |
The final conclusion is that both the 100-pound rock and the 1-pound metal arrow will hit the ground at the same time because the time taken to fall depends only on the height from which they are dropped and the acceleration due to gravity, not on their masses.
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There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
The motion of a particle is described in the velocity vs. time graph shown above. Over the nine-second interval shown, we can say that the speed of the particle…
A projectile has the least speed at what point in its path?
A cart starts from rest and accelerates uniformly at \(4.0 \, \text{m/s}^2\) for \(5.0 \, \text{s}\). It next maintains the velocity it has reached for \(10 \, \text{s}\). Then it slows down at a steady rate of \(2.0 \, \text{m/s}^2\) for \(4.0 \, \text{s}\). What is the final speed of the car?
On a strange, airless planet, a ball is thrown downward from a height of \( 17 \, \text{m} \). The ball initially travels at \( 15 \, \text{m/s} \). If the ball hits the ground in \( 1 \, \text{s} \), what is this planet’s gravitational acceleration?
The motions of a car and a truck along a straight road are represented by the velocity–time graphs in the figure. The two vehicles are initially alongside each other at time \(t = 0\). At time \(T\), what is true of the distances traveled by the vehicles since time \(t = 0\)?
A coin is dropped from a hot air-balloon that is \(250 \, \text{m}\) above the ground rising at \(11 \, \text{m/s}\) upwards. For the coin, assume up is positive and find the following:
Mary and Sally are in a foot race. When Mary is \( 22 \) \( \text{m} \) from the finish line, she has a speed of \( 4.0 \) \( \text{m/s} \) and is \( 5.0 \) \( \text{m} \) behind Sally, who has a speed of \( 5.0 \) \( \text{m/s} \). Sally thinks she has an easy win and, during the remaining portion of the race, decelerates at a constant rate of \( 0.40 \) \( \text{m/s}^2 \) until she reaches the finish line. What constant acceleration must Mary maintain during the remaining portion of the race if she wishes to cross the finish line side-by-side with Sally?
The graph above represents the motion of an object traveling in a straight line as a function of time. What is the average speed of the object during the first four seconds?
What does displacement mean in the context of motion?
It takes the same time since both objects experience the same acceleration \( 9.81 \, \text{m/s}^2 \).
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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] |
Metric Prefixes
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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