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| Derivation/Formula | Reasoning |
|---|---|
| \[m_w v_{x,w} + m_s v_{x,s} = 0\] | Conservation of horizontal momentum; external horizontal forces are negligible during the push. |
| \[v_{x,s} = -\frac{m_w}{m_s} v_{x,w}\] | Algebraically solve for the son’s final velocity \(v_{x,s}\). |
| \[v_{x,s} = -\frac{70}{35}(0.55)\] | Substitute \(m_w = 70\,\text{kg}\), \(m_s = 35\,\text{kg}\), and \(v_{x,w} = 0.55\,\text{m/s}\). |
| \[\boxed{v_{x,s} = -1.1\,\text{m/s}}\] | The negative sign indicates motion opposite to the woman; speed is \(1.1\,\text{m/s}\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[J = m_s (v_{x,s} – v_i)\] | Impulse–momentum theorem with \(v_i = 0\). |
| \[J = 35(-1.1 – 0)\] | Insert values for the son. |
| \[|J| = 38.5\,\text{Ns}\] | Magnitude of impulse. |
| \[F_{\text{avg}} = \frac{|J|}{\Delta t}\] | Average force equals impulse divided by the interaction time \(\Delta t\). |
| \[F_{\text{avg}} = \frac{38.5}{0.60} = 64\,\text{N}\] | Compute using \(\Delta t = 0.60\,\text{s}\). |
| \[\boxed{F_{\text{avg}} = 64\,\text{N}}\] | Magnitude of the force the mother exerts on the son. |
| Derivation/Formula | Reasoning |
|---|---|
| \[F_{s \rightarrow w} = -F_{w \rightarrow s}\] | Newton’s third law: forces between two bodies are equal in magnitude and opposite in direction. |
| \[\boxed{|F_{s \rightarrow w}| = |F_{w \rightarrow s}| = 64\,\text{N}}\] | The mother experiences a \(64\,\text{N}\) force directed opposite to the force on the son. |
| Derivation/Formula | Reasoning |
|---|---|
| \[a = \mu_k g\] | Kinetic-friction force \(\mu_k m g\) divided by mass gives a deceleration independent of mass. |
| \[\Delta x = \frac{v_i^2}{2a}\] | Stopping distance for constant deceleration. |
| \[\frac{\Delta x_s}{\Delta x_w} = \left(\frac{v_{x,s}}{v_{x,w}}\right)^2\] | Because both skaters have the same \(a = \mu_k g\), we can set the ratio of distances equal to the ratio of velocities (proportional analysis using the equation in step two). |
| \[\frac{\Delta x_s}{\Delta x_w} = \left(\frac{1.1}{0.55}\right)^2 = 4\] | Insert their initial speeds. The son’s stopping distance is \(4\) times greater than his mother’s stopping distance. |
| \[\boxed{\Delta x_s = 4(7.0) = 28\,\text{m}}\] | Multiply the woman’s \(7.0\,\text{m}\) by the ratio to find the son’s stopping distance. |
Just ask: "Help me solve this problem."
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?
Two students hold a large bed sheet vertically between them. A third student, who happens to be the star pitcher on the school baseball team, throws a raw egg at the center of the sheet. Explain why the egg does not break when it hits the sheet, regardless of its initial speed.
A \(15 \, \text{g}\) marble moves to the right at \(3.5 \, \text{m/s}\) and makes an elastic head-on collision with a \(22 \, \text{g}\) marble. The final velocity of the \(22 \, \text{g}\) marble is \(2.0 \, \text{m/s}\) to the right, and the final velocity of the \(15 \, \text{g}\) marble is \(5.4 \, \text{m/s}\) to the left. What was the initial velocity of the \(22 \, \text{g}\) marble?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
A \(1200 \, \text{kg}\) car moving at \(15.6 \, \text{m/s}\) suddenly collides with a stationary car of mass \(1500 \, \text{kg}\). If the two vehicles lock together, what is their combined velocity immediately after the collision?
\(1.1\,\text{m/s}\)
\(64\,\text{N}\)
\(\text{equal magnitude, opposite direction}\)
\(28\,\text{m}\)
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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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