AP Physics

Unit 1 - Vectors and Kinematics




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Step Derivation/Formula Reasoning
(a) How high does the ball rise?
1 Use the kinematic equation: v_f^2 = v_i^2 + 2a d This equation relates the final velocity, initial velocity, acceleration, and distance traveled.
2 0 = (22.5 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)d At the highest point, the final velocity v_f = 0 m/s. The initial velocity v_i = 22.5 m/s and acceleration a = -9.8 m/s^2 (gravity acting downwards).
3 d = \frac{(22.5 \, \text{m/s})^2}{2 \cdot 9.8 \, \text{m/s}^2} \approx 25.8 \, \text{m} Solve for d, the height reached by the ball.
4 d \approx 25.8 \, \text{m} The maximum height the ball reaches.
(b) How long does it take for the ball to reach its highest point?
1 Use the kinematic equation: v_f = v_i + at This equation relates the final velocity, initial velocity, acceleration, and time.
2 0 = 22.5 \, \text{m/s} + (-9.8 \, \text{m/s}^2) t At the highest point, v_f = 0 m/s. The initial velocity v_i = 22.5 m/s and acceleration a = -9.8 m/s^2.
3 t = \frac{22.5 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 2.30 \, \text{s} Solve for t, the time taken to reach the highest point.
4 t \approx 2.30 \, \text{s} The time taken to reach the maximum height.
(c) How long does the ball remain in the air?
1 \text{Total time} = 2 \times t_{\text{up}} Time to go up is equal to time to come down because distances and accelerations are the same.
2 2 \times 2.30 \, \text{s} \approx 4.60 \, \text{s} Double the time to reach the highest point to find total time in the air.
3 \text{Total time} \approx 4.60 \, \text{s} The total time the ball remains in the air.
(d) How fast was it going just before it is caught?
1 Use symmetry: v_{\text{final}} = -v_{\text{initial}} Due to symmetry, speed upon returning to original height equals initial speed but opposite in direction.
2 \text{Speed} = 22.5 \, \text{m/s} The magnitude of the speed is the same.
3 \text{Speed} = 22.5 \, \text{m/s} The speed just before being caught.
(e) What is the velocity and acceleration of the ball at the highest point?
1 \text{Velocity} = 0 \, \text{m/s} At the highest point, the velocity is zero as the ball changes direction.
2 \text{Acceleration} = -9.8 \, \text{m/s}^2 Acceleration due to gravity remains constant throughout the motion.
3 \text{Velocity} = 0 \, \text{m/s}, \text{Acceleration} = -9.8 \, \text{m/s}^2 Velocity and acceleration at the highest point.

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  1. 25.8 \, m
  2. 2.3 \, s
  3. 4.6 \, s
  4. 22.5 \, m/s
  5. 0 \, m/s , \, -9.8 \, m/s^2

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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}}



Power of Ten




















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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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