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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( p_{x} = m_{1} v_{i} = 1000 \;\text{kg}\times20\;\text{m/s} = 20000 \;\text{kg}\cdot\text{m/s} \) | This is the momentum in the east (\(x\)) direction for the first car. |
| 2 | \( p_{y} = m_{2} v_{i} = 2000 \;\text{kg}\times15\;\text{m/s} = 30000 \;\text{kg}\cdot\text{m/s} \) | This is the momentum in the north (\(y\)) direction for the second car. |
| 3 | \( p = \sqrt{p_{x}^{2}+p_{y}^{2}} = \sqrt{20000^{2}+30000^{2}} = \sqrt{400\times10^{6}+900\times10^{6}} = \sqrt{1300\times10^{6}} \) | This computes the magnitude of the total momentum after the collision. |
| 4 | \( m_{\text{total}} = m_{1} + m_{2} = 1000 \;\text{kg} + 2000 \;\text{kg} = 3000 \;\text{kg} \) | This is the combined mass of the two cars after the collision. |
| 5 | \( v_{i} = \frac{p}{m_{\text{total}}} = \frac{\sqrt{1300\times10^{6}}}{3000} = \frac{1000\sqrt{1300}}{3000} = \frac{\sqrt{1300}}{3} \;\text{m/s} \) | This gives the speed immediately after the collision (the initial speed for the skid), where \(v_{i}\) is used consistently. |
| 6 | \( \theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}}\right) = \tan^{-1}\left(\frac{30000}{20000}\right) = \tan^{-1}(1.5) \) | This determines the direction of the combined velocity relative to east. (Numerically, \(\tan^{-1}(1.5)\) is approximately \(56.3^\circ\) north of east.) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( a = \mu_{k}g = 0.9\times9.8 = 8.82 \;\text{m/s}^{2} \) | This calculates the deceleration due to kinetic friction acting on the combined cars. |
| 2 | \( \Delta x = \frac{v_{i}^{2}}{2a} = \frac{\left(\frac{\sqrt{1300}}{3}\right)^{2}}{2\times0.9\times9.8} = \frac{\frac{1300}{9}}{17.64} \) | We use the kinematic relation for stopping distance when the final velocity is \(v_{x}=0\); here, \(\Delta x\) is the skid distance. |
| 3 | \( \Delta x = \frac{1300}{9\times17.64} \approx 8.19 \;\text{m} \) | This is the calculated skid distance after substituting the numerical values. |
| 4 | \( \boxed{\Delta x \approx 8.19 \;\text{m} \quad\text{and}\quad \theta \approx 56.3^\circ \;\text{north of east}} \) | This row presents the final answers: the cars skid approximately \(8.19\) meters in a direction approximately \(56.3^\circ\) north of east. |
Just ask: "Help me solve this problem."
Consider a ball thrown up from the surface of the earth into the air at an angle of \( 30^\circ \) above the horizontal. Air resistance is negligible. The ball’s acceleration just after release is most nearly
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
An object is thrown straight upward at 64 m/s.
On a strange, airless planet, a ball is thrown downward from a height of \( 17 \, \text{m} \). The ball initially travels at \( 15 \, \text{m/s} \). If the ball hits the ground in \( 1 \, \text{s} \), what is this planet’s gravitational acceleration?
A body starting from rest moves along a straight line under the action of a constant force. After traveling a distance \( d \) the speed of the body is \( v \). The speed of the body when it has travelled a distance \( \dfrac{d}{2} \) from its initial position is
\(\Delta x \approx 8.19 \;\text{m}\) and \(\theta \approx 56.3^\circ\;\text{north of east}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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