| 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 horizontal speed. One ball moves right at \(v_x\), the other left at \(v_x\). Since they head directly toward each other, the gap closes at \(v_x + v_x = 2v_x\). |
| 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 | \[\boxed{\text{Yes}}\] | Both balls have identical vertical motion, so their heights are always the same. With the canyon open below, they meet horizontally at the midpoint when the gap closes. |
| 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 (upward positive). |
| 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 | \[y = 50 – 61.25 = -11.25 = -\frac{45}{4}\ \text{m}\] | Height is \(11.25\) m below launch level. |
| 6 | \[\boxed{t \approx 3.54\, \text{s}},\qquad \boxed{y = -11.25\, \text{m}}\] | Final numerical results. |
| 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 | \[d_{\text{sep}} = 100 – 30\sqrt{2} \approx 57.57\] | Simplify to exact and approximate forms. |
| 4 | \[\boxed{d_{\text{sep}} \approx 57.6\, \text{m}}\] | Horizontal separation 1.5 s after launch. |
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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 \( 27^\circ \).
Which of the following statements about the acceleration due to gravity is TRUE?
A plane, 220 meters high, is dropping a supply crate to an island below. It is traveling with a horizontal velocity of 150 m/s. At what horizontal distance must the plane drop the supply crate for it to land on the island? Use [katex] g = 9.81 \, m/s^2[/katex].
Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with \( \theta_A \) larger than \( \theta_B \).
In a lab experiment, a ball is rolled down a ramp so that it leaves the edge of the table with a horizontal velocity [katex]v[/katex]. Assume there are no frictional forces. If the table has a height [katex]h[/katex] above the ground, how far away from the edge of the table, a distance [katex]x[/katex], does the ball land?
A soccer ball is kicked horizontally off an \( 85 \) \( \text{m} \) high cliff at a speed of \( 34 \) \( \text{m/s} \). What is the ball’s final speed when it hits the ground below?
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
Two balls are launched at the same speed. Ball A is launched at an angle of \( 45^{\circ} \) and Ball B is launched at an angle of \( 60^{\circ} \). Which one reaches a higher point?
Suppose the water at the top of Niagara Falls has a horizontal speed of \( 2.7 \, \text{m/s} \) just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a \( 75^\circ \) angle below the horizontal?
Seo-Jun throws a ball to her friend Zuri. The ball leaves Seo-Jun’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground with an initial speed \( \vec{v}_{s,0} = 12 \) \( \text{m/s} \) at an angle of \( \theta = 25^\circ \) with respect to the horizontal. Zuri catches the ball at a height of \( h = 1.5 \) \( \text{m} \) above the ground.
After catching the ball, Zuri throws it back to Seo-Jun. The ball leaves Zuri’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground. The ball is moving with a speed of \( 15 \) \( \text{m/s} \) when it reaches a maximum height of \( 5.8 \) \( \text{m} \) above the ground.
At what height \( h’ \) above the ground will the ball be when the return throw reaches Seo-Jun?
\(\text{Yes}\)
\(t = \frac{5\sqrt{2}}{2}\,\text{s},\ y = -11.25\,\text{m}\)
\(\Delta x \approx 57.6\,\text{m}\)
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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] |
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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