AP Physics Unit

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

GQ

A 4 kg mass is traveling at 10 m/s to the right when it collides elastically with a stationary 7 kg mass. The 7 kg mass then travels at 2 m/s at an angle of 22° below the horizontal. What is the velocity of the 4 kg mass?

Step Formula Derivation Reasoning
1 m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 Conservation of momentum
2 4,kg \cdot 10,m/s + 7,kg \cdot 0,m/s = 4,kg \cdot v_{1x} + 7,kg \cdot 2,m/s \cdot \cos(22^\circ) Plugging in given values and decomposing the 7 kg mass’s velocity into horizontal component.
3 40 = 4v_{1x} + 13.08 Calculating the horizontal momentum contribution from the 7 kg mass post-collision.
4 4v_{1x} = 26.92 Solve for the 4 kg mass’s horizontal velocity component.
5 v_{1x} = 6.73,m/s Calculating the horizontal velocity of the 4 kg mass.
6 m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y} Conservation of momentum in the vertical direction. Since the initial vertical momentum is 0, the final combined vertical momentum must also be 0.
7 0 = 4,kg \cdot v_{1y} + 7,kg \cdot 2,m/s \cdot \sin(22^\circ) Recognizing that initial vertical velocities are 0 and calculating the vertical component for the 7 kg mass.
8 v_{1y} = -1.04,m/s Calculating the vertical velocity of the 4 kg mass (negative indicates opposite direction to the 7 kg mass’s vertical component).
9 v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} and \theta = \arctan\left(\frac{v_{1y}}{v_{1x}}\right) Combining horizontal and vertical components to find magnitude and direction of the 4 kg mass’s velocity.
10 v_1 \approx \sqrt{6.73^2 + (-1.04)^2} Plugging in horizontal and vertical components.
11 v_1 \approx 6.81,m/s Calculating the magnitude of velocity.
12 \theta \approx \arctan\left(\frac{-1.04}{6.73}\right) Calculating the direction of the velocity.
13 \theta \approx -8.8^\circ Determining the angle below the horizontal for the 4 kg mass’s velocity.

6.81 m/s at 8.8° above the horizontal

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6.81 m/s at 8.8° above the horizontal

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Nerd-Notes.com
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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