Apply \(F_{\text{net}} = m \cdot a\) to each isolated block using your calculated system acceleration from Page 1:

Block A (\(m_1 = 3\text{ kg}\)) Directly Pushed

A (3 kg) F = 20 N F_B on A

1. Calculations of Net Force: \(F_{\text{net}, A}\)

\(F_{\text{net}, A} =\) \(\text{N}\)

2. Contact Force Calculation: \(F_{B \text{ on } A}\)

Force equation: \(F_{\text{push}} - F_{B \text{ on } A} = F_{\text{net}, A}\)

\(F_{B \text{ on } A} =\) \(\text{N}\)

Block B (\(m_2 = 2\text{ kg}\)) Pushed via Contact

B (2 kg) F_A on B Frictionless

1. Calculations of Net Force: \(F_{\text{net}, B}\)

\(F_{\text{net}, B} =\) \(\text{N}\)

2. Contact Force Calculation: \(F_{A \text{ on } B}\)

Force equation: \(F_{A \text{ on } B} = F_{\text{net}, B}\)

\(F_{A \text{ on } B} =\) \(\text{N}\)

✓

Calculations of Net Force: System Verification

\(F_{\text{net}, A} + F_{\text{net}, B} =\) \(+\) \(=\) (\(\equiv F_{\text{push}} = 20\text{ N}\))

3rd Law Pair Check

The Universal Multi-Mass Problem Solving Protocol

Step 1: System Level

Add all masses (\(m_{\text{tot}} = \Sigma m\)). Ignore internal forces. Solve for system acceleration: \(a = \frac{F_{\text{ext, net}}}{m_{\text{tot}}}\).

Step 2: Assign \(a\)

Every mass in the connected group shares this exact acceleration (\(a_i = a_{\text{sys}}\)).

Step 3: Component Level

Isolate any single object. Calculate its required \(F_{\text{net}} = m_i \cdot a\). Solve for internal contact force.

Concept Check: Push applied from the right onto Block B instead?

The total mass (\(5\text{ kg}\)) and acceleration remain identical. However, Block B now propels Block A with contact force \(F_{\text{net}, A} = m_A \cdot a = 3\text{ kg} \times a_{\text{sys}}\). Internal contact forces depend directly on which masses are being propelled ahead!

Key Rule: System acceleration is uniform; net force distributes in direct proportion to mass. Newton's Laws • Dynamics Unit

3

Concept 3: Free-Body Diagrams for Each Mass

For an ideal massless cord, tension \(T\) is uniform throughout. Writing Newton's Second Law along each object's motion:

Mass 1 (\(m_1\)) • Ascending Net Force Up

m₁ T m₁g

Equation of Motion: \(T - m_1 g = m_1 a\)

Solving for Tension: \(T = m_1(g + a)\)

Mass 2 (\(m_2\)) • Descending Net Force Down

m₂ T m₂g

Equation of Motion: \(m_2 g - T = m_2 a\)

Solving for Tension: \(T = m_2(g - a)\)

T

The Universal Cord Tension Formula

Substituting \(a = \frac{m_2 - m_1}{m_1 + m_2}g\) into either equation yields: \(T = \frac{2 m_1 m_2}{m_1 + m_2} g\)

Harmonic Mean

Guided Calculation Practice: Given \(m_1 = 2.0\text{ kg}\), \(m_2 = 3.0\text{ kg}\), \(g = 10\text{ m/s}^2\)

Show all steps & units

1. Total Mass & Net Force

Calculate \(m_{\text{tot}}\) and \(F_{\text{net, ext}}\):

\(m_{\text{tot}} =\) \(\text{kg}\)

\(F_{\text{net}} =\) \(\text{N}\)

2. Calculations for Acceleration

Apply \(a = \frac{F_{\text{net}}}{m_{\text{tot}}}\):

\(a_{\text{sys}} =\) \(\text{m/s}^2\)

3. Calculations of Tension

Verify \(T\) on either mass:

\(T =\) \(\text{N}\)

Variant: The Half-Atwood (Table Pulley System)

If \(m_1\) sits on a frictionless tabletop and \(m_2\) hangs over the edge, only the hanging weight provides driving force: \(F_{\text{net, ext}} = m_2 g\). The total mass remains \((m_1 + m_2)\), yielding acceleration \(a = \frac{m_2 g}{m_1 + m_2}\).

Key Takeaway: In Atwood machines, gravity drives the difference; both masses share the inertia. Newton's Laws • Dynamics Unit

Turn to Page 2 for Atwood Machine Solutions, Cord Tension Derivations & Rubric Coupled Mass Dynamics • Answer Key

Teacher Resource • Complete Solutions AP Physics 1

Coupled Mass Dynamics Answer Key: Part 2

Atwood Machine Solutions

Page 2 of 2

Guide 2 Solutions: Atwood Machine System Acceleration & Tension Analysis

Max Points: 10 pts

1 Atwood System Expressions (Page 1 of Student Guide) 3 Points Total

Total System Mass (\(m_{\text{total}}\)) [1 pt]

\(m_{\text{total}} = m_1 + m_2\)

Both masses contribute to total inertia.

Net Driving Force (\(F_{\text{net, ext}}\)) [1 pt]

\(F_{\text{net}} = (m_2 - m_1)g\)

Difference between opposing weights.

System Acceleration (\(a_{\text{sys}}\)) [1 pt]

\(a = \frac{m_2 - m_1}{m_1 + m_2}g\)

Master unwrapped coordinate formula.

2 Guided Numerical Calculations Solutions (Page 2) 5 Points Total

Given Parameters: \(m_1 = 2.0\text{ kg}\), \(m_2 = 3.0\text{ kg}\), \(g = 10\text{ m/s}^2\)

1. Mass & Net Force [1.5 pt]

\(m_{\text{tot}} = 2.0 + 3.0 = \mathbf{5.0\text{ kg}}\)

\(F_{\text{net}} = (3.0 - 2.0)(10) = \mathbf{10\text{ N}}\)

Full credit requires both quantities correct with units.

2. Acceleration [1.5 pt]

\(a = \frac{F_{\text{net}}}{m_{\text{tot}}} = \frac{10\text{ N}}{5.0\text{ kg}}\)

\(a_{\text{sys}} = \mathbf{2.0\text{ m/s}^2}\)

Check that student divided by total mass (\(5.0\text{ kg}\)), not just \(m_2\).

3. Cord Tension [2.0 pt]

\(T = m_1(g + a)\)

\(= 2.0(10 + 2.0) = \mathbf{24\text{ N}}\)

Or: \(3.0(10 - 2.0) = 24\text{ N}\)

Accept either mass equation or universal formula.

3 Conceptual Rigor: Tension Bounds Check [2 Points] Rubric Criteria

Physical Intuition Check: \(20\text{ N} < 24\text{ N} < 30\text{ N}\)

Verify that student recognizes why tension must be strictly bounded between the two gravitational weights: \(m_1 g = 20\text{ N}\) (upward acceleration requires \(T > 20\text{ N}\)) and \(m_2 g = 30\text{ N}\) (downward acceleration requires \(T < 30\text{ N}\)). An answer of \(T = 25\text{ N}\) (the average) shows faulty static equilibrium thinking.

AP Physics 1 Free-Response Scoring Alignment

System Object Identification: Students must clearly specify whether Newton's 2nd law is being applied to the combined system or a single isolated component before writing equations.

Newton's 3rd Law Consistency: In contact forces (\(F_{A \text{ on } B} = F_{B \text{ on } A}\)) and cord tension, internal interaction forces must be equal in magnitude and opposite in direction.

Classroom Resource • Complete Coupled Mass Dynamics Answer Key Coupled Mass Dynamics • Answer Key