AP Physics Unit

Unit 1 - Vectors and Kinematics

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Mathematical

FRQ

You kick a soccer ball with an initial velocity directed 53° above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25 m away, and it takes the ball 2.1 seconds to pass directly over the wall.

  1. (a) Calculate the initial velocity of the ball. (2 points)
  2. (b) Determine the horizontal range of the ball. (2 points)
  3. (c) By what vertical distance does the ball clear the wall of the building? (3 points)
  1. 19.8 m/s
  2. 32.2 m
  3. 5.52 m
0

First, calculate the initial velocity of the ball.

Step Formula Derivation Reasoning
1 y = v_{0y} t – \frac{1}{2} g t^2 Vertical motion equation for height (y), initial vertical velocity (v_{0y}), time (t), and acceleration due to gravity (g).
2 v_{0y} = \frac{y + \frac{1}{2} g t^2}{t} Solve for initial vertical velocity (v_{0y}).
3 v_{0x} = \frac{d}{t} Horizontal velocity (v_{0x}) is constant, where d is the distance to the wall.
4 v_{0} = \sqrt{v_{0x}^2 + v_{0y}^2} Initial velocity magnitude using Pythagorean theorem, combining horizontal and vertical components.

Given values:

  • Wall height y = 7.2 , \text{m}
  • Distance to wall d = 25 , \text{m}
  • Time to pass over wall t = 2.1 , \text{s}
  • Acceleration due to gravity g = -9.81 , \text{m/s}^2

Next, determine the horizontal range of the ball.

Step Formula Derivation Reasoning
1 R = v_{0x} T Horizontal range (R), where T is the total time of flight.
2 \Delta y = v_{oy}t \frac{1}{2}gt^2 Total time of flight from launch to landing on rood, using symmetry of projectile motion. T = 2.65 seconds.

Finally, calculate the vertical distance the ball clears the wall.

Step Formula Derivation Reasoning
1 h_{clear} = y_{peak} – y_{wall} Vertical clearance (h_{clear}) is the difference between the peak height (y_{peak}) and wall height (y_{wall}).
2 y_{peak} = \frac{v_{0y}^2}{2g} Peak height calculation using the initial vertical velocity. Peak height = 12.7 m.

 

The calculations yield the following results:

  1. Initial velocity of the ball: \boxed{19.8 , \text{m/s}}
  2. Horizontal range of the ball: \boxed{32.2 , \text{m}}
  3. Vertical distance by which the ball clears the wall: \boxed{5.52 , \text{m}}

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  1. 19.8 m/s
  2. 32.2 m
  3. 5.52 m

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KinematicsForces
\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}}
ConstantDescription
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}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

  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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