| Derivation/Formula | Reasoning |
|---|---|
| \[K = \dfrac{p^2}{2m}\] | Kinetic energy expressed in terms of momentum \(p\) and mass \(m\). |
| \[p = \sqrt{2mK}\] | Solving the previous relation for momentum. |
| \[\dfrac{p_c}{p_b} = \sqrt{\dfrac{m_c}{m_b}}\] | Since the cardinal and baseball share the same \(K\), the \(2K\) factor cancels when taking the ratio. |
| \[\dfrac{p_c}{p_b} = \sqrt{\dfrac{3.80 \times 10^{-2}\,\text{kg}}{0.150\,\text{kg}}}\] | Substituting the given masses \(m_c = 0.038\,\text{kg}\) and \(m_b = 0.150\,\text{kg}\). |
| \[\dfrac{p_c}{p_b} = \sqrt{0.2533} \approx 0.503\] | Evaluating the square root. |
| \[\boxed{\dfrac{p_c}{p_b} \approx 0.503}\] | The cardinal’s momentum magnitude is about half that of the baseball. |
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A \(81 \, \text{kg}\) student dives off a \(45 \, \text{m}\) tall bridge with an \(18 \, \text{m}\) long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be \(2.0 \, \text{m}\) above the water?
A \(90 \, \text{kg}\) individual is cycling up a hill inclined at \(30^\circ\) on a \(12 \, \text{kg}\) bicycle. The hill is quite steep, and the coefficient of static friction is \(0.85\). The cyclist ascends \(12 \, \text{m}\) up the hill and then pauses at the summit. They then start descending from rest and travel \(9 \, \text{m}\) before firmly applying the brakes, causing the wheels to lock.
Two identical arrows, one with \( 2 \) times the speed of the other, are fired into a bale of hay. Assuming the hay exerts a constant “frictional” force on the arrows, the faster arrow will penetrate how much farther than the slower arrow?
A \(6 \, \text{kg}\) cube rests against a compressed spring with a force constant of \(1{,}800 \, \text{N/m}\), initially compressed by \(0.3 \, \text{m}\). Upon release, the cube slides on a horizontal surface with a kinetic friction coefficient of \(\mu_k = 0.12\) for \(3 \, \text{m}\), then ascends a \(12^\circ\) slope, stopping after \(4.5 \, \text{m}\). Determine the coefficient of kinetic friction on the slope.
A \(100 \, \text{kg}\) person is riding a \(10 \, \text{kg}\) bicycle up a \(25^\circ\) hill. The hill is long and the coefficient of static friction is \(0.9\). The person rides \(10 \, \text{m}\) up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of \(7 \, \text{m}\) before squeezing hard on the brakes locking the wheels, how much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is \(0.65\)?
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
A proton (mp = 1.67 x10-27 kg) is being accelerated along a straight line at 3.6 ×1015 m/s2 in a machine. The proton has an initial speed of 2.4 x107 m/s and travels 3.5 cm.
An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
Two masses \(m_1\) and \(4m_1\) are on an incline. Both surfaces have the same coefficient of kinetic friction. Both objects start from rest at the same height. Which mass has the largest speed at the bottom?
A rubber ball bounces off of a wall with an initial speed \(v\) and reverses its direction so its speed is \(v\) right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?
\(0.503\)
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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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