Show a short (30 s) clip or animation of an ice skater or diver performing a somersault: their limbs twist and flail wildly, but a red dot marking their center of mass traces a smooth parabola through the air. Ask students: 'The skater is spinning and contorting — how can any single point on her body follow such a clean, predictable curve?' Take 2–3 quick hypotheses. Do NOT resolve it; tell them by the end of class they will be able to explain exactly why that red dot obeys projectile motion even as the rest of the body writhes. Bridge: the red dot is the center of mass, and today we learn the rule that governs it.

1. 6m

   What a system is and why we choose one

   Content

   In physics we don't analyze 'the universe' — we choose a system, an imaginary dashed boundary around the objects we care about. Everything inside is part of the system; everything outside is the environment. The choice is ours, and it determines which forces count as internal (between parts of the system) and which count as external (from environment onto system). Consider two blocks on a frictionless track connected by a compressed spring. If the system is 'block 1 + spring + block 2,' the spring forces are internal. If the system is 'block 1 only,' the same spring force is now external. The physics doesn't change — only the bookkeeping does.

   Delivery

   Emphasize that 'system' is a CHOICE, not a physical thing — this is the modeling move that trips students up. Ask: 'If I pick just block 1 as the system, what happens to the spring force in my accounting?' (It becomes external.) Preview that the choice matters because Newton's second law will apply to the system's center of mass with only external forces on the right side. Foreshadow the exploding-cart demo they'll see later — the smart system choice makes the problem trivial.

2. 6m

   Locating the center of mass — weighted average

   Content

   The center of mass is the mass-weighted average position of all the particles in a system: x_cm = (m₁x₁ + m₂x₂ + … + mₙxₙ)/(m₁ + m₂ + … + mₙ). Worked example: two blocks on a track, m₁ = 2.0 kg at x₁ = 0.20 m and m₂ = 6.0 kg at x₂ = 1.00 m. Then x_cm = (2.0·0.20 + 6.0·1.00)/(2.0 + 6.0) = (0.40 + 6.00)/8.0 = 0.80 m. Notice it sits closer to the 6.0 kg block — three-quarters of the way from the lighter mass to the heavier one, because it's three times heavier. For continuous objects (rods, plates), the same idea holds but the sum becomes an integral; for AP Physics 1 we work with discrete objects or symmetric shapes where symmetry gives the answer. Key geometric consequences: (1) the center of mass shifts toward the heavier region — it is NOT the geometric center of an uneven object; (2) for a ring, a donut, or an L-shape the center of mass can sit in empty space, outside the material entirely, because it is a mathematical average, not a piece of matter.

   Delivery

   Work the two-block example on the board with students calling out the numbers. Then ask: 'Where would x_cm be if I doubled m₂ to 12 kg?' (Even closer to m₂ — students should feel the pull toward the heavier mass.) Explicitly confront the two misconceptions here: point to the visual of a ring with its center of mass at the hole's center — no matter there! — and to the L-plate with the cm marked off-material. Say plainly: 'Center of mass is a MATH point. It doesn't have to be where the stuff is.'

3. 8m

   Newton's second law for a system: ΣF_ext = M·a_cm

   Content

   Here is the payoff. When you apply Newton's second law to a whole system, every internal force cancels because internal forces come in Newton's-third-law pairs (block A pushes block B, block B pushes block A back — equal and opposite, they sum to zero inside the system). What survives on the left side is only the net EXTERNAL force. The result: ΣF_ext = M·a_cm, where M is the total mass of the system and a_cm is the acceleration of the center of mass. Consequences: (1) if ΣF_ext = 0, then a_cm = 0 — the center of mass moves at constant velocity (or stays at rest) no matter what wild internal motion is happening. Two skaters push off each other on ice: they fly apart, but their combined center of mass stays put. A firecracker explodes mid-flight: the fragments scatter, but the center of mass keeps sailing along the SAME parabola the whole firework was on. (2) The diver from our hook: gravity is the only external force during flight (air resistance neglected), so a_cm = g downward, and the center of mass traces a parabola — the arms and legs twist about it but can't change its path. Worked example: A 60 kg astronaut and a 30 kg tool float at rest in space. The astronaut throws the tool at 4.0 m/s to the right. What is the astronaut's velocity? System = astronaut + tool; ΣF_ext = 0, so total momentum stays zero, and v_cm stays zero. Then 60·v_astronaut + 30·(+4.0) = 0 → v_astronaut = −2.0 m/s (2.0 m/s to the left). The center of mass didn't budge.

   Delivery

   Walk the derivation reasoning slowly: 'Sum all forces on all parts. Internal ones cancel in pairs. Only externals remain.' Write ΣF_ext = M·a_cm large. Then hammer the two big consequences with the firecracker and skater examples. Attack the misconception head-on: 'Two carts push apart with a spring, no friction — does the center of mass move? Vote.' Many students say yes. Show why it CAN'T: no external horizontal force, so a_cm = 0. Emphasize that internal forces reshape the system but never move its center of mass.

1. 20m

   Exploding-cart investigation: does the center of mass move?Lab

   Setup and run (teacher): Place the track level. Put cart A (lighter, mass m₁ ≈ 0.50 kg) on the left half and cart B (heavier, mass m₂ ≈ 1.00 kg loaded) on the right half, touching with the spring plunger compressed between them. Position one motion sensor at each end of the track aimed inward. Zero the sensors when the carts are stationary in contact. Release the plunger — carts fly apart. Students record v₁ and v₂ from the motion-sensor graphs (or photogate speeds) as each cart moves at nearly constant velocity right after release. Groups repeat with a different mass ratio (e.g., swap loads so m₂ ≈ 1.50 kg). Targets SP 1 Creating Representations, SP 2 Mathematical Routines, and SP 3 Scientific Questioning and Argumentation. Students represent the system with a boundary sketch, compute predicted v_cm before release and after release, and argue from their data whether internal spring forces changed the motion of the center of mass. Walk around and check: (1) that groups draw the system boundary AROUND both carts so the spring force is internal; (2) that they carry units through the weighted-average calculation; (3) that they connect a_cm = 0 to the observed |v_cm| ≈ 0. Expect small nonzero v_cm (0.02–0.08 m/s) from track friction and measurement error — treat that as evidence of small external forces, not a failure. Debrief in the last 2 min: 'What external force would you need to make the center of mass actually accelerate?' (A push from outside, friction, gravity along a tilt.) Student handout: Part 1 — Set up your system Draw the two carts on the track. Draw a dashed loop around BOTH carts — that loop is your system boundary. Label: - m₁ = ______ kg (mass of cart A, measured) - m₂ = ______ kg (mass of cart B, measured) - The spring force between the carts is internal or external? ______ - Neglecting friction, the net EXTERNAL horizontal force on the system is ΣF_ext = ______ N Part 2 — Predict Before release, both carts are at rest. Compute the center-of-mass velocity of the system: - v_cm (before) = (m₁v₁ + m₂v₂)/(m₁ + m₂) = ______ m/s Using ΣF_ext = M·a_cm, predict v_cm AFTER the spring releases: - Prediction: v_cm (after) = ______ m/s - Justification (one sentence): ____________________ Part 3 — Measure Release the plunger. From the motion-sensor graphs, read each cart's speed just after separation (before it hits a stop): - v₁ (cart A, with sign — left is negative) = ______ m/s - v₂ (cart B, with sign) = ______ m/s - Measured v_cm (after) = (m₁v₁ + m₂v₂)/(m₁ + m₂) = ______ m/s Part 4 — Compare and argue - Predicted v_cm (after) = ______ m/s - Measured v_cm (after) = ______ m/s - Percent difference: ______ % Claim, evidence, reasoning (3–4 sentences): Did the internal spring force move the center of mass of the system? Use your numbers and Newton's third law to justify your answer. If the measured v_cm was not exactly zero, name a specific external force that could explain the discrepancy. Part 5 — Extension Repeat with a different mass ratio. Which cart moved faster — the heavier or the lighter? By what factor? Show that m₁|v₁| ≈ m₂|v₂| (which is exactly what v_cm = 0 demands).

   Materials

   • Two low-friction dynamics carts with a spring-loaded plunger on one (or a compressed spring between them)
   • Dynamics track (2 m) with end stops
   • Assorted masses (200 g, 500 g, 1000 g) to load one cart
   • Balance or electronic scale
   • Meter stick and masking tape
   • Two motion sensors OR two photogates (one on each side)
   • Data-collection software (LoggerPro / Capstone / Pasco)
   • Student handout (printed below)

   Example outputs

   • m₁ = 0.50 kg, m₂ = 1.00 kg; measured v₁ = −0.62 m/s, v₂ = +0.31 m/s; v_cm (after) = (0.50·(−0.62) + 1.00·(+0.31))/1.50 = 0.00 m/s. Claim: the internal spring force did not move the center of mass. Evidence: v_cm (after) ≈ 0, matching v_cm (before) = 0. Reasoning: the spring forces on the two carts form a Newton's-third-law pair internal to the system, so ΣF_ext ≈ 0 and a_cm = 0.
   • With m₂ loaded to 1.50 kg: v₁ = −0.75 m/s, v₂ = +0.25 m/s; v_cm = (0.50·(−0.75) + 1.50·(+0.25))/2.00 = 0.00 m/s. Cart A moved 3× faster than cart B because it was ⅓ the mass — mass ratio inversely tracks speed ratio, exactly what v_cm = 0 requires.

   No-equipment fallback

   Provide printed motion-sensor data (velocity-vs-time graphs for two carts before and after a spring release, with mass values given). Students read v₁ and v₂ off the graphs and complete Parts 1–4 above using the printed data instead of live measurements.

2. 15m

   Whiteboard problem set — locating the center of mass and ranking accelerations

   Students work in pairs on three short AP-style items. Give 3–4 min per item, then a 1-min whip-around where two pairs present. Targets SP 1 Creating Representations (Problem 1 — number-line sketch), SP 2 Mathematical Routines (Problems 1 & 2 — calculation and ranking), and SP 3 Scientific Questioning and Argumentation (Problem 3 — free-response justification). Student problem set: Problem 1 — Weighted average (SP 1, SP 2) Three point masses sit on the x-axis: - m₁ = 1.0 kg at x₁ = 0.0 m - m₂ = 3.0 kg at x₂ = 2.0 m - m₃ = 2.0 kg at x₃ = 5.0 m a) Sketch a number line and mark all three masses to scale. Estimate by eye where the center of mass should be, and mark it with a triangle (▲). b) Calculate x_cm. Show the substitution. c) Is x_cm closer to m₂ or m₃? Explain in one sentence why the answer makes sense. Problem 2 — Internal vs external ranking (SP 2, SP 3) A 2.0 kg block and a 6.0 kg block sit on a frictionless horizontal table connected by a compressed spring. The system is (block 1 + spring + block 2). In each scenario below, decide whether ΣF_ext on the system is zero or nonzero, and whether the center of mass accelerates. - Scenario A: Spring released, blocks fly apart on the frictionless table. - Scenario B: A hand pushes the 2.0 kg block to the right with 10 N while the spring is still compressed. - Scenario C: The whole apparatus is on a frictionless ramp tilted at 20°, spring still compressed. For each, write: ΣF_ext = ____, a_cm = ____ (zero / nonzero, and direction if nonzero). Problem 3 — Free-response justification (SP 3) A firework of total mass 2.0 kg is launched and follows a parabolic path. At the top of its arc, it explodes into two fragments of mass 0.5 kg and 1.5 kg. Air resistance is negligible. In 3–5 sentences, justify what happens to the CENTER OF MASS of the fragment system after the explosion. Reference (a) whether the explosion forces are internal or external, (b) Newton's second law for the system, and (c) the resulting path of the center of mass. Walk around and check: In Problem 2C, students often say ΣF_ext = 0 because 'nothing external is pushing' — remind them gravity along the ramp IS external. Expected answers below.

   Materials

   • Small whiteboards and markers (one per pair)
   • Printed problem cards (content below)

   Example outputs

   • Problem 1: x_cm = (1.0·0.0 + 3.0·2.0 + 2.0·5.0)/(1.0+3.0+2.0) = (0 + 6.0 + 10.0)/6.0 = 2.67 m. Closer to m₂ (2.0 m) because m₂ has the largest mass and pulls the average toward its position.
   • Problem 2: A) ΣF_ext = 0 (frictionless, gravity balanced by normal), a_cm = 0. B) ΣF_ext = 10 N right, a_cm = 10/8.0 = 1.25 m/s² right. C) ΣF_ext = Mg sin20° = 8.0·9.8·0.342 = 26.8 N down the ramp, a_cm = g sin20° = 3.35 m/s² down the ramp.
   • Problem 3: The explosion forces between the two fragments are INTERNAL to the fragment system and cancel in Newton's-third-law pairs. The only external force is gravity, unchanged by the explosion. By ΣF_ext = M·a_cm, the center of mass continues to accelerate downward at g and traces the SAME parabolic path the intact firework would have followed. The fragments individually take different paths, but their mass-weighted average position stays on the original parabola until a piece hits the ground.

1. Two point masses lie on the x-axis: a 4.0 kg mass at x = 1.0 m and a 1.0 kg mass at x = 6.0 m. What is the x-coordinate of the center of mass? (SP 2 Mathematical Routines)

   calculationx_cm = (4.0·1.0 + 1.0·6.0)/(4.0 + 1.0) = (4.0 + 6.0)/5.0 = 2.0 m. Note it lies closer to the heavier 4.0 kg mass, as expected.

2. A 70 kg student and a 50 kg student stand at rest on frictionless ice, facing each other. The 70 kg student pushes the 50 kg student, who slides away at 2.0 m/s. Which statement about the center of mass of the two-student system is correct? A) It accelerates in the direction of the push. B) It moves at 2.0 m/s in the direction the 50 kg student moves. C) It remains at rest because ΣF_ext = 0 on the system. D) It moves in the direction of the heavier student at a slower speed. (SP 3 Scientific Argumentation)

   multiple choiceC. The push between the students is an internal force pair; there is no net external horizontal force (frictionless ice), so ΣF_ext = 0 and v_cm stays 0. The 70 kg student must recoil at (50/70)·2.0 ≈ 1.43 m/s in the opposite direction to keep v_cm = 0.

3. A wooden ring lies flat on a table. A student claims, 'The center of mass of the ring can't be at the middle of the hole, because there's no material there — mass has to be located where there's something to have mass.' In 2–3 sentences, evaluate this claim. (SP 3 Scientific Argumentation)

   short answerThe claim is incorrect. The center of mass is a mass-weighted AVERAGE position, not a physical piece of matter — it is a mathematical point that summarizes where the mass is distributed. For a symmetric ring the average of all the mass elements sits at the geometric center of the ring, in the hole, even though no matter is there. The same is true for donuts, boomerangs, and L-shaped plates.

4. A 3.0 kg cart and a 1.0 kg cart are held together on a frictionless track with a compressed spring between them. When released, the 1.0 kg cart moves right at 6.0 m/s. Determine (a) the velocity of the 3.0 kg cart and (b) the velocity of the center of mass of the two-cart system after release. Justify part (b) using ΣF_ext. (SP 2, SP 3)

   short answer(a) The system starts at rest so v_cm = 0. Setting m₁v₁ + m₂v₂ = 0: 3.0·v + 1.0·(+6.0) = 0 → v = −2.0 m/s (2.0 m/s to the left). (b) v_cm = (3.0·(−2.0) + 1.0·(+6.0))/4.0 = 0 m/s. Justification: the spring force is internal to the system and the track is frictionless, so ΣF_ext = 0. By ΣF_ext = M·a_cm, a_cm = 0, and since v_cm was 0 before release it remains 0 after.

• ☐Two low-friction dynamics carts (one with spring-loaded plunger)
• ☐2 m dynamics track with end stops
• ☐Assorted cart masses: 200 g, 500 g, 1000 g
• ☐Balance or electronic scale (0.01 kg resolution)
• ☐Meter stick and masking tape
• ☐Two motion sensors OR photogate pair
• ☐Laptop with LoggerPro / Capstone data-collection software
• ☐Small whiteboards and dry-erase markers (one set per pair)
• ☐Printed student handouts for the exploding-cart lab
• ☐Printed whiteboard problem cards