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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[m_1 = 4\,\text{kg},\; m_2 = 7\,\text{kg},\; v_i = 10\,\text{m/s}\] | Identify the masses and the initial velocity \(v_i\) of the \(4\,\text{kg}\) mass. The \(7\,\text{kg}\) mass is initially at rest. |
| 2 | \[p_{x,i}=m_1 v_i = 4(10)=40,\; p_{y,i}=0\] | Calculate the initial momentum components. Motion is purely horizontal to the right, so the vertical component is zero. |
| 3 | \[v_{x2}=2\cos22^{\circ},\; v_{y2}=-2\sin22^{\circ}\] | Resolve the \(7\,\text{kg}\) mass’s given final speed (\(2\,\text{m/s}\)) into horizontal and vertical components. The vertical component is negative (below the horizontal). |
| 4 | \[v_{x2}\approx1.854,\; v_{y2}\approx-0.749\] | Numerical evaluation of the trigonometric components. |
| 5 | \[4 v_{x1}+7 v_{x2}=40\] | Apply conservation of momentum in the \(x\)-direction: total initial \(p_x\) equals total final \(p_x\). |
| 6 | \[4 v_{y1}+7 v_{y2}=0\] | Apply conservation of momentum in the \(y\)-direction: initial \(p_y\) is zero, so the final \(p_y\) must also be zero. |
| 7 | \[v_{x1}=\frac{40-7 v_{x2}}{4}\] | Solve the \(x\)-momentum equation for the unknown horizontal component \(v_{x1}\) of the \(4\,\text{kg}\) mass. |
| 8 | \[v_{x1}=\frac{40-7(1.854)}{4}\approx6.755\,\text{m/s}\] | Substitute \(v_{x2}\) and compute \(v_{x1}\). |
| 9 | \[v_{y1}=-\frac{7 v_{y2}}{4}\] | Rearrange the \(y\)-momentum equation to isolate the vertical component \(v_{y1}\) of the \(4\,\text{kg}\) mass. |
| 10 | \[v_{y1}=-\frac{7(-0.749)}{4}\approx1.311\,\text{m/s}\] | Insert \(v_{y2}\) and calculate \(v_{y1}\). The result is positive, meaning the mass moves upward after the collision. |
| 11 | \[v_x=\sqrt{v_{x1}^2+v_{y1}^2}\] | Use the Pythagorean relation to find the magnitude \(v_x\) of the final velocity of the \(4\,\text{kg}\) mass. |
| 12 | \[v_x=\sqrt{(6.755)^2+(1.311)^2}\approx6.88\,\text{m/s}\] | Compute the numerical value of the speed. |
| 13 | \[\theta=\tan^{-1}\!\left(\frac{v_{y1}}{v_{x1}}\right)\] | Determine the direction angle \(\theta\) measured above the horizontal. |
| 14 | \[\theta=\tan^{-1}\!\left(\frac{1.311}{6.755}\right)\approx11^{\circ}\] | Evaluate the inverse tangent to find the angle. |
| 15 | \[\boxed{v_x\approx6.9\,\text{m/s},\;\theta\approx11^{\circ}\,\text{above horizontal}}\] | Present the final boxed answer: the speed and its angle relative to the horizontal. |
Just ask: "Help me solve this problem."
A fisherman is standing in the back of his small fishing boat (the mass of the fisherman is the same as the mass of the boat) and he is a few meters from shore. He is done fishing so he starts walking towards the shore so he can get off the boat. What happens to the boat and the fisherman? Select all that apply and assume there is no friction between the boat and the water.
From the figure above, determine which characteristic fits this collision best.
Consider the following cases of inelastic collisions.
Case (1) – A car moving at \(75 \, \text{mph}\) collides with another car of equal mass moving at \(75 \, \text{mph}\) in the opposite direction and comes to a stop.
Case (2) A car moving at \(75 \, \text{mph}\) hits a stationary steel wall and rolls back.
The collision time is the same for both cases. In which of these cases would result in the greatest impact force?
A \(3800 \, \text{kg}\) open railroad car coasts along with a constant speed of \(8.60 \, \text{m/s}\) along a level track. Snow begins to fall vertically and fills the car at a rate of \(3.50 \, \text{kg/min}\). Ignoring friction with the tracks, what is the speed of the car after \(90 \, \text{min}\)?
Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?
\(6.9\,\text{m/s}\)
\(11^{\circ}\text{ above horizontal}\)
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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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