| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v = v_0 + at \) | Use the kinematic equation for vertical motion. Here, \(v\) is the final velocity at the max height (which is 0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity, acting downward), and \(t\) is the time. |
| 2 | \( 0 = v_0 – g \times 0.586 \) | Set the final velocity \(v\) at the maximum height to 0, and solve for \(v_0\). The acceleration due to gravity \(g\) is \(9.8 \, \text{m/s}^2\). |
| 3 | \( v_0 = g \times 0.586 \) | Rearrange to solve for \(v_0\) |
| 4 | \( v_0 = 9.8 \times 0.586 \) | Substitute the values of \(g\) and \(t\) |
| 5 | \( v_0 = 5.74 \, \text{m/s} \) | Final answer for the initial speed. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v^2 = v_0^2 + 2a \Delta x \) | Use the kinematic equation to relate velocity, acceleration, and displacement. Here, \(v\) is the final velocity at the max height (0), \(v_0\) is the initial velocity, \(a\) is the acceleration (gravity), and \(\Delta x\) is the displacement. |
| 2 | \( 0 = (5.74)^2 – 2 \cdot 9.8 \cdot \Delta x \) | Substitute \(v = 0\), \(v_0 = 5.74 \, \text{m/s}\), and \(a = 9.8 \, \text{m/s}^2\) |
| 3 | \( 2 \cdot 9.8 \cdot \Delta x = (5.74)^2 \) | Rearrange to solve for \(\Delta x\). |
| 4 | \( \Delta x = \frac{(5.74)^2}{2 \cdot 9.8} \) | Solve for \(\Delta x\). |
| 5 | \( \Delta x = 1.68 \, \text{m} \) | Compute the displacement \(\Delta x\). |
| 6 | \(\text{Max height} = 3.25 + 1.68 = 4.93 \, \text{m} \) | Since the ice cube was initially 3.25 m above the ground, add this to \(\Delta x\) to get the maximum height. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\Delta y = v_0 t + \frac{1}{2} a t^2\) | Use the kinematic equation for vertical displacement, where \(\Delta y\) is the change in height, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(t\) is the time. |
| 2 | \(\Delta y = 0\) at maximum height, then use \(y = 4.93 \, \text{m}\) | The maximum height the ice cube reaches is previously calculated as 4.93 m. |
| 3 | \( -4.93 \, \text{m} = – \frac{1}{2} \cdot 9.8 \cdot t^2\) | Solve for the time taken to fall from the maximum height to the ground. Note the displacement (\(\Delta y\)) is negative when falling down. |
| 4 | \(t = \sqrt{\frac{2 \cdot 4.93}{9.8}}\) | Substitute the values and solve for \(t\). |
| 5 | \(t = \sqrt{\frac{9.86}{9.8}} = \sqrt{1.01} \approx 1.005 \, \text{s}\) | Calculate the square root to find the fall time. |
| 6 | Total time = \(0.586 \, \text{s} (up) + 1.005\, \text{s} (down) \approx 1.59 \, \text{s}\) | Add the time going up (0.586 s) to the time coming down (1.005 s) to get the total time. |
| 7 | \( t_{\text{total}} = 1.59 \, \text{s} \) | Final time taken for the ice cube to reach the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity, \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement (4.93 m). |
| 2 | \(v_y^2 = 0 + 2 \cdot 9.8 \cdot 4.93\) | Substitute the values into the equation. |
| 3 | \( v_y^2 = 96.508 \) | Calculate the right side of the equation. |
| 4 | \( v_y = \sqrt{96.508}\) | Take the square root to solve for \(v_y\). |
| 5 | \( v_y \approx 9.82 \, \text{m/s}\) | Final speed of the ice cube when it reaches the ground. |
| 6 | \( v_{\text{final}} \approx 9.82 \, \text{m/s} \) | The boxed final answer for the speed of the ice cube when it hits the ground. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_y^2 = v_{max}^2 + 2 a \Delta y\) | Use the kinematic equation relating velocity, acceleration, and displacement, where \(v_y\) is the final velocity (7.00 m/s), \(v_{max}\) is the velocity at maximum height (0 m/s), \(a\) is the acceleration due to gravity, and \(\Delta y\) is the displacement from the maximum height. |
| 2 | \( (7.00)^2 = 0 + 2 \cdot 9.8 \cdot \Delta y \) | Substitute the given speed and constants. |
| 3 | \( 49 = 19.6 \cdot \Delta y\) | Solve for \(\Delta y\) by isolating it on one side of the equation. |
| 4 | \( \Delta y = \frac{49}{19.6}\) | Rearrange to solve for \(\Delta y\). |
| 5 | \(\Delta y = 2.5 \, \text{m}\) | Final height change from the maximum height when the ice cube reaches a speed of 7.00 m/s downward. |
| 6 | \( \text{Height above ground} = 4.93 – 2.5 \) | Subtract the downwards displacement from the maximum height to get the current height above the ground. |
| 7 | \( \text{Height above ground} = 2.43 \, \text{m}\) | Final height of the ice cube above the ground when traveling at 7.00 m/s downward. |
| 8 | \( \text{Height} = 2.43 \, \text{m} \) | Boxed final answer. |
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Does the odometer of a car measure a scalar or a vector quantity? What about the speedometer?
A stone is thrown vertically upward with a speed of \( 24.0 \) \( \text{m/s} \).
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
A car starts from rest and accelerates uniformly over a time of 5 seconds for a distance of 100 m. Determine the acceleration of the car.
A horizontal \( 300 \) \( \text{N} \) force pushes a \( 40 \) \( \text{kg} \) object across a horizontal \( 10 \) \( \text{meter} \) frictionless surface. After this, the block slides up a \( 20^\circ \) incline. Assuming the incline has a coefficient of kinetic friction of \( 0.4 \), how far along the incline will the object slide?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
A car accelerates from rest with an acceleration of \( 3.5 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp \( 50 \, \text{m} \) ahead and takes \( 0.6 \, \text{s} \) to react. After reacting, the driver hits the brakes, which slow the car with an acceleration of \( 7.2 \, \text{m/s}^2 \). Unfortunately, the driver does not stop in time and goes off the \( 3 \, \text{m} \) high ramp that is angled at \( 27^\circ \).
Two objects are dropped from rest from the same height. Object \( A \) falls through a distance \( d_A \) during a time \( t \), and object \( B \) falls through a distance \( d_B \) during a time \( 2t \). If air resistance is negligible, what is the relationship between \( d_A \) and \( d_B \)?
Note answers may vary by \( \pm 0.2 \).
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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] | |
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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