| 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. |
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The displacement \( x \) of an object moving in one dimension is shown above as a function of time \( t \). The acceleration of this object must be
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Can an object have \( 0 \) velocity and nonzero acceleration at the same time? Give two examples.
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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