Open by playing a 60-second clip (or describing, if no video) of a car crash test: a modern car hits a wall, the hood folds up, the airbags fire, the passenger dummy's head decelerates over ~100 ms. Then pose the paradox: 'The car is destroyed. Why is that GOOD news for the passenger?' Take 2-3 hands. Steer the discussion toward the idea that the passenger's momentum has to go to zero no matter what — the only variable engineers control is HOW LONG that takes. Introduce today's driving question: 'How can we design a bumper that stops a moving cart with the smallest possible peak force?' Tell students they will use a force probe on a wall to actually MEASURE what different bumper designs do, and iterate through at least two builds. Emphasize: today the destroyed bumper is the WIN.

1. 10m

   Impulse and the impulse-momentum theorem

   Content

   Momentum is p = mv, measured in kg·m/s. When a net force acts on an object for a time Δt, it changes the object's momentum by an amount called the impulse: J = FΔt = Δp. This is just Newton's second law rearranged — ΣF = ma becomes ΣF = Δp/Δt, so ΣFΔt = Δp. Units check: N · s = (kg·m/s²)(s) = kg·m/s ✓. Worked example: a 0.50 kg cart traveling at 1.2 m/s hits a wall and stops. Its momentum change is Δp = m(v − v₀) = 0.50 kg × (0 − 1.2 m/s) = −0.60 kg·m/s, so the impulse delivered by the wall on the cart is 0.60 N·s directed opposite the motion. That impulse is FIXED by the cart's mass and speed — no bumper design can change it.

   Delivery

   Anchor students on the idea that Δp is set by the collision itself, not by the bumper. Work the cart example on the board slowly, sign convention and all — honors students should be able to reproduce it. Ask: 'If the cart hits the wall at the same speed but a bumper is added, what changes and what does not?' The answer you want: Δp does not change, but Δt (and therefore F) does. This is the pivot the whole lesson rests on. Head off the misconception that a bumper reduces Δp — it does not, unless it makes the collision inelastic in a different way (which we hold constant today by always stopping the cart).

2. 10m

   Force-time tradeoff: why cushions beat rigid walls

   Content

   Rearrange the impulse equation for average force: F = Δp/Δt. For a fixed Δp, force and contact time are inversely related — double the stopping time and you halve the average force. On a force-vs-time graph the AREA under the curve equals the impulse. A rigid collision produces a tall, narrow spike; a cushioned collision produces a shorter, wider bump. Both have the same area (same Δp), but the cushioned one has a much lower peak force. Worked comparison using the same 0.60 N·s impulse from the previous example: a rigid stop over Δt = 0.020 s gives an average force of 0.60/0.020 = 30 N; a cushioned stop over Δt = 0.10 s gives 0.60/0.10 = 6.0 N — five times smaller. That's the entire principle behind crumple zones, airbags, gymnasium mats, running shoes, and playground mulch.

   Delivery

   The two force-vs-time curves on the slide are the whole conceptual payoff — make sure every student can articulate 'same area, different peak.' Ask them to predict: 'If I double the contact time, what happens to the peak force?' (halves). 'If I make the bumper twice as stiff, what happens?' (peak force goes up). Pre-empt the very common misconception that a stiffer, harder bumper protects better — students' intuition says 'stronger material = safer,' but the physics says the opposite: rigidity shortens Δt and raises F. Also head off the idea that airbags work because they are soft — they work because they extend Δt (and spread force over area, lowering pressure on the skull).

3. 10m

   Engineering design: criteria, constraints, and iteration

   Content

   Engineering design turns a physics principle into a testable object. It has four moves. (1) Criteria — the measurable goals: today, minimize peak force on the cart while still stopping it. (2) Constraints — the limits: bumper attaches to the front of the cart, total added mass ≤ 30 g, materials only from the supplied bin, must fit inside a 10 cm × 5 cm × 5 cm envelope. (3) Test — collect force-vs-time data with the wall-mounted force probe at a fixed release point on the ramp so incoming speed is repeatable. (4) Iterate — read the data, identify what limited performance (peak too high? bumper bottomed out? cart bounced?), change ONE variable, and retest. Two important honors-level ideas: change one variable at a time so causal claims are valid, and remember that a bumper that dissipates energy (crushes, tears, deforms permanently) will outperform one that stores and returns energy (a spring bounces the cart back, doubling Δp).

   Delivery

   Emphasize the phrase 'change one variable per iteration' — sloppy iteration is the most common failure mode in this lab. Ask: 'What would it mean if your cart bounces backward off the bumper?' (Δp is now larger than mv₀, so the impulse is bigger, and peak force is likely higher — bouncy is bad.) Show students where the force probe is mounted, how to zero it, and where the release line on the ramp is. Set the expectation of at least two iterations with data-based reasoning, not just 'this one felt better.'

1. 65m

   Bumper Design Lab: Build, Measure, IterateLab

   Groups of 3. Each group builds two bumper iterations and collects force-vs-time data for each. Setup (teacher, before class): Mount the force probe on a rigid wall or heavy block at the end of each track. Mark a release line on the ramp so every trial starts at the same height (aim for about 1.0-1.3 m/s at the bottom — verify once with a motion sensor). Zero each probe. Preload the data-collection file to sample at ≥ 1000 Hz over a 0.5 s window triggered at 0.5 N. Student handout — hand this out at the start of the activity: Bumper Design Lab Goal: Design a bumper attached to the front of a dynamics cart that MINIMIZES the peak force recorded by the wall-mounted force probe when the cart is released from the marked line on the ramp. Criteria - Minimize peak force (N) — this is the score - Cart must fully stop at the wall (no bounce-back past a line 10 cm behind the wall) Constraints - Bumper mass ≤ 30 g (weigh it — record the value) - Bumper must fit inside a 10 cm × 5 cm × 5 cm envelope - Use ONLY materials from the supplied bin - Cart must be released from the marked line — no pushing Change only ONE variable between Iteration 1 and Iteration 2. You must be able to name that variable. Part 1 — Predict (5 min) 1. Cart mass m = ______ kg 2. Estimated speed at bottom of ramp v₀ ≈ ______ m/s (your teacher will give a class value) 3. Predicted Δp for the collision = ______ kg·m/s 4. If the collision lasts 0.05 s, the predicted average force is ______ N. Show your work. 5. Sketch, in words, the bumper you will build first and explain WHY you think it will lower the peak force. Reference contact time in your answer. Part 2 — Iteration 1: Build and Test (20 min) - Build Bumper 1. Weigh it and record mass: ______ g - Attach to the front of the cart with tape (tape counts toward the 30 g) - Release the cart from the marked line 3 times. Record for each trial: - Peak force (N): trial 1 ______ trial 2 ______ trial 3 ______ - Contact duration Δt (width of the pulse at the baseline, in s): ______ ______ ______ - Did the cart bounce back past the line? Y / N - Compute the average peak force across the 3 trials: ______ N - From your F-vs-t graph, estimate the impulse (area under the curve) for trial 2: ______ N·s - Does this impulse match your predicted Δp within ~20%? If not, suggest why. Part 3 — Analyze and Redesign (10 min) Answer BEFORE building Iteration 2: 1. What was the peak force limited by? Pick one and defend it in a sentence: - Bumper too stiff (short Δt) - Bumper bottomed out (crushed fully before cart stopped) - Cart bounced (Δp larger than expected) - Bumper too soft (cart penetrated to hard cart body) 2. State the ONE variable you will change in Iteration 2 and predict its effect on peak force. Use the equation F = Δp/Δt in your reasoning. Part 4 — Iteration 2: Build and Test (15 min) - Build Bumper 2 (change only the one variable). Mass: ______ g - Run 3 trials as before. Record peak force and Δt for each. - Average peak force, Iteration 2: ______ N - Percent change from Iteration 1: ______ % Part 5 — Claim, Evidence, Reasoning (15 min) Write 4-6 sentences. - Claim: Which bumper better protects the cart, and by how much? - Evidence: Cite specific peak force values and Δt values from your data. - Reasoning: Use FΔt = Δp explicitly. Explain why Δp was approximately the same for both iterations and why F differed. - Refinement: If you had one more iteration, what would you change and what peak force would you predict? Safety reminder: keep hands clear of the space between the cart and the force probe — moving carts can pinch fingers, and springs under compression can release suddenly. Teacher moves during the lab: - First 10 min: circulate to confirm every group has release line, zeroed probe, and a first sketch before they touch materials. - Middle 30 min: watch for groups changing two variables at once — stop them and make them pick one. - Watch for bouncy designs (springs, tightly wrapped rubber bands). Ask those groups to compute Δp assuming the cart rebounds at half its incoming speed — they should see Δp goes up by 50%. - Last 20 min: push groups from 'ours was lower' to quantitative CER with equation reasoning.

   Materials

   • Low-friction dynamics cart (~0.5 kg) per group
   • Ramp or track with fixed release line
   • Wall-mounted force probe (Vernier/PASCO) + interface + laptop
   • Graphing software (Logger Pro, Capstone, or equivalent)
   • Balance (0.1 g resolution)
   • Meter stick and calipers
   • Masking tape, rubber bands, paper clips
   • Bumper materials bin per group: cotton balls, foam sheet, bubble wrap, cardboard strips, drinking straws, popsicle sticks, small springs, sponge pieces, tissue paper, mini marshmallows

   Example outputs

   • Iteration 1 (bare cart with tape only): peak F = 28.4 N averaged over 3 trials, Δt ≈ 0.018 s. Iteration 2 (three stacked cotton balls in a cardboard cup, 8 g): peak F = 6.2 N, Δt ≈ 0.085 s. CER: 'Δp was the same (~0.55 kg·m/s from m = 0.50 kg and v₀ ≈ 1.1 m/s). Iteration 2's Δt was ~4.7× longer, so F dropped by ~4.6× — matches FΔt = Δp.'
   • Iteration 1 (small spring): peak F = 22 N, cart bounced back 30 cm. Iteration 2 (spring replaced with crumpled tissue behind foam): peak F = 9 N, no rebound. CER cites that the bouncy spring made Δp larger (cart reversed direction), so both Δp and F/Δt worked against them; the crushable design absorbed energy plastically and extended Δt.
   • Iteration 1 (thick foam block, 25 g): peak F = 11 N, Δt ≈ 0.06 s, but bumper bottomed out. Iteration 2 (same foam + straws oriented axially as a crumple structure): peak F = 4.8 N, Δt ≈ 0.11 s. Refinement proposed: add a second stage of softer material behind the straws to smooth the tail of the pulse.

   No-equipment fallback

   If force probes are unavailable, run the lab with an egg or a raw tomato taped to the front of the cart and measure stopping distance d (from wall contact to final rest) instead of force. Peak force can be estimated from work-energy: F ≈ ½mv₀²/d. Groups still iterate twice, compute F for each iteration, and complete the same CER.

1. A 1200 kg car traveling at 15 m/s crashes head-on into a wall and comes to rest. (a) Calculate the impulse delivered to the car. (b) If the crumple zone extends the collision time from 0.040 s to 0.15 s compared to a rigid front end, calculate the average force on the car in each case. (c) Explain, using FΔt = Δp, why the crumple zone protects the passengers.

   calculation(a) Δp = m(v − v₀) = 1200 × (0 − 15) = −18000 kg·m/s, so impulse magnitude = 1.8 × 10⁴ N·s. (b) Rigid: F = Δp/Δt = 18000/0.040 = 4.5 × 10⁵ N. Crumple zone: F = 18000/0.15 = 1.2 × 10⁵ N. (c) Δp is fixed at 1.8 × 10⁴ N·s because it depends only on the car's mass and speed. The crumple zone extends Δt by a factor of ~3.75, so from F = Δp/Δt the average force (and therefore the deceleration a = F/m felt by the passengers) drops by the same factor — about 3.75× smaller. Lower peak force means less injury.

2. Two bumper designs are tested on identical carts at identical speeds. Design A's force-vs-time graph is a tall narrow spike peaking at 40 N and lasting 0.02 s. Design B's is a shorter wider bump peaking at 10 N and lasting 0.08 s. Which design better protects the cart, and what MUST be true about the areas under the two curves? Justify.

   short answerDesign B better protects the cart because its peak force (10 N) is one-fourth of Design A's (40 N), so the peak deceleration is smaller. The AREAS under the two curves must be approximately equal because both designs stop the same cart from the same speed — so both deliver the same impulse (Δp is fixed by mass and initial velocity). Design B spreads that same impulse over a longer contact time, which is exactly why its peak force is lower: F = Δp/Δt.

3. A student claims: 'To make the safest bumper, we should use the stiffest, hardest material we can find because it can withstand the biggest force.' Identify the misconception and correct it using the impulse-momentum theorem.

   short answerThe misconception is that a stiffer bumper protects better. In fact, a stiffer material stops the cart in a SHORTER contact time Δt. Since Δp is fixed by the collision, F = Δp/Δt means a smaller Δt produces a LARGER peak force on the cart. Effective safety devices (crumple zones, airbags, padding) do the opposite — they deform or compress to EXTEND Δt, lowering peak force. The bumper being 'able to withstand' force is irrelevant; what matters is the force transferred to the protected object.

4. Which statement best explains how an airbag reduces injury in a car crash?

   multiple choiceC. It increases the time and area over which the occupant decelerates, lowering peak force and pressure. (Distractors: A) It reduces the occupant's change in momentum — WRONG, Δp is set by the crash. B) It absorbs all of the kinetic energy so the occupant does not decelerate — WRONG, the occupant still decelerates to zero. D) It pushes the occupant backward with an equal and opposite force — WRONG, this misapplies Newton's third law.)

• ☐Dynamics carts (one per group, ~0.5 kg)
• ☐Tracks or ramps with marked release line
• ☐Wall-mounted force probes + interfaces + laptops with graphing software
• ☐Balances (0.1 g resolution)
• ☐Meter sticks and calipers
• ☐Masking tape
• ☐Bumper materials bin per group: cotton balls, foam sheet, bubble wrap, cardboard, straws, popsicle sticks, small springs, sponges, tissue paper, mini marshmallows
• ☐Printed student handout (Parts 1–5)
• ☐Optional: motion sensor to verify incoming cart speed
• ☐Whiteboard or projector for the two force-vs-time curves