Part a:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = y_0 + v_0 t – \frac{1}{2} g t^2[/katex] | This is the kinematic equation for vertical position, where [katex]y_0[/katex] is the initial position, [katex]v_0[/katex] is the initial velocity, [katex]g[/katex] is the acceleration due to gravity, and [katex]t[/katex] is the time. |
| 2 | Substitute [katex]y = 0[/katex], [katex]y_0 = 800\ \text{m}[/katex], [katex]v_0 = 65\ \text{m/s}[/katex], [katex]g = 9.8\ \text{m/s}^2[/katex] | The object hits the ground when [katex]y=0[/katex]. It is thrown upward from a height of 800 m and the initial velocity is upward, hence positive. |
| 3 | [katex]0 = 800 + 65t – 4.9t^2[/katex] | Rearrange the substituted equation and simplify [katex]g/2[/katex] from [katex]9.8/2[/katex] to [katex]4.9[/katex]. |
| 4 | [katex]4.9t^2 – 65t – 800 = 0[/katex] | This is a quadratic equation in the form of [katex]at^2 + bt + c = 0[/katex]. |
| 5 | [katex]t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/katex] | Apply the quadratic formula to solve for [katex]t[/katex]. Here, [katex]a = 4.9[/katex], [katex]b = -65[/katex], and [katex]c = -800[/katex]. |
| 6 | [katex]t = \frac{-(-65) \pm \sqrt{(-65)^2 – 4 \cdot 4.9 \cdot (-800)}}{2 \cdot 4.9}[/katex] | Plug in the values of [katex]a[/katex], [katex]b[/katex], and [katex]c[/katex]. |
| 10 | [katex]t \approx 21.03\ \text{s} \textbf{ (positive root)}[/katex] | Discard negative time and keep the physically meaningful positive root, being the actual time the object takes to hit the ground. |
Part b:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = v_0 – gt[/katex] | Use the kinematic equation for velocity in vertical motion, taking downward as the negative direction. |
| 2 | [katex]v = 65 – 9.8 \times 21.04[/katex] | Substitute the values of [katex]g[/katex] and [katex]t[/katex] into the equation. |
| 3 | [katex]v \approx -141.14\ \text{m/s}[/katex] | Subtract to find the final velocity, where negative indicates direction downwards. |
Part c:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v \propto v_0[/katex] | The final velocity is proportional to the initial velocity when all other factors remain the same. |
| 2 | [katex]v_{\text{final, new}} = 2v_0 – gt[/katex] | Note this is the equation we used in part b to find the final speed. Time, however, is not the same in the two scenarios. So unless you have found time with the double the initial speed, do NOT use this equation again. |
| 3 | [katex]v_{\text{final, new}} = {2v_0}^2 + 2g\Delta \theta[/katex] | Use this kinematic equation instead. |
| 4 | [katex]{v_{\text{final, new}}}^2 = = 180.54 \, m/s[/katex] | Plug in values and solve for the new final speed. |
| 5 | [katex] \frac{v_{\text{final, new}}}{v_{\text{final, orginal}}[/katex] | Find the factor the final velocity has increased by. |
| 6 | [katex] \frac{180.54}{141.14} \approx 1.28 [/katex] | Plug in values and solve. |
| 7 | Factor = 1.28 | If the initial velocity doubles the final velocity of the ball increases by a factor of 1.28. |
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A car accelerates from rest with an acceleration of \( 4.3 \, \text{m/s}^2 \) for a time of \( 6.8 \, \text{s} \). The car then slows to a stop with an acceleration of \( 5.1 \, \text{m/s}^2 \). What is the total distance traveled by the car?
A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
Which of the following statements about the acceleration due to gravity is TRUE?
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?
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What total distance will be covered after the 3rd second?
In which of the following is the rate of change of the particle’s momentum zero?
A whiffle ball is tossed straight up, reaches a highest point, and falls back down. Air resistance is not negligible. Which of the following statements are true?
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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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