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UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) The minimum velocity needed to hit the spaceship. | ||
| 1 | Use the kinematic equation to find minimum initial velocity \( v_0 \) to reach height \( h = 1500 \, \text{m} \):
\( v^2 = v_0^2 – 2 g h \) |
At maximum height, final velocity \( v = 0 \). |
| 2 | Solve for \( v_0 \):
\( 0 = v_0^2 – 2 g h \) |
Rearranged equation to solve for \( v_0 \). |
| 3 | Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h = 1500 \, \text{m} \):
\( v_0 = \sqrt{2 \times 9.8 \times 1500} \) |
Calculated the minimum initial velocity. |
| (b) Time taken to reach maximum height when launched at 180 m/s. | ||
| 4 | Use the equation for velocity at maximum height \( v = v_0 – g t \):
At maximum height, \( v = 0 \). So: |
Set final velocity to zero to solve for time \( t \). |
| 5 | Solve for \( t \):
\( t = \dfrac{180}{9.8} \approx 18.37 \, \text{s} \) |
Calculated time to reach maximum height. |
| (c) Maximum height the rocket reaches when launched at 180 m/s. | ||
| 6 | Use the kinematic equation:
\( h_{\text{max}} = \dfrac{v_0^2}{2 g} \) |
Calculated maximum height using initial speed. |
| 7 | Compute \( h_{\text{max}} \):
\( h_{\text{max}} = \dfrac{32,400}{19.6} \approx 1,653.06 \, \text{m} \) |
Found the maximum height reached. |
| (d) Velocity and time of impact with the spacecraft. | ||
| 8 | Use vertical motion equation to find time \( t \) when \( y = 1500 \, \text{m} \):
\( y = v_0 t – \dfrac{1}{2} g t^2 \) |
Set up equation for position at impact height. |
| 9 | Rearrange into quadratic form:
\( -4.9 t^2 + 180 t – 1500 = 0 \) |
Prepared equation for quadratic formula. |
| 10 | Use quadratic formula \( t = \dfrac{-b \pm \sqrt{b^2 – 4 a c}}{2 a} \):
\( a = 4.9 \), \( b = -180 \), \( c = 1500 \) |
Calculated discriminant \( D \). |
| 11 | Solve for \( t \):
\( t = \dfrac{-(-180) \pm \sqrt{3,000}}{2 \times 4.9} \) |
Found two times when rocket is at \( 1500 \, \text{m} \). |
| 12 | Choose the most likely ascending time \( t = 12.78 \, \text{s} \) (on way up). | The rocket hits the spacecraft on its ascent. |
| 13 | Calculate velocity at impact:
\( v = v_0 – g t \) |
Determined velocity upon reaching the spacecraft. |
Just ask: "Help me solve this problem."
In which of the following is the rate of change of the particle’s momentum zero?
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A car moving at 30 m/s makes a head-on collision with a stone wall. From what height would the car have to fall in order to make an equally hard collision with the ground?
In a 4.0-kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6.0 minutes. What is the average speed of the runner? Use standard units: m/s.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?
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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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