Open with a real-world stopping-power question: a 7.3 kg bowling ball rolls down the lane at 8.0 m/s; a 0.145 kg baseball is pitched at 45 m/s (≈100 mph). Ask students, on a scrap of paper, which one would be harder to stop, and to defend the answer with a number. Take a quick show of hands, then reveal both quantities: p_ball = 58 kg·m/s vs p_baseball = 6.5 kg·m/s (bowling ball wins), but KE_ball = 234 J vs KE_baseball = 147 J (bowling ball still wins, but by much less). This tees up the whole lesson: mass and velocity trade off differently for p than for KE, and 'heavy' does not automatically mean 'more of everything.' Do NOT resolve the KE vs p distinction yet — flag it as a puzzle we'll unpack.

1. 6m

   Defining linear momentum p = mv

   Content

   Linear momentum is defined as p = mv, where m is the object's mass in kilograms and v is its velocity in m/s. Because velocity is a vector, momentum is a vector quantity — it has both magnitude and direction, and its direction is exactly the direction the object is moving. The SI unit is the kg·m/s. Every moving object carries momentum at every instant; it is a state property of the motion, not something that only appears in a collision. Worked example: a 1200 kg car moving east at 15 m/s has p = (1200 kg)(15 m/s) = 18,000 kg·m/s directed east. If the same car sits at rest, p = 0. If it moves west at 15 m/s, p = 18,000 kg·m/s directed west (or −18,000 kg·m/s with east as positive).

   Delivery

   Emphasize the vector nature from the first sentence — say 'momentum points where the object is going.' Have students copy p = mv with an arrow over p and v in their notes to reinforce vectorhood. Walk the car example on the fly and prompt: 'What changes if I reverse the direction?' Pre-empt the misconception that momentum only exists during collisions: point at a student's water bottle sitting still and ask what its momentum is (zero), then have them 'give it momentum' by sliding it. The vector-arrow visual on the slide shows p as an arrow attached to a moving object — use it to make the direction concrete.

2. 5m

   Sign, direction, and components

   Content

   Because p is a vector, we must pick a positive direction before assigning a sign. On a 1-D track with 'right' chosen as positive, a 0.50 kg cart moving right at 2.0 m/s has p = +1.0 kg·m/s; the SAME cart moving left at 2.0 m/s has p = −1.0 kg·m/s. The negative sign does NOT mean negative mass — mass is always positive — it means the velocity (and therefore p) points opposite to the chosen positive axis. In 2-D, momentum has components: pₓ = mvₓ and p_y = mv_y. For a 2.0 kg puck moving at 5.0 m/s at 37° above the +x axis: pₓ = (2.0)(5.0)(cos 37°) = 8.0 kg·m/s, and the y-component is (2.0)(5.0)(sin 37°) = 6.0 kg·m/s. The magnitude is |p| = 10 kg·m/s at 37° above +x.

   Delivery

   Insist students state a positive direction BEFORE writing any number — 'no axis, no sign.' The number-line slide shows a cart at the origin with arrows left and right; use it to ask students to call out the sign for each direction. Then walk the 2-D puck example live, computing pₓ and p_y and checking with the Pythagorean magnitude. Head off the misconception that a negative p implies negative mass by explicitly saying: 'The sign is a direction label, not a property of the object.'

3. 5m

   Momentum vs. kinetic energy — same inputs, different physics

   Content

   Momentum p = mv and kinetic energy KE = ½mv² share the inputs m and v, but behave very differently. Momentum is a vector, linear in v, with units kg·m/s; kinetic energy is a scalar, quadratic in v, with units J. Consider two objects with equal KE: object A has m = 4.0 kg at v = 2.0 m/s → KE = 8.0 J, p = 8.0 kg·m/s. Object B has m = 1.0 kg at v = 4.0 m/s → KE = 8.0 J, p = 4.0 kg·m/s. Same KE, but A has twice the momentum. Reverse it: two objects with equal p can have very different KE. Doubling v doubles p but quadruples KE. Reversing v flips the sign of p but leaves KE unchanged (v² is positive).

   Delivery

   Work the A/B example on the fly and let students verify with calculators. Draw the punchline out loud: 'Equal KE does not imply equal p, and vice versa.' The split-panel visual on the slide shows a scalar 'blob' for KE vs. a directed arrow for p on the same moving object — use it to reinforce that KE has no direction while p does. This is where the hook's puzzle resolves: the bowling ball dominates in p because p is linear in v, so the baseball's 45 m/s can't overcome its tiny mass; in KE the v² helps the baseball close the gap. Pre-empt the misconception 'momentum and KE are the same thing' by saying it out loud and then killing it with the numbers.

4. 4m

   The mass-velocity trade-off

   Content

   Because p = mv is linear in both m and v, a small fast object can carry the same momentum as a large slow one. If m₁v₁ = m₂v₂, the two objects have equal-magnitude momentum vectors even though one is 'heavy and slow' and the other is 'light and fast.' Example: a 0.010 kg bullet at 400 m/s has p = 4.0 kg·m/s; an 80 kg student walking at 0.05 m/s also has p = 4.0 kg·m/s. Same p, very different feel — because their KEs differ by a factor of 10⁴. This defeats the common shortcut 'heavier means more momentum': you must multiply m and v, not just eyeball the mass.

   Delivery

   The side-by-side arrow visual on the slide shows a small fast object and a large slow object with momentum arrows of equal length — that visual literally is this beat's punchline. Ask students to invent their own pair with equal p (m₁v₁ = m₂v₂) before moving on — cold-call two examples. Warn: the same trick does NOT work for KE.

1. 28m

   Cart Lab + Ranking Task: p = mv as a VectorLab

   Groups of 3 rotate through a two-part investigation. Circulate and check that every group states a positive direction BEFORE recording signs, and that units appear on every number. Targets SP 1 Creating Representations (vector diagrams and p-vs-v graph), SP 2 Mathematical Routines (calculating p and KE, comparing), and SP 3 Scientific Questioning and Argumentation (justifying claims from data). Student handout: Part 1 — Ranking task (8 min, no equipment) For each object below, compute p = mv and KE = ½mv². Take motion to the right as positive. - Object A: 0.145 kg baseball moving right at 40 m/s - Object B: 7.3 kg bowling ball moving right at 3.0 m/s - Object C: 60 kg student walking left at 1.2 m/s - Object D: 1500 kg car moving right at 0.10 m/s - Object E: 0.020 kg bullet moving left at 350 m/s Fill in the table: - Object A: p = ______ kg·m/s, KE = ______ J - Object B: p = ______ kg·m/s, KE = ______ J - Object C: p = ______ kg·m/s, KE = ______ J - Object D: p = ______ kg·m/s, KE = ______ J - Object E: p = ______ kg·m/s, KE = ______ J Then answer: 1. Rank the objects by magnitude of momentum, greatest to least: ______ 2. Rank the objects by kinetic energy, greatest to least: ______ 3. The two rankings are different. In 2-3 sentences, explain WHY using the equations p = mv and KE = ½mv². (SP 3) 4. Which object has negative momentum? Why is its momentum negative even though mass cannot be negative? (SP 3) Part 2 — Cart measurement (18 min, motion sensor) You will measure the momentum of a cart at several different speeds and confirm the linear relationship p = mv. Setup: - Level the track. Place the motion sensor at one end pointing along the track. Define 'away from the sensor' as the positive direction. - Mass the empty cart on the balance. Cart mass m = ______ kg. Procedure: 1. Give the empty cart a gentle push away from the sensor. Record the steady velocity from the flat portion of the v-t graph. 2. Repeat for 5 total pushes at DIFFERENT speeds (roughly 0.2, 0.4, 0.6, 0.8, 1.0 m/s). 3. For each run, compute p = mv. 4. Add a 0.50 kg bar to the cart (new mass = ______ kg). Give ONE push at roughly 0.5 m/s and record v and p. 5. Remove the bar. Push the cart TOWARD the sensor once at ≈0.5 m/s and record the (signed) velocity and momentum. Data table (empty cart): - Run 1: v = ______ m/s, p = ______ kg·m/s - Run 2: v = ______ m/s, p = ______ kg·m/s - Run 3: v = ______ m/s, p = ______ kg·m/s - Run 4: v = ______ m/s, p = ______ kg·m/s - Run 5: v = ______ m/s, p = ______ kg·m/s Loaded cart: v = ______ m/s, p = ______ kg·m/s Toward sensor: v = ______ m/s, p = ______ kg·m/s Analysis (SP 1 & SP 2): 1. Plot p (y-axis) vs. v (x-axis) for the 5 empty-cart runs. Draw a best-fit line. 2. What is the slope of your line? Slope = ______ (units: ______) 3. Compare the slope to the measured cart mass. Are they equal within uncertainty? Explain why they should be. (SP 3) 4. The loaded-cart point should NOT lie on the empty-cart best-fit line. Where does it lie relative to the line, and why? (SP 3) 5. The 'toward the sensor' run has a __________ velocity and therefore a __________ momentum. State the positive direction you chose and use it to justify the sign. (SP 3) Every number needs a unit and a sign.

   Materials

   • Dynamics cart (low-friction) — 1 per group
   • Extra cart masses (0.25 kg and 0.50 kg bars) — 1 set per group
   • Motion sensor (Vernier/PASCO) with interface and laptop
   • Dynamics track, ~1.2 m
   • Electronic balance (0.01 kg resolution)
   • Calculator
   • Student handout (printed, 1 per student)

   Example outputs

   • Part 1: Object A p = +5.8 kg·m/s, KE = 116 J. Object B p = +21.9 kg·m/s, KE = 32.9 J. Object C p = −72 kg·m/s, KE = 43.2 J. Object D p = +150 kg·m/s, KE = 7.5 J. Object E p = −7.0 kg·m/s, KE = 1225 J. Rank by |p|: D > C > B > E > A. Rank by KE: E > A > C > B > D. The rankings differ because p is linear in v and KE is quadratic in v — the bullet's high speed dominates in KE (v² term) but not in p, while the car's huge mass dominates in p but its tiny v² wipes out its KE. Object C and Object E have negative momentum because their velocities point in the chosen negative direction; the sign is a direction label, not a property of the mass.
   • Part 2: For a 0.512 kg cart, sample data v = 0.20, 0.41, 0.58, 0.79, 1.02 m/s gives p = 0.102, 0.210, 0.297, 0.404, 0.522 kg·m/s. Slope of p vs. v ≈ 0.51 kg, matching the cart's mass — confirming p = mv with mass as the slope. The loaded cart (m ≈ 1.01 kg) at v = 0.50 m/s gives p = 0.505 kg·m/s, which lies ABOVE the empty-cart line because the same velocity now corresponds to more momentum. The toward-sensor run gives v = −0.48 m/s and p = −0.246 kg·m/s: negative because motion is opposite the chosen positive direction.

   No-equipment fallback

   Skip Part 2's live measurement. Instead, provide the following pre-recorded data set for the empty cart (m = 0.512 kg): v = 0.20, 0.41, 0.58, 0.79, 1.02 m/s. Students compute p for each, plot p vs. v by hand on graph paper, determine the slope, and answer the same analysis questions. Add a 6th 'toward sensor' data point: v = −0.48 m/s.

1. A 0.50 kg cart moves to the LEFT at 3.0 m/s. Taking rightward as positive, what is the cart's momentum?

   multiple choice−1.5 kg·m/s. (Correct: mass is 0.50 kg, velocity is −3.0 m/s in the chosen convention, so p = (0.50)(−3.0) = −1.5 kg·m/s. The negative sign comes from the direction, not from negative mass.) Targets SP 2.

2. Object X has mass 2.0 kg and speed 6.0 m/s. Object Y has mass 8.0 kg and speed 2.0 m/s. Which statement is correct? A) X has greater |p| and greater KE. B) Y has greater |p| and greater KE. C) X has greater KE but Y has greater |p|. D) X has greater |p| but Y has greater KE.

   multiple choiceC. p_X = 12 kg·m/s, p_Y = 16 kg·m/s → Y has greater |p|. KE_X = ½(2.0)(6.0)² = 36 J, KE_Y = ½(8.0)(2.0)² = 16 J → X has greater KE. This is the equal-inputs-different-physics point: KE's v² term rewards speed, while p is linear in both. Targets SP 2 and SP 3.

3. A 0.20 kg puck slides across frictionless ice at 5.0 m/s at an angle of 53° above the +x axis. Determine pₓ, p_y, and the magnitude |p|. Then explain in one sentence why treating momentum as a vector matters here.

   calculationpₓ = (0.20)(5.0)(cos 53°) = (0.20)(5.0)(0.60) = 0.60 kg·m/s. p_y = (0.20)(5.0)(sin 53°) = (0.20)(5.0)(0.80) = 0.80 kg·m/s. |p| = √(0.60² + 0.80²) = 1.0 kg·m/s, at 53° above +x. Momentum must be tracked as a vector because its components are conserved independently in later interactions — reducing p to a scalar 'amount' would erase the direction information needed to predict motion. Targets SP 1 and SP 2.

4. Two students argue: Alice says 'A heavier object always has more momentum.' Bob says 'Two objects with the same kinetic energy must have the same momentum.' In 2-3 sentences each, refute both claims using p = mv and KE = ½mv². Give a numerical counterexample for at least one.

   short answerAlice is wrong because p = mv depends on BOTH mass and velocity; a light fast object can out-momentum a heavy slow one. Counterexample: a 0.010 kg bullet at 400 m/s has p = 4.0 kg·m/s, larger than a 100 kg person standing nearly still (p ≈ 0). Bob is wrong because p is linear in v while KE is quadratic in v; equal KE does not force equal m|v|. Counterexample: 4.0 kg at 2.0 m/s and 1.0 kg at 4.0 m/s both have KE = 8.0 J, but their momenta are 8.0 kg·m/s and 4.0 kg·m/s respectively. Targets SP 3.

• ☐Dynamics cart, 1 per group (of 3 students)
• ☐Set of cart masses (0.25 kg and 0.50 kg bars), 1 set per group
• ☐Dynamics track (≈1.2 m), 1 per group
• ☐Motion sensor + interface + laptop, 1 per group
• ☐Electronic balance (0.01 kg), 1 per group or shared
• ☐Printed 2-part student handout, 1 per student
• ☐Graph paper (for no-equipment fallback)
• ☐Calculators
• ☐Whiteboard markers for group work