| Derivation / Formula | Reasoning |
|---|---|
| \[m_1 = 2\,\text{kg},\; m_2 = 3\,\text{kg},\; v_{i1}=4\,\text{m/s},\; v_{i2}=0\] | Define masses and initial velocities. Object 1 (\(m_1\)) moves east; object 2 (\(m_2\)) is stationary. |
| \[m_1 v_{x1} \cos 15^{\circ}+ m_2 v_{x2} \cos 38^{\circ}=m_1 v_{i1}\]\[m_1 v_{x1} \sin 15^{\circ}- m_2 v_{x2} \sin 38^{\circ}=0\] | Apply conservation of linear momentum in the \(x\)- (east) and \(y\)- (north) directions. Positive \(y\) is north, so the southward component is negative. |
| \[v_{x1}=\frac{m_2 \sin 38^{\circ}}{m_1 \sin 15^{\circ}}\,v_{x2}\] | Solve the \(y\)-momentum equation for \(v_{x1}\) in terms of \(v_{x2}\). |
| \[v_{x2}=\frac{m_1 v_{i1}}{m_1\left(\dfrac{m_2 \sin 38^{\circ}}{m_1 \sin 15^{\circ}}\right)\!\cos 15^{\circ}+m_2 \cos 38^{\circ}}\] | Substitute the expression for \(v_{x1}\) into the \(x\)-momentum equation and solve algebraically for \(v_{x2}\). |
| \[v_{x2}\approx 0.86\,\text{m/s}\] | Insert \(m_1=2\,\text{kg},\;m_2=3\,\text{kg},\;v_{i1}=4\,\text{m/s}\) and evaluate with \(\sin 15^{\circ},\;\cos 15^{\circ},\;\sin 38^{\circ},\;\cos 38^{\circ}.\) |
| \[v_{x1}=\frac{m_2 \sin 38^{\circ}}{m_1 \sin 15^{\circ}}\,v_{x2}\approx 3.08\,\text{m/s}\] | Use the relation from the third row and the calculated \(v_{x2}\) to find \(v_{x1}.\) |
| \[\boxed{\,v_{x1}=3.1\,\text{m/s at }15^{\circ}\text{ N of E}\,}\] | Final speed and direction for the \(2\,\text{kg}\) object. |
| \[\boxed{\,v_{x2}=0.86\,\text{m/s at }38^{\circ}\text{ S of E}\,}\] | Final speed and direction for the \(3\,\text{kg}\) object. |
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A \(0.025 \, \text{kg}\) golf ball moving at \(18.0 \, \text{m/s}\) crashes through the window of a house in \(5.0 \times 10^{-4} \, \text{s}\). After the crash, the ball continues in the same direction with a speed of \(10.0 \, \text{m/s}\).
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?
A bullet at speed \( v_0 \) trikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true?
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?
A block of mass \(M_1\) travels horizontally with a constant speed \(v_0\) on a plateau of height \(H\) until it comes to a cliff. A toboggan of mass \(M_2\) is positioned on level ground below the cliff. The center of the toboggan is a distance \(D\) from the base of the cliff.
Two blocks connected to a compressed spring move right at speed \( v \). After releasing the spring, the left block moves left at speed \( v_2 \), the right block moves right. What is the center speed of the blocks then?
An egg dropped on the road usually beaks, while one dropped on the grass usually does not break because for the egg dropped on the grass:
On a frictionless horizontal air table, puck A (with mass \( 0.249 \) \( \text{kg} \)) is moving toward puck B (with mass \( 0.375 \) \( \text{kg} \)), which is initially at rest. After the collision, puck A has velocity \( 0.115 \) \( \text{m/s} \) to the left, and puck B has velocity \( 0.645 \) \( \text{m/s} \) to the right.
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
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
\(v_{x1}=3.1\,\text{m/s at }15^{\circ}\text{ N of E}\)
\(v_{x2}=0.86\,\text{m/s at }38^{\circ}\text{ S of E}\)
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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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