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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_x = v_i \cos 45^{\circ} = \frac{20}{\sqrt{2}}\] | Horizontal component of each ball’s velocity. |
| 2 | \[v_{\text{rel}} = 2 v_x\] | Relative speed because they move toward each other. |
| 3 | \[t_{\text{collide}} = \frac{100}{v_{\text{rel}}} = \frac{100}{2 v_x} = \frac{100}{20\sqrt{2}} = \frac{5\sqrt{2}}{2}\] | Time needed to close the 100 m gap. |
| 4 | \[t_{\text{flight}} = \frac{2 v_i \sin 45^{\circ}}{g} = \frac{2(20/\sqrt{2})}{9.8}\] | Time for either ball to return to launch height. |
| 5 | \[t_{\text{flight}} \approx 2.89\, \text{s} < t_{\text{collide}}\] | They remain airborne after this instant since the canyon is open below. |
| 6 | \[\boxed{\text{Yes}}\] | Identical vertical motion guarantees equal heights; horizontal positions coincide while they are still in the air. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = t_{\text{collide}} = \frac{5\sqrt{2}}{2}\] | Use time found in part (a). |
| 2 | \[v_{y0} = v_i \sin 45^{\circ} = \frac{20}{\sqrt{2}}\] | Initial vertical velocity. |
| 3 | \[y = v_{y0} t – \frac{1}{2} g t^2\] | Kinematic equation for vertical displacement. |
| 4 | \[y = \left(\frac{20}{\sqrt{2}}\right)\left(\frac{5\sqrt{2}}{2}\right) – \frac{1}{2}(9.8)\left(\frac{5\sqrt{2}}{2}\right)^2\] | Substitute numerical values. |
| 5 | \[y = 50 – 61.25 = -11.25\] | Height is 11.25 m below launch level. |
| 6 | \[\boxed{t = 3.54\, \text{s}},\qquad \boxed{y = -11.25\, \text{m}}\] | Final numerical results (two-decimal accuracy). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = v_x (1.5) = \frac{20}{\sqrt{2}}(1.5)\] | Horizontal distance each ball travels in 1.5 s. |
| 2 | \[d_{\text{sep}} = 100 – 2\Delta x = 100 – 2\left(\frac{20}{\sqrt{2}}(1.5)\right)\] | Subtract both covered distances from the canyon width. |
| 3 | \[d_{\text{sep}} = 100 – 30\sqrt{2} \approx 57.6\] | Simplify to exact and approximate forms. |
| 4 | \[\boxed{d_{\text{sep}} \approx 57.6\, \text{m}}\] | Horizontal separation 1.5 s after launch. |
Just ask: "Help me solve this problem."
During projectile motion (neglecting air resistance), what is the vertical acceleration at the highest point?
A rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?
A toy car moves off the edge of a table that is \(1.25 \, \text{m}\) high. If the car lands \(0.40 \,\text{m}\) from the base of the table…
An eagle is flying horizontally at \(6 \, \text{m/s}\) with a fish in its claws. It accidentally drops the fish.
A drinking fountain projects water at an initial angle of \( 50^ \circ \) above the horizontal, and the water reaches a maximum height of \( 0.150 \) \( \text{m} \) above the point of exit. Assume air resistance is negligible.
\(\text{Yes}\)
\(t = \frac{5\sqrt{2}}{2}\,\text{s},\ y = -11.25\,\text{m}\)
\(\Delta x \approx 57.6\,\text{m}\)
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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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