AP Physics Unit

Unit 1 - Vectors and Kinematics

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GQ

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

33.23 m

0
  1. Calculate the time taken for the rocket to reach the wall.
  2. Calculate the vertical height of the rocket at this time.
  3. Subtract the height of the wall from this vertical height to find the clearance.
Step Formula Derivation Reasoning
1.1 v_{x0} = v_0 \cos(\theta) Calculate the horizontal component of the initial velocity. v_0 is the launch speed and \theta is the launch angle.
1.2 v_{x0} = 75.0 , \text{m/s} \times \cos(60.0^\circ) Plug in v_0 = 75.0 , \text{m/s} and \theta = 60.0^\circ.
1.3 t = \frac{d}{v_{x0}} Calculate the time t it takes for the rocket to reach the wall. d is the horizontal distance to the wall.
1.4 t = \frac{27.0 , \text{m}}{v_{x0}} Plug in d = 27.0 , \text{m}.
2.1 v_{y0} = v_0 \sin(\theta) Calculate the vertical component of the initial velocity.
2.2 v_{y0} = 75.0 , \text{m/s} \times \sin(60.0^\circ) Plug in v_0 = 75.0 , \text{m/s} and \theta = 60.0^\circ.
2.3 y = v_{y0} t – \frac{1}{2} g t^2 Use the kinematic equation for vertical motion. g is the acceleration due to gravity.
2.4 y = v_{y0} t – \frac{1}{2} (9.8 , \text{m/s}^2) t^2 Plug in g = 9.8 , \text{m/s}^2.
2.5 y = v_{y0} \times \frac{27.0 , \text{m}}{v_{x0}} – \frac{1}{2} (9.8 , \text{m/s}^2) \left( \frac{27.0 , \text{m}}{v_{x0}} \right)^2 Substitute t from step 1.4.
3.1 \text{Clearance} = y – 11.0 , \text{m} Subtract the wall height from the rocket’s height to find the clearance.

We will now calculate these steps.

The rocket clears the top of the wall by approximately 33.23 , \text{m}.

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33.23 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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