Open with a 45-second clip or still image of a large NFL lineman (≈ 140 kg) running at 4.0 m/s about to collide with a smaller running back (≈ 90 kg) sprinting the opposite way at 8.0 m/s. Ask the class to VOTE by hand: after they collide and lock together, which way does the pile move — toward the lineman's original direction, toward the running back's, or does it stop? Most students vote 'toward the lineman' because he's heavier. Reveal the momenta: p_lineman = (140)(+4.0) = +560 kg·m/s, p_back = (90)(−8.0) = −720 kg·m/s, so the pile moves in the RUNNING BACK's direction at about −0.70 m/s. The 'lighter' player wins the momentum battle. Tell students this is the puzzle we crack today: momentum is not about who is bigger — it's about mass AND velocity, and the TOTAL is what nature keeps constant.

1. 10m

   Momentum is a vector: p = mv

   Content

   Momentum is defined as p = mv, a vector quantity with the same direction as the velocity, measured in kg·m/s. In one dimension we track direction with a sign — rightward positive, leftward negative. A 1500 kg car moving at +20 m/s has p = +30,000 kg·m/s; the same car at −20 m/s has p = −30,000 kg·m/s. Total momentum of a system is the vector sum of every object's momentum: for two carts on a track, Σp = m₁v₁ + m₂v₂ where each velocity carries its own sign. Worked example: cart A is 0.50 kg moving at +0.60 m/s, cart B is 1.0 kg moving at −0.20 m/s. p_A = +0.30 kg·m/s, p_B = −0.20 kg·m/s, so Σp = +0.10 kg·m/s (net rightward). Notice the heavier cart does NOT carry the larger momentum here — velocity matters just as much.

   Delivery

   Anchor students on the vector nature — sign IS direction in 1-D, and dropping a sign is the single biggest source of wrong answers today. Do the two-cart sum on the board with them, then ask 'which cart has more momentum?' to force them to compute, not eyeball. Pre-empt the misconception that the heavier object always has more momentum: point back at the football hook, where the lighter player carried more.

2. 10m

   Impulse–momentum theorem: J = FΔt = Δp

   Content

   Newton's second law can be rewritten as ΣF = Δp/Δt, so ΣFΔt = Δp. The quantity FΔt is called impulse, symbol J, with units N·s that are equivalent to kg·m/s. Impulse is the AREA under a force-vs-time graph. Worked example: a 0.15 kg baseball comes in at −40 m/s and leaves the bat at +45 m/s. Δp = m(v_final − v_initial) = 0.15(45 − (−40)) = +12.75 kg·m/s. If the bat is in contact for 0.90 ms = 9.0 × 10⁻⁴ s, the average force is F = Δp/Δt = 12.75 / 9.0 × 10⁻⁴ ≈ 1.4 × 10⁴ N. The force-vs-time graph the slide shows is a spike whose SHADED AREA equals that 12.75 kg·m/s impulse.

   Delivery

   Emphasize that impulse gives you a bridge between forces (what Newton's laws talk about) and momentum change (what today's standard talks about). Have students note the sign of Δv — the ball reversed direction, so the change is much bigger than 45 − 40. Ask 'what happens to the required force if we double the contact time?' Answer: it halves. This sets up why airbags, crumple zones, and follow-through in sports work, which they'll see next class in HS-PS2-3.

3. 10m

   Why momentum is conserved: Newton's third law

   Content

   During a collision between objects 1 and 2, object 1 pushes object 2 with force F, and by Newton's third law object 2 pushes back with force −F. Both forces act for the SAME contact time Δt, so the impulses are equal and opposite: J on 2 = +FΔt, J on 1 = −FΔt. Therefore Δp₂ = −Δp₁, meaning the momentum GAINED by one is exactly the momentum LOST by the other. Adding: Δp₁ + Δp₂ = 0, so Σp_after = Σp_before. This argument only requires that the collision forces are internal to the system — if a net EXTERNAL force (like a shove from outside, or unbalanced friction) acts during the collision, that external impulse changes Σp. Real cart tracks are engineered flat and low-friction precisely to make external forces negligible over the millisecond of collision.

   Delivery

   This is the conceptual heart of the standard — spend time here. Ask students explicitly: 'When a truck hits a bike, is the force on the bike larger?' Most will say yes. Walk them through Newton's third law: forces are equal and opposite, but ACCELERATIONS are wildly unequal because masses differ. The bike's Δv is huge; the truck's Δv is tiny; but mΔv is the SAME magnitude for both. That's the misconception fix. Then define what makes today's lab a 'closed system': flat track, brief collision, negligible friction impulse.

4. 8m

   Elastic vs. inelastic collisions

   Content

   Momentum is conserved in EVERY collision where ΣF_external = 0 — elastic OR inelastic. What differs is kinetic energy. In an elastic collision (e.g., magnetic-bumper carts, hard steel balls), KE_before = KE_after. In an inelastic collision some KE converts to heat, sound, and permanent deformation, so KE_after < KE_before. In a PERFECTLY inelastic collision the objects stick together (Velcro-bumper carts, tackled football players) and share one final velocity. Worked example: a 1.0 kg cart at +2.0 m/s hits and sticks to a 1.0 kg cart at rest. Momentum: (1.0)(2.0) + 0 = (2.0)v_f → v_f = +1.0 m/s. KE before = ½(1.0)(2.0)² = 2.0 J. KE after = ½(2.0)(1.0)² = 1.0 J. Half the KE is GONE — momentum is still perfectly conserved.

   Delivery

   Hammer the point: 'momentum conserved' does NOT imply 'energy conserved.' Ask the class where the missing 1.0 J went. Expect answers: heat in the Velcro, sound of the stick, deformation. Warn against the common belief that inelastic collisions somehow violate conservation of momentum — they don't. This distinction is often on the CST/CAST-style items in the assessment.

1. 50m

   Collision Cart Lab: Elastic and Inelastic on a TrackLab

   Groups of 3. Each group runs three trials at three stations they cycle through: (1) elastic collision, (2) perfectly inelastic collision, (3) unequal-mass inelastic. They record masses and pre/post velocities, compute Σp and ΣKE for each trial, and decide whether momentum and/or KE are conserved. Before starting, groups LEVEL the track using the bubble level (a tilted track adds an external gravitational impulse and will visibly break conservation). Show them how to zero the motion sensor and set positive direction toward sensor B. Student handout: Collision Cart Lab — Conservation of Momentum Setup checklist - Track is level (bubble centered) checked before every trial - Motion sensors zeroed; +x direction is toward sensor B - Cart masses measured on the balance and recorded to 0.01 kg Part 1 — Elastic collision (magnetic bumpers face each other) - Cart 1 mass m₁ = ______ kg - Cart 2 mass m₂ = ______ kg - Cart 2 starts at rest at center of track. Push cart 1 gently toward it. - Record velocities from the sensor graphs at the instant just before and just after contact. - v₁ before = ______ m/s - v₂ before = 0 m/s - v₁ after = ______ m/s - v₂ after = ______ m/s - Compute: - Σp before = ______ kg·m/s - Σp after = ______ kg·m/s - % difference in momentum = ______ % - KE before = ______ J - KE after = ______ J - % difference in KE = ______ % Part 2 — Perfectly inelastic collision (Velcro tabs facing each other) - Same masses, same starting geometry. Push cart 1 into a stationary cart 2. They should stick. - v₁ before = ______ m/s - v combined after = ______ m/s - Σp before = ______ kg·m/s - Σp after = ______ kg·m/s (use combined mass m₁ + m₂) - % difference in momentum = ______ % - KE before = ______ J - KE after = ______ J - % of KE lost = ______ % Part 3 — Unequal-mass inelastic (load cart 2 with a 500 g bar) - m₁ = ______ kg, m₂ (loaded) = ______ kg - v₁ before = ______ m/s, cart 2 at rest - v combined after = ______ m/s - Predict v_combined BEFORE running the trial using conservation of momentum: predicted v = ______ m/s - Actual v combined = ______ m/s - % difference between prediction and measurement = ______ % Analysis questions — answer in complete sentences 1) In which parts was momentum conserved (within about 5%)? In which was kinetic energy conserved? 2) Explain any part where % difference in momentum exceeded 5%. What external force is the most likely culprit? 3) In Part 2, where did the missing KE go? Name two specific destinations. 4) Use Newton's third law to explain why, during the split-second of contact in Part 2, the impulse on cart 1 must be equal and opposite to the impulse on cart 2. Teacher circulation: watch for groups who forget to include the sign of v₁_after in the elastic trial (cart 1 usually bounces backward — negative). Also watch for the sneaky error of using m₁ instead of m₁ + m₂ in the after-momentum of the Velcro trial. Expect % differences of 2–6% in momentum and 30–55% KE loss in perfectly inelastic. If a group's momentum is off by 15%+, first check the track level.

   Materials

   • Dynamics track (leveled)
   • 2 dynamics carts per group (one with magnetic bumper, one with Velcro tab)
   • 2 motion sensors (one at each end of track) OR 2 photogates
   • Digital balance
   • Lab computer with data-collection software (LoggerPro / Capstone / Pasco Sparkvue)
   • Extra masses (200 g and 500 g bars)
   • Bubble level

   Example outputs

   • Part 1 sample: m₁ = 0.50 kg, m₂ = 0.50 kg, v₁_before = +0.42 m/s, v₁_after = −0.02 m/s, v₂_after = +0.41 m/s. Σp before = +0.210 kg·m/s, Σp after = +0.195 kg·m/s, 7% difference (slight friction impulse). KE before = 0.0441 J, KE after = 0.0421 J → nearly elastic.
   • Part 2 sample: m₁ = m₂ = 0.50 kg, v₁_before = +0.60 m/s, v_combined = +0.29 m/s. Σp before = +0.30 kg·m/s, Σp after = (1.00)(0.29) = +0.29 kg·m/s (3% diff). KE before = 0.090 J, KE after = 0.042 J → 53% of KE lost to heat/sound in the Velcro.
   • Part 3 sample: m₁ = 0.50 kg, m₂ = 1.00 kg, v₁_before = +0.80 m/s. Predicted v = 0.50(0.80)/1.50 = +0.27 m/s. Measured +0.26 m/s (4% diff).

   No-equipment fallback

   Use the PhET simulation 'Collision Lab' at https://phet.colorado.edu/en/simulations/collision-lab. Set 'elasticity' to 100% for Part 1, 0% for Parts 2 and 3, and complete the same handout with sim-read velocities. The analysis questions are unchanged.

2. 15m

   Exploding-Spring Cart Demo & Whiteboard AnalysisLab

   Whole-class demo, then group whiteboards. Place the light and heavy cart side by side at the center of the track, plunger compressed between them. Release. They fly apart. Motion sensors capture v_light and v_heavy. On each group's whiteboard, they must: 1) Predict which cart has the larger SPEED — light or heavy — and justify. 2) After the demo, record: m_light = ______ kg, m_heavy = ______ kg, v_light = ______ m/s, v_heavy = ______ m/s. 3) Compute p_light and p_heavy. What should they add to? Why? 4) Compute the KE of each cart. Where did that KE come from — it wasn't there before the release? 5) Answer: 'Was this a closed system? Justify using external forces.' Expected result: the lighter cart moves ~2× faster, but momenta are equal and opposite, summing to ≈ 0. KE came from the elastic PE stored in the compressed spring. Debrief on why the SAME mechanism (Newton's third law + equal contact time) makes the impulses equal and opposite, whether the interaction is a collision or an explosion.

   Materials

   • 2 dynamics carts of different mass (one loaded to ~2× the other)
   • Spring-plunger mechanism between carts (or trigger-release)
   • Motion sensors or two photogates
   • Whiteboards + markers per group

   Example outputs

   • m_light = 0.50 kg, m_heavy = 1.00 kg. After release: v_light = +0.60 m/s, v_heavy = −0.29 m/s. p_light = +0.30 kg·m/s, p_heavy = −0.29 kg·m/s. Σp ≈ +0.01 kg·m/s (~3% of individual value) — conserved.
   • KE_light = 0.090 J, KE_heavy = 0.042 J, total 0.13 J — all supplied by the compressed spring's elastic PE.

   No-equipment fallback

   Show a short (30 s) video of an air-track cart explosion or use PhET Collision Lab in 'explosion' mode (set both velocities to 0 and 'elasticity' to 100%, then click 'explode'). Groups do the same whiteboard analysis with sim values.

1. A 1200 kg car moving east at 15 m/s collides with and sticks to a 1800 kg truck moving west at 10 m/s. What is the velocity of the wreckage immediately after the collision? (Take east as positive.)

   calculationΣp before = (1200)(+15) + (1800)(−10) = 18,000 − 18,000 = 0 kg·m/s. Σp after = (1200 + 1800) v_f = 3000 v_f. Set equal: 3000 v_f = 0 → v_f = 0 m/s. The wreckage is momentarily at rest — the momenta canceled exactly.

2. Two carts collide on a frictionless track. Cart A (0.40 kg) moves right at +0.50 m/s; cart B (0.60 kg) is at rest. After the collision A moves right at +0.10 m/s and B moves right at +0.267 m/s. Was this collision elastic, inelastic, or perfectly inelastic? Justify with BOTH momentum and kinetic energy calculations.

   short answerMomentum: Σp before = (0.40)(0.50) + 0 = 0.20 kg·m/s. Σp after = (0.40)(0.10) + (0.60)(0.267) = 0.040 + 0.160 = 0.20 kg·m/s. ✓ conserved. Kinetic energy: KE before = ½(0.40)(0.50)² = 0.050 J. KE after = ½(0.40)(0.10)² + ½(0.60)(0.267)² = 0.002 + 0.0214 ≈ 0.0234 J. Momentum is conserved but KE dropped from 0.050 J to 0.023 J (about 53% lost). The carts did NOT stick together, so it is INELASTIC (not perfectly inelastic, not elastic).

3. Which statement about a head-on collision between a heavy truck and a light car is TRUE?

   multiple choiceCorrect answer: C. A) The truck exerts a larger force on the car than the car does on the truck. — WRONG (Newton's third law: forces are equal and opposite.) B) Momentum is conserved only if the collision is elastic. — WRONG (Momentum is conserved in ALL collisions with no net external force.) C) The magnitudes of the impulses on the car and truck are equal, and their momentum changes are equal and opposite. — CORRECT. D) The car experiences a larger change in momentum because it is lighter. — WRONG (Δp is equal in magnitude for both; the CAR just has larger Δv because its mass is smaller.)

• ☐Dynamics track (one per group of 3)
• ☐2 dynamics carts per group (magnetic bumper + Velcro bumper attachments)
• ☐Motion sensors or photogates (2 per track)
• ☐Digital balance (shared, 1 per 2 groups)
• ☐Bubble level (1 per track)
• ☐Extra mass bars: 200 g and 500 g (1 set per group)
• ☐Spring-plunger cart pair for exploding demo (1 for whole class)
• ☐Lab computer with data-collection software (LoggerPro, Capstone, or SparkVue)
• ☐Whiteboards + dry-erase markers (1 per group)
• ☐Printed lab handouts (1 per student)
• ☐Calculator (student-supplied)
• ☐Projector for slide deck