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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = v_0y t – \frac{1}{2} g t^2[/katex] for both balls | These are the equations of motion in the vertical direction (y) for the two balls, assuming upward motion is positive. The first term in both equations represents the initial vertical velocity component times time (t), and the second term is due to the acceleration due to gravity (g) acting downward. |
| 2 | [katex]v_0y[/katex] up = [katex]v_0[/katex], [katex]v_0y[/katex] angle = [katex]v_0 \cos(45^\circ) = \frac{v_0}{\sqrt{2}}[/katex] | The initial vertical velocities. For the straight up throw, the entire velocity is vertical, hence [katex]v_0y[/katex] up = [katex]v_0[/katex]. For the angled throw, the vertical component is [katex]v_0 \cos(45^\circ)[/katex] since the projection is at 45°. |
| 3 | Total time in air for each ball can be calculated using the equation: [katex]t_\text{total} = \frac{2v_{0y}}{g}[/katex] | The time of flight until each ball returns to the original level of projection can be determined from the formula, substituting in each ball’s initial vertical speed. This accounts for the time to ascend to the peak and to descend back to the ground level. |
| 4 | [katex]t_\text{straight up} = \frac{2v_0}{g}[/katex], [katex]t_\text{at angle} = \frac{2(v_0/\sqrt{2})}{g} = \frac{\sqrt{2}v_0}{g}[/katex] | Conclusion: [katex]t_\text{straight up} > t_\text{at angle}[/katex]. The time taken for each is different due to differences in their initial vertical velocities. |
| 5 | [katex] PE + KE_i = KE_f [/katex]
[katex] mgh + \frac{1}{2}mv^2 = KE_f [/katex] Height and total speeds are the same in both cases regardless of direction. |
Since mechanical energy is conserved (ignoring air resistance), the final kinetic energy will be equal to kinetic + potential energy of the ball as it was thrown. Since both balls start with the same energy (same height and speed), the final kinetic energies of both balls must be the same. This would imply identical, final speeds. |
| 6 | Answer: (b) | Both with the same speed, based on the conservation of mechanical energy, (b) as the correct statement. |
Just ask: "Help me solve this problem."
A textbook is launched up with a speed of 20 m/s, at an angle of 36°, from a 12 m high roof.
A person is making homemade ice cream. She exerts a force of magnitude 23 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.25 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.7 s, what is the average power being expended?
How does the speed v1 of a block m reaching the bottom of slide 1 compare with v2, the speed of a block 2m reaching the end of slide 2? The blocks are released from the same height.
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
A stone is thrown horizontally at \(8.0 \, \text{m/s}\) from a cliff \(80 \,\text{m}\) high. How far from the base of the cliff will the stone strike the ground?
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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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