| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for displacement. |
| 2 | Set \( y = 0 \), \( y_0 = 19.6 \) m, \( a = -9.8 \) m/s². | Define the variables. |
| 3 | For Ball A (downward): \( v_i = -14.7 \) m/s | Initial velocity downward is negative. |
| 4 | \( 0 = 19.6 -14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 5 | \( 4.9 t^2 +14.7 t -19.6 = 0 \) | Rearrange into standard quadratic form. |
| 6 | Solve for \( t \): \( t = 1 \) s | Find the positive root of the quadratic equation. |
| 7 | For Ball B (upward): \( v_i = +14.7 \) m/s | Initial velocity upward is positive. |
| 8 | \( 0 = 19.6 +14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 9 | \( -4.9 t^2 +14.7 t +19.6 = 0 \) | Simplify equation. |
| 10 | \( 4.9 t^2 -14.7 t -19.6 = 0 \) | Multiply both sides by -1. |
| 11 | Solve for \( t \): \( t = 4 \) s | Find the positive root of the quadratic equation. |
| 12 | \( \Delta t = t_{\text{Ball B}} – t_{\text{Ball A}} = 4 \, \text{s} – 1 \, \text{s} = 3 \, \text{s} \) | Calculate the difference in time. |
Answer: The difference in time the balls spend in the air is 3 seconds.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( v = v_i + a t \) | Use the kinematic equation for velocity. |
| 2 | For Ball A: \( v = -14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(1 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball A. |
| 3 | For Ball B: \( v = +14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(4 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball B. |
Answer: Each ball strikes the ground with a velocity of -24.5 m/s downward.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for position. |
| 2 | For Ball A: \( y_{\text{A}} = 19.6 + (-14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball A at \( t = 0.8 \) s. |
| 3 | \( y_{\text{A}} = 19.6 -11.76 -3.136 = 4.704 \, \text{m} \) | Simplify to find \( y_{\text{A}} \). |
| 4 | For Ball B: \( y_{\text{B}} = 19.6 + (+14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball B at \( t = 0.8 \) s. |
| 5 | \( y_{\text{B}} = 19.6 +11.76 -3.136 = 28.224 \, \text{m} \) | Simplify to find \( y_{\text{B}} \). |
| 6 | \( \Delta y = y_{\text{B}} – y_{\text{A}} = 28.224 – 4.704 = 23.52 \, \text{m} \) | Calculate the distance between the balls. |
Answer: The balls are 23.52 meters apart 0.800 seconds after they are thrown.
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A ball is thrown straight up with a speed of \( 30 \) \( \text{m/s} \), and air resistance is negligible.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
An object travels along a path shown above, with changing velocity as indicated by vectors \( A \) and \( B \). Which vector best represents the net acceleration of the object from time \( t_A \) to \( t_B \)?
A body starting from rest moves along a straight line under the action of a constant force. After traveling a distance \( d \) the speed of the body is \( v \). The speed of the body when it has travelled a distance \( \dfrac{d}{2} \) from its initial position is
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 45^\circ \). It lands on a slope \( 5 \) \( \text{m} \) below the launch height. On landing, it rebounds vertically with \( 80\% \) of its speed and falls straight down from there. Find the total time from launch to final impact at the base of the slope.
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
Mary and Sally are in a foot race. When Mary is \( 22 \) \( \text{m} \) from the finish line, she has a speed of \( 4.0 \) \( \text{m/s} \) and is \( 5.0 \) \( \text{m} \) behind Sally, who has a speed of \( 5.0 \) \( \text{m/s} \). Sally thinks she has an easy win and, during the remaining portion of the race, decelerates at a constant rate of \( 0.40 \) \( \text{m/s}^2 \) until she reaches the finish line. What constant acceleration must Mary maintain during the remaining portion of the race if she wishes to cross the finish line side-by-side with Sally?
A skater glides across the ice at a constant \( 6 \) \( \text{m/s} \). After \( 4 \) \( \text{s} \), friction gradually slows them down until they come to rest in \( 6 \) \( \text{s} \). They pause for \( 2 \) \( \text{s} \), then push off in the opposite direction, steadily gaining speed for \( 5 \) \( \text{s} \). Draw the velocity vs. time graph.
A ball is dropped from a window \(10 \, \) above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
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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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