Open with a real safety statistic: a car at 60 mph does not take twice the stopping distance of a car at 30 mph — it takes about four times the distance. Ask students why highway speed limits jump so cautiously (25 → 35 → 55 → 65) rather than doubling freely. Take 2–3 predictions on the board (expect answers like 'more time to react' or 'more force needed'). Do NOT resolve — tell them by the end of class they will have derived the exact reason from a single equation. Tie the phenomenon to today's target: kinetic energy scales with v², not v.

1. 6m

   Defining Translational Kinetic Energy

   Content

   Translational kinetic energy is the energy an object carries because its center of mass is moving through space. It is defined as KE = ½mv², where m is the object's mass in kilograms and v is its instantaneous speed in meters per second. The result is measured in joules, where 1 J = 1 kg·m²/s². Kinetic energy is a scalar quantity — it has magnitude only, no direction, and it is never negative because v is squared. A 2.0 kg cart moving at 3.0 m/s has KE = ½(2.0 kg)(3.0 m/s)² = 9.0 J. If that same cart moves in the opposite direction at 3.0 m/s, its KE is still 9.0 J — direction is discarded when we square the speed.

   Delivery

   Emphasize the three things that determine KE: mass, speed, and NOTHING else about direction. Walk through the 9.0 J calculation slowly, saying the units out loud so students see kg·m²/s² collapse to J. Pre-empt the misconception that reversing direction makes KE negative — ask 'what happens to (−3)² ?' and let them answer 4. Anchor: KE ≥ 0, always.

2. 7m

   The Quadratic Dependence on Speed

   Content

   Because v is squared, kinetic energy grows much faster than speed. Doubling v multiplies KE by 2² = 4. Tripling v multiplies KE by 3² = 9. Halving v cuts KE to one-quarter. Consider a 1500 kg car: at 15 m/s, KE = ½(1500)(15)² = 168{,}750 J ≈ 1.7 × 10⁵ J. At 30 m/s (double the speed), KE = ½(1500)(30)² = 675{,}000 J ≈ 6.75 × 10⁵ J — exactly four times as much. This is why the 60-mph car in the hook takes four times the stopping distance: the brakes must remove four times the energy. A graph of KE versus v is a parabola, curving upward with an ever-steepening slope; a graph of KE versus v² is a straight line through the origin with slope ½m.

   Delivery

   This is the load-bearing beat of the lesson — spend real time here. Ask students to predict the KE ratio if speed triples BEFORE you show it (target: 9×). Point at the parabola on the slide and trace how a small increase in v at high speeds produces a huge jump in KE. Head off the misconception that KE ∝ v directly — have students verbally state 'square the ratio of the speeds' as the rule. Connect back to the hook: brakes dissipate KE, so 4× KE means 4× stopping distance at constant braking force.

3. 4m

   Mass Matters Too — But Only Linearly

   Content

   Kinetic energy scales linearly with mass: doubling m doubles KE at the same speed. A 70 kg sprinter running at 10 m/s carries KE = ½(70)(10)² = 3500 J. A 140 kg lineman running at the same 10 m/s carries 7000 J — twice as much, because his mass is twice as large but the v² factor is identical. However, a light-and-fast object can easily out-carry a heavy-and-slow one: a 0.045 kg golf ball at 70 m/s has KE = ½(0.045)(70)² ≈ 110 J, more than a 5.0 kg bowling ball at 6.0 m/s, which has only 90 J. Because of v², speed usually wins.

   Delivery

   Emphasize the asymmetry — mass is linear, speed is quadratic. Ask 'which would you rather be hit by: the golf ball or the bowling ball?' Let students argue, then compute. This puzzles them productively. Reinforce that KE ranking requires plugging in numbers — you cannot rank by speed alone or mass alone.

4. 3m

   Energy Bar Charts as a Representation

   Content

   An energy bar chart is a single stacked bar showing how much energy of each type a system has at one instant. For a cart moving on a level track with only translational KE, the bar has one column labeled KE whose height is proportional to ½mv². When the cart's speed doubles, the bar quadruples in height — not doubles. Bar charts are a required AP representation (SP 1) and help students SEE the nonlinear scaling before they commit to conservation-of-energy problems next class.

   Delivery

   Reinforce that bar heights are proportional, not exact — students often try to draw perfectly to scale. What matters is the RATIO: a 2× speed change means a 4× taller bar. Preview: tomorrow we add work and potential energy bars.

1. 25m

   Photogate Cart Lab: Testing KE ∝ v²Lab

   Set up: Each group gets a track inclined at a fixed low angle (about 5–8°), a photogate mounted near the bottom, two carts of different masses, and a balance. Groups release the SAME cart from four different heights along the ramp (measured as distance up the track: 10 cm, 20 cm, 30 cm, 40 cm) to produce four different exit speeds v at the photogate. Then they swap to the heavier cart and repeat one trial for comparison. This targets SP 1 (build a KE-vs-v² graph), SP 2 (compute KE from measured v and m), and SP 3 (justify the ½mv² model from data). Walk around and check: (a) students measure the flag width and enter it correctly, (b) they compute v = flag width / photogate time, (c) their KE-vs-v² graph is linear and the slope matches ½m within ~15%. Student handout: Part 1 — Setup and measurements - Mass your assigned cart on the balance. Record: m = ______ kg - Measure the width of the photogate flag: w = ______ m - Set the track angle so the cart rolls smoothly but not too fast. Part 2 — Data collection (same cart, four release points) Release the cart from rest at each distance d up the track. Record the photogate time t. Compute v = w / t. Compute KE = ½mv². - Trial 1: d = 0.10 m, t = ______ s, v = ______ m/s, v² = ______ m²/s², KE = ______ J - Trial 2: d = 0.20 m, t = ______ s, v = ______ m/s, v² = ______ m²/s², KE = ______ J - Trial 3: d = 0.30 m, t = ______ s, v = ______ m/s, v² = ______ m²/s², KE = ______ J - Trial 4: d = 0.40 m, t = ______ s, v = ______ m/s, v² = ______ m²/s², KE = ______ J Part 3 — Graph and analyze 1. Plot KE (y-axis) versus v² (x-axis). Draw a best-fit line. 2. Determine the slope of your line: slope = ______ J·s²/m² 3. According to KE = ½mv², the slope should equal ½m. Compute ½m for your cart: ½m = ______ kg 4. Compare: % difference = ______ % Part 4 — Second cart comparison Swap to the second cart (different mass). Record: m₂ = ______ kg. Release from d = 0.30 m. - t = ______ s, v = ______ m/s, KE = ______ J - Compare to your Trial 3 KE from the first cart. Which cart carries more KE? Explain in one sentence why, referencing both mass and speed. Part 5 — Argument (SP 3) In 2–3 sentences, argue from your data whether KE is proportional to v or to v². Cite at least one specific pair of trials as evidence. Do not release carts until the photogate is armed and the catch pad is in place at the end of the track.

   Materials

   • Dynamics carts (2 different masses per group, e.g., 0.50 kg and 1.00 kg)
   • Ramp or track with adjustable incline
   • Photogate with picket fence flag or single-flag timer
   • Meter stick
   • Digital balance (0.01 kg resolution)
   • Calculator
   • Student handout (one per student)

   Example outputs

   • Trial 2 sample: m = 0.50 kg, d = 0.20 m, t = 0.042 s (flag w = 0.025 m) → v = 0.60 m/s, v² = 0.36 m²/s², KE = 0.090 J. Trial 4: v ≈ 0.85 m/s, KE ≈ 0.18 J — roughly double the KE for √2× the speed, consistent with v² scaling.
   • Part 5 argument: 'Doubling v² from 0.18 to 0.36 m²/s² doubled KE from 0.045 to 0.090 J — a linear relationship on the KE-vs-v² plot. If KE were proportional to v, doubling v² (which is only √2× v) should have multiplied KE by √2 ≈ 1.4, not 2. The data supports KE ∝ v².'
   • Part 3 slope check: measured slope = 0.24 J·s²/m², ½m = 0.25 kg, % difference ≈ 4% — within experimental uncertainty, confirming KE = ½mv².

1. A 1200 kg car increases its speed from 10 m/s to 30 m/s. By what factor does its translational kinetic energy increase? (A) 3 (B) 6 (C) 9 (D) 27

   multiple choice(C) 9. KE ∝ v², and the speed ratio is 30/10 = 3, so the KE ratio is 3² = 9. Targets SP 2 Mathematical Routines.

2. Rank the following objects from LEAST to GREATEST translational kinetic energy. Show KE for each. - Object 1: m = 4.0 kg, v = 5.0 m/s - Object 2: m = 2.0 kg, v = 8.0 m/s - Object 3: m = 8.0 kg, v = 3.0 m/s

   calculationKE₁ = ½(4.0)(5.0)² = 50 J KE₂ = ½(2.0)(8.0)² = 64 J KE₃ = ½(8.0)(3.0)² = 36 J Ranking (least → greatest): Object 3 (36 J) < Object 1 (50 J) < Object 2 (64 J). Note that Object 3 has the greatest mass but the lowest KE because v² dominates. Targets SP 2 and SP 3.

3. A student claims: 'Cart A moves east at 4.0 m/s and Cart B moves west at 4.0 m/s, so their kinetic energies are opposite in sign and cancel out.' Explain what is wrong with this claim in 2–3 sentences, citing the definition of KE.

   short answerKinetic energy is a scalar defined by KE = ½mv², where v is squared. Squaring any speed — positive or negative direction — gives a positive number, so KE is always ≥ 0 and has no direction. Both carts have identical, positive kinetic energies; energies do not cancel by direction the way momenta do. Targets SP 3 Scientific Argumentation.

4. An AP student measures a 0.500 kg cart's speed at the bottom of a ramp to be 1.20 m/s. Determine the cart's translational kinetic energy, with correct units and appropriate significant figures.

   calculationKE = ½(0.500 kg)(1.20 m/s)² = ½(0.500)(1.44) = 0.360 J. Report to 3 significant figures because the given data has 3 sig figs. Targets SP 2 Mathematical Routines.

• ☐Dynamics carts, at least 2 masses per group
• ☐Track/ramp with adjustable incline
• ☐Photogate + picket-fence flag (or single flag) per group
• ☐Digital balance
• ☐Meter stick
• ☐Calculators
• ☐Printed student handout (photogate lab)
• ☐Whiteboard/projector for slide deck