PHYS 210A Exam 1 Study Notes

Exam Format and Materials

You are allowed a one-sided 8x10 sheet of formulas only (no written solutions, no definitions, no prefixes, etc.). This sheet will be collected with the exam.
- Place your name at the top of the sheet.
Use the Section Summary in the back of the book as a starting point for your formula sheet.
Duration: You will be given the entire class time for the exam.
Exam medium: Blue Scantron. The instructor will not provide or accept other scantrons, so bring the blue one.
Scratch paper: Available for use; you will hand in all scratch work at the end along with the exam packet and the scantron.
Calculator: Allowed, but not the calculator on your cell phone.
Topics: Exam covers Chapters 1, 2, and 3.
At the end: Hand in exam packet, scantron, formula sheet, and all scratch paper.
Focus areas on the exam (from the guide):
- Know which units go with which physical quantity.
- Know which symbol goes with which variable (e.g., v for velocity).
- Know how to convert units.
- Know the prefixes.

Identify and map quantities to units and symbols.
Distinguish between scalar and vector quantities; identify vectors.
Solve for:
- Displacement, velocity, and acceleration in 1D (one dimension).
- Displacement, velocity, and acceleration in 2D (two dimensions).
Recognize which kinematic equations to apply and how to solve for the unknown quantity.
Recognize when to use Pythagoras’ Theorem and when to use trigonometry.
Understand projectile motion and related calculations.
Build and interpret a consistent set of equations from the given data.
Connect exam topics to foundational principles of motion (kinematics).
Consider ethical, philosophical, or practical implications of measurement and modeling in physics when relevant.

Symbols: Ensure you know common symbols for physical quantities (e.g., v for velocity, r for position, a for acceleration).
Units: Be able to associate each physical quantity with its unit (e.g., velocity in m/s, acceleration in m/s^2).
Unit conversion: Practice converting between unit systems (e.g., SI units).
Significant figures: Identify appropriate significant figures in measurements and calculations.
Scientific notation: notation form and conversion between formats.

Be able to match prefixes to scales (as part of the homework and in-class practice):
- Prefixes listed in the guide (note: the transcript shows potential typographical errors; standard prefixes are discussed below).
Standard prefixes (for reference, though the transcript includes misprints):
- Deci: 10^{-1}
- Micro: 10^{-6}
- Milli: 10^{-3}
- Kilo: 10^{3}
The practice items in the transcript include:
- “The unit for velocity is __.”
- “Match the appropriate prefix with the scale.”
- A set of prefix-value pairs:
- 1) Deci 10^(-1) (note: transcript shows “103” which appears to be a typographical error)
- 2) Micro 10^(-6) (transcript shows “10-1”, likely a misprint)
- 3) Milli 10^(-3) (transcript shows “10-6” which is inconsistent with standard prefixes)
- 4) Kilo 10^3 (transcript shows “10-3” which is inconsistent with standard prefixes)
Practical note: treat the given items as presented, but be aware that some numbers in the transcript appear mis-typed; rely on standard prefixes as needed for calculations.
Practice tasks referenced: Convert 10 mm to meters; Convert 25 km/hr to SI units; Determine the conversion factor used to convert a given quantity and place it as a ratio.

1D kinematics (constant acceleration):
- Displacement: x(t) = x0 + v0 t + frac{1}{2} a t^2
- Velocity: v(t) = v_0 + a t
2D kinematics (vector motion):
- Position: oldsymbol{r}(t) = oldsymbol{r}0 + oldsymbol{v}0 t + frac{1}{2} oldsymbol{a} t^2
- Components: x(t) = x0 + v{0x} t + frac{1}{2} ax t^2,\ y(t) = y0 + v{0y} t + frac{1}{2} ay t^2
Recognize when to apply each model based on where motion occurs (straight-line vs two-dimensional trajectories).

Pythagoras’ Theorem to relate components to resultant magnitudes:
- If you know components, magnitude is |oldsymbol{v}| = \sqrt{vx^2 + vy^2}
Trigonometry to resolve vectors:
- Horizontal and vertical components of velocity: vx = v \, \cos\theta, \quad vy = v \, \sin\theta
- For projectile motion without air resistance, horizontal velocity is constant: v_x = ext{constant}

Projectile motion assumes constant acceleration due to gravity in the vertical direction, g \,=\, 9.80\ \mathrm{m\,s^{-2}} (downward).
Equations separate into components: horizontal (x) and vertical (y):
- x(t) = x0 + v{0x} t (since a_x = 0)
- y(t) = y0 + v{0y} t - \tfrac{1}{2} g t^2
Range, maximum height, and time of flight can be derived using these component equations.

Chapter 1:
- Review all problems on the lecture slides.
- Textbook Ch1: Problems & Exercises #17.
- The unit for velocity is ____.
- Match the appropriate prefix with the scale.
Chapter 2:
- Review all problems on the lecture slides.
- Textbook Ch 2: Problems & Exercises #9, 16, 18, 23, 41.
- Achieve HW#2, Problems #7, 9, 12, 13.
Chapter 3:
- Review all problems on the lecture slides.
- Review all problems from HW#3.
- Textbook Ch 3: #21, 25, 27, 29.
Note: The exam problems may be similar or variations of these problems.

Problem (paraphrased): A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of v_0 = 13.0\ \mathrm{m/s}. The rock misses the edge of the cliff as it falls back to earth. Calculate the position of the rock 1.00\ \mathrm{s} after it is thrown.
Known/Unknown (as listed in the prompt):
- a. y0 — Known (often set to 0 if the cliff edge is the origin)\n - b. y — Unknown (position after 1.00 s)\n - c. v0 — Known (13.0\ \mathrm{m/s})\n - d. v — Unknown (velocity at t = 1.00 s)\n - e. a — Known (gravity, a = -g = -9.80\ \mathrm{m\,s^{-2}})\n - f. t — Known (1.00 s)
Which equation is best to use? (Kinematic equation in vertical motion with constant acceleration):
- Best choice: y = y0 + v0 t + \tfrac{1}{2} a t^2
Final calculation (using the above values):
- Put numbers: y = y0 + v0 t + \tfrac{1}{2} a t^2 = 0 + 13.0(1.00) + \tfrac{1}{2}(-9.80)(1.00)^2
- Compute: y = 13.0 - 4.90 = 8.10\ \mathrm{m}
Answer choices (from the prompt):
- a. 4.05 b. 8.10 c. 3.26 d. 2.54
Correct answer: \text{b) } 8.10\ \mathrm{m}
Unit of the final answer: \mathrm{m}

1D displacement: x(t) = x0 + v0 t + \tfrac{1}{2} a t^2
1D velocity: v(t) = v_0 + a t
2D/3D displacement: \boldsymbol{r}(t) = \boldsymbol{r}0 + \boldsymbol{v}0 t + \tfrac{1}{2} \boldsymbol{a} t^2
Component form (2D): x(t) = x0 + v{0x} t + \tfrac{1}{2} ax t^2,\ y(t) = y0 + v{0y} t + \tfrac{1}{2} ay t^2
Projectile-specific: horizontal velocity constant, vx = v{0x}, vertical motion under gravity: y(t) = y0 + v{0y} t - \tfrac{1}{2} g t^2
Vector magnitude from components: |oldsymbol{v}| = \sqrt{vx^2 + vy^2}
Vector components: vx = v \cos\theta,\quad vy = v \sin\theta
Gravitational acceleration: g \approx 9.80\ \mathrm{m\,s^{-2}}

Some prefixes/typographical entries in the transcript appear to be misprints (e.g., prefix mappings like "Deci 103" and "Kilo 10-3"). Use standard prefix values for study and calculations, and note these discrepancies when reviewing the material.
The provided Chapter references (Ch1, Ch2, Ch3 problems and HW numbers) should be treated as concrete targets for practice, even though the exact problems on the exam may vary.
The sample problem (Example 2.14) demonstrates the process of identifying knowns/unknowns, selecting the appropriate equation, and performing a straightforward calculation to obtain the numerical result and unit.