AP Physics

Unit 1 - Vectors and Kinematics




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(a) How far does the textbook travel horizontally after it is released?

Step Derivation/Formula Reasoning
1 v_{0x} = v_0 \cos(\theta) Calculate the initial horizontal velocity. Use the initial speed and the angle of projection.
2 v_{0y} = v_0 \sin(\theta) Calculate the initial vertical velocity. Use the initial speed and the angle of projection.
3 v_{0x} = 20 \cos(36^\circ) \approx 16.2 \, \text{m/s} Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the horizontal velocity formula.
4 v_{0y} = 20 \sin(36^\circ) \approx 11.8 \, \text{m/s} Substitute v_0 = 20 \, \text{m/s} and \theta = 36^\circ into the vertical velocity formula.
5 y = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} Use the equation of motion in the vertical direction. The textbook is moving under gravity.
6 0 = 12 + 11.8 t – \frac{1}{2} \cdot 9.8 t^2 Set the displacement y to zero because we are calculating the time t when the textbook reaches the ground. g = 9.8 \text{m/s}^2 .
7 4.9 t^2 – 11.8 t – 12 = 0 Simplify the quadratic equation to solve for t .
8 t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} Use the quadratic formula where a = 4.9 , b = -11.8 , and c = -12 .
9 t \approx 3.18 \, \text{s} Solve the equation and take the positive root. This is the time the textbook stays in the air.
10 x = v_{0x} \cdot t Calculate the horizontal distance the textbook travels. Use the horizontal velocity and the time.
11 x \approx 16.2 \times 3.18 \approx 51.5 \, \text{m} Substitute v_{0x} = 16.2 \, \text{m/s} and t = 3.18 \, \text{s} into the horizontal distance formula to get the final answer.
\text{The horizontal distance traveled is approximately } 51.5 \, \text{m}

(b) What is the book’s velocity (speed and direction) when it reaches the ground?

Step Derivation/Formula Reasoning
1 v_y = v_{0y} – g t Calculate the final vertical velocity using the initial vertical velocity, gravitational acceleration, and time.
2 v_y = 11.8 – 9.8 \times 3.18 \approx -19.4 \, \text{m/s} Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , and t = 3.18 \, \text{s} into the vertical velocity formula.
3 v_{\text{total}} = \sqrt{v_{x}^2 + v_y^2} Calculate the magnitude of the total velocity using the Pythagorean theorem.
4 v_{\text{total}} \approx \sqrt{16.2^2 + (-19.4)^2} \approx 25.3 \, \text{m/s} Substitute v_{x} = 16.2 \, \text{m/s} and v_y = -19.4 \, \text{m/s} into the total velocity formula.
5 \theta = \tan^{-1} \left(\frac{v_y}{v_x}\right) Calculate the direction of the velocity. Use the inverse tangent to find the angle.
6 \theta \approx \tan^{-1} \left(\frac{-19.4}{16.2}\right) \approx -50.1^\circ The vector points 50.1° below the x-axis.
\text{The velocity when the book reaches the ground is approximately } 25.3 \, \text{m/s} \text{ at } -50.1^\circ

(c) What is the book’s maximum height above the ground?

Step Derivation/Formula Reasoning
1 v_y = 0 The vertical velocity at the maximum height is zero.
2 v_y = v_{0y} – g t Use the vertical motion equation to find the time to reach maximum height.
3 0 = 11.8 – 9.8 t Set final vertical velocity v_y = 0 and solve for t .
4 t = \frac{11.8}{9.8} \approx 1.20 \, \text{s} Solving the equation gives the time to reach maximum height.
5 H = v_{0y} t – \frac{1}{2} g t^2 + \text{initial height} Use the vertical motion equation to find the maximum height.
6 H \approx 11.8 \times 1.2 – \frac{1}{2} \times 9.8 \times (1.2)^2 + 12 Substitute v_{0y} = 11.8 \, \text{m/s} , g = 9.8 \, \text{m/s}^2 , t = 1.2 \, \text{s} , and initial height = 12 m.
7 H \approx 19.1 \, \text{m} Calculate the maximum height above the ground.
\text{The maximum height above the ground is approximately } 19.1 \, \text{m}

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a) 51.5 m

b) 25.3 m/s at 50.1° below the horizontal

c) 19.1 m

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\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: \text{5 km}

  2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

  3. Perform the multiplication: \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}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}



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  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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