AP Physics Unit

Unit 1 - Vectors and Kinematics

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Mathematical

FRQ

A coin is dropped from a hot air-balloon that is 250 m above the ground rising at 11 m/s upwards. For the coin, find the following:

  1. The maximum height reached (in meters)
  2. The position, 4.20 seconds after being released. (Assume up is positive)
  3. The velocity 4.20 seconds after being released. (Assume up is positive)
  4. The time ( in seconds) before it hits the ground

  1. 256.2 m
  2. 210 m
  3. -30.16 m/s
  4. 8.4 s

1. Maximum Height Reached by the Coin

Step Formula Derivation Reasoning
1 v^2 = u^2 + 2as Kinematic equation for motion, with v as final velocity, u as initial velocity, a as acceleration, and s as displacement.
2 0 = (11, \text{m/s})^2 – 2 \times 9.81, \text{m/s}^2 \times s At maximum height, final velocity v = 0, initial velocity u = 11, \text{m/s} upward, acceleration a = -9.81, \text{m/s}^2 (gravity acts downward).
3 Solve for s Calculate the displacement s.
2. Position 4.20 Seconds After Being Released

Step Formula Derivation Reasoning
1 s = ut + \frac{1}{2}at^2 Kinematic equation for displacement.
2 s = 11, \text{m/s} \times 4.20, \text{s} – \frac{1}{2} \times 9.81, \text{m/s}^2 \times (4.20, \text{s})^2 Substitute values for u, a, and t (time after release).

3. Velocity 4.20 Seconds After Being Released

Step Formula Derivation Reasoning
1 v = u + at Kinematic equation for velocity.
2 v = 11, \text{m/s} – 9.81, \text{m/s}^2 \times 4.20, \text{s} Substitute values for u, a, and t.

4. Time Before Coin Hits the Ground

Step Formula Derivation Reasoning
1 s = ut + \frac{1}{2}at^2 Kinematic equation for displacement.
2 -250, \text{m} = 11, \text{m/s} \times t – \frac{1}{2} \times 9.81, \text{m/s}^2 \times t^2 Substitute values for s (downward displacement), u, and a. Solve for t.

Let’s perform the calculations for each part.

The results for each part are as follows:

  1. Maximum height reached by the coin: \boxed{256.17, \text{meters}}
  2. Position of the coin 4.20 seconds after being released: \boxed{209.68, \text{meters}} above the ground
  3. Velocity of the coin 4.20 seconds after being released: \boxed{-30.20, \text{m/s}} (downward, as indicated by the negative sign)
  4. Time before the coin hits the ground: \boxed{8.35, \text{seconds}}

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  1. 256.2 m
  2. 210 m
  3. -30.16 m/s
  4. 8.4 s

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Nerd-Notes.com
KinematicsForces
\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2F = m \cdot a
v = v_i + a \cdot tF_g = \frac{G \cdot m_1 \cdot m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu \cdot N
R = \frac{v_i^2 \cdot \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{m \cdot v^2}{r}KE = \frac{1}{2} m \cdot v^2
a_c = \frac{v^2}{r}PE = m \cdot g \cdot h
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m \cdot v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum m \cdot r^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k \cdot 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}
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 (kg·m/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (N·m)}
I (Moment of Inertia)\text{kilogram meter squared (kg·m}^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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