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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[p_{\text{initial}} = 0\] | Astronaut and wrench are initially at rest relative to the mothership, so the total momentum is \(0\). |
| 2 | \[m_w v_w + m_J v_J = 0\] | Apply conservation of linear momentum after the wrench is thrown. Here \(m_w\) and \(v_w\) are the wrench’s mass and velocity, while \(m_J\) and \(v_J\) are Jennifer’s mass and velocity. |
| 3 | \[v_J = -\frac{m_w v_w}{m_J}\] | Algebraically solve for Jennifer’s velocity \(v_J\). The negative sign indicates she moves toward the ship (opposite the wrench). |
| 4 | \[v_J = -\frac{(2.00\,\text{kg})(20\,\text{m/s})}{200\,\text{kg}}\] | Insert \(m_w = 2.00\,\text{kg}\), \(v_w = 20\,\text{m/s}\), and \(m_J = 200\,\text{kg}\). |
| 5 | \[v_J = -0.20\,\text{m/s}\] | Calculate Jennifer’s speed. The magnitude is \(0.20\,\text{m/s}\) toward the ship. |
| 6 | \[\Delta x = v t\] | Use uniform motion because no external forces act horizontally. Here \(\Delta x = 100\,\text{m}\) and \(v = 0.20\,\text{m/s}\). |
| 7 | \[t = \frac{\Delta x}{|v|} = \frac{100\,\text{m}}{0.20\,\text{m/s}}\] | Solve for the time \(t\), using the magnitude of Jennifer’s velocity. |
| 8 | \[t = 500\,\text{s}\] | Compute the time required for Jennifer to reach the ship. |
| 9 | \[\boxed{500\,\text{s}}\] | Final boxed answer. |
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A bowling ball moving with speed \(v\) collides head-on with a stationary tennis ball. The collision is elastic and there is no friction. The bowling ball barely slows down. What is the speed of the tennis ball after the collision?
A boy of mass \( m \) and a girl of mass \( 2m \) are initially at rest at the center of a frozen pond. They push each other so that she slides to the left at speed \( v \) across the frictionless ice surface and he slides to the right. What is the total work done by the children?
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at \( 18 \) \( \text{m/s} \). The third piece has \( 2.5 \) times the mass as the other two.
A block with mass \( m \) slides at speed \( v_0 \) on a smooth surface and hits a stationary block with mass \( M \). They stick together and move at speed \( \frac{v_0}{3} \). Find \( M \) in terms of \( m \).
A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed of the boat
A mass \( m_1 \) traveling with an initial velocity \( v \) has an elastic collision with a mass \( m_2 \) that is initially at rest.
A \(70 \, \text{kg}\) woman and her \(35 \, \text{kg}\) son are standing at rest on an ice rink. They push against each other for a time of \(0.60 \, \text{s}\), causing them to glide apart. The speed of the woman immediately after they separate is \(0.55 \, \text{m/s}\). Assume that during the push, friction is negligible compared with the forces the people exert on each other.
A golf club exerts an average horizontal force of \(1000 \, \text{N}\) on a \(0.045 \, \text{kg}\) golf ball that is initially at rest on the tee. The club is in contact with the ball for \(1.8 \, \text{milliseconds}\). What is the speed of the golf ball just as it leaves the tee?
From the figure above, determine which characteristic fits this collision best.
You are lying in bed and want to shut your bedroom door. You have a bouncy “superball” and a blob of clay, both with the same mass \( m \). Which one would be more effective to throw at your door to close it?
\(500\,\text{s}\)
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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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