AP Physics

Unit 5 - Linear Momentum

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Step Derivation / Formula Reasoning
1 \[m_1 = 4\,\text{kg},\; m_2 = 7\,\text{kg},\; v_i = 10\,\text{m/s}\] Identify the masses and the initial velocity \(v_i\) of the \(4\,\text{kg}\) mass. The \(7\,\text{kg}\) mass is initially at rest.
2 \[p_{x,i}=m_1 v_i = 4(10)=40,\; p_{y,i}=0\] Calculate the initial momentum components. Motion is purely horizontal to the right, so the vertical component is zero.
3 \[v_{x2}=2\cos22^{\circ},\; v_{y2}=-2\sin22^{\circ}\] Resolve the \(7\,\text{kg}\) mass’s given final speed (\(2\,\text{m/s}\)) into horizontal and vertical components. The vertical component is negative (below the horizontal).
4 \[v_{x2}\approx1.854,\; v_{y2}\approx-0.749\] Numerical evaluation of the trigonometric components.
5 \[4 v_{x1}+7 v_{x2}=40\] Apply conservation of momentum in the \(x\)-direction: total initial \(p_x\) equals total final \(p_x\).
6 \[4 v_{y1}+7 v_{y2}=0\] Apply conservation of momentum in the \(y\)-direction: initial \(p_y\) is zero, so the final \(p_y\) must also be zero.
7 \[v_{x1}=\frac{40-7 v_{x2}}{4}\] Solve the \(x\)-momentum equation for the unknown horizontal component \(v_{x1}\) of the \(4\,\text{kg}\) mass.
8 \[v_{x1}=\frac{40-7(1.854)}{4}\approx6.755\,\text{m/s}\] Substitute \(v_{x2}\) and compute \(v_{x1}\).
9 \[v_{y1}=-\frac{7 v_{y2}}{4}\] Rearrange the \(y\)-momentum equation to isolate the vertical component \(v_{y1}\) of the \(4\,\text{kg}\) mass.
10 \[v_{y1}=-\frac{7(-0.749)}{4}\approx1.311\,\text{m/s}\] Insert \(v_{y2}\) and calculate \(v_{y1}\). The result is positive, meaning the mass moves upward after the collision.
11 \[v_x=\sqrt{v_{x1}^2+v_{y1}^2}\] Use the Pythagorean relation to find the magnitude \(v_x\) of the final velocity of the \(4\,\text{kg}\) mass.
12 \[v_x=\sqrt{(6.755)^2+(1.311)^2}\approx6.88\,\text{m/s}\] Compute the numerical value of the speed.
13 \[\theta=\tan^{-1}\!\left(\frac{v_{y1}}{v_{x1}}\right)\] Determine the direction angle \(\theta\) measured above the horizontal.
14 \[\theta=\tan^{-1}\!\left(\frac{1.311}{6.755}\right)\approx11^{\circ}\] Evaluate the inverse tangent to find the angle.
15 \[\boxed{v_x\approx6.9\,\text{m/s},\;\theta\approx11^{\circ}\,\text{above horizontal}}\] Present the final boxed answer: the speed and its angle relative to the horizontal.

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\(6.9\,\text{m/s}\)
\(11^{\circ}\text{ above horizontal}\)

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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