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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_i = 14.8\,\text{km s}^{-1} = 1.48 \times 10^{4}\,\text{m s}^{-1}\] | Convert the given initial speed to \(\text{m s}^{-1}\) for SI consistency. |
| 2 | \[m_i = 1\,312\,\text{kg},\quad m_f = 1\,312\,\text{kg} – 150\,\text{kg} = 1\,162\,\text{kg}\] | Initial mass \(m_i\) includes fuel; after burning \(150\,\text{kg}\) the final mass is \(m_f\). |
| 3 | \[J = F \Delta t = 156 \times 10^{3}\,\text{N} \times 220\,\text{s} = 3.432 \times 10^{7}\,\text{N s}\] | Impulse \(J\) equals constant thrust multiplied by burn time. |
| 4 | \[J = m_f v_x – m_i v_i\] | Impulse–momentum theorem: the external impulse equals the change in momentum of the probe (rocket exhaust treated as external force producing thrust). |
| 5 | \[v_x = \frac{J + m_i v_i}{m_f}\] | Re-arrange the equation in Step 4 to isolate the final velocity \(v_x\). |
| 6 | \[v_x = \frac{3.432 \times 10^{7} + (1.312 \times 10^{3})(1.48 \times 10^{4})}{1.162 \times 10^{3}}\] | Substitute \(J\), \(m_i\), \(v_i\), and \(m_f\) with their numerical values. |
| 7 | \[v_x \approx 4.625 \times 10^{4}\,\text{m s}^{-1} = 46.2\,\text{km s}^{-1}\] | Evaluate the expression; convert back to \(\text{km s}^{-1}\). |
| 8 | \[\boxed{\,46.2\,\text{km s}^{-1}\,}\] | Final speed of the space probe. |
Just ask: "Help me solve this problem."
A block of mass M1 travels horizontally with a constant speed v0 on a plateau of height H until it comes to a cliff. A toboggan of mass M2 is positioned on level ground below the cliff. The center of the toboggan is a distance D from the base of the cliff.
A student presses a \( 0.5 \) \( \text{kg} \) book against the wall. If the \( \mu_s \) between the book and the wall is \( 0.2 \), what force must the student apply to hold the book in place?
A 20 g piece of clay moving at a speed of 50 m/s strikes a 500 g pendulum bob at rest. The length of a string is 0.8 m. After the collision the clay-bob system starts to oscillate as a simple pendulum.
A box is sliding down an incline at a constant speed of \( 2 \) \( \text{m s}^{-1} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \) \( \text{N} \). What is the force of gravity acting on the box?
A 2kg object slides east at 4 m/s and collides with a stationary 3 kg object. After the collision, the 2 kg object is traveling at an unknown velocity at 15° north of east and the 3 kg object is traveling at 38° south of east. What is each object’s final velocity?
\(46.2\,\text{km s}^{-1}\)
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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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