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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."
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
A tennis ball is thrown straight up with an initial speed of \( 22.5 \, \text{m/s} \). It is caught at the same distance above ground.
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?
A whiffle ball is tossed straight up, reaches a highest point, and falls back down. Air resistance is not negligible. Which of the following statements are true?
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball \( B \) passes ball \( A \):
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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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