Materials & Prep
Prepare two low-friction carts with Velcro or magnets, one added mass, a meterstick, and a board or projected slide. Write the sign convention + right and - left on the board. No special setup is needed beyond a clear tabletop or track. Confirm local procedures for safe use of carts and added masses.
Opening prediction
DisplayA 1 kg cart moving right at 4 m/s hits a 3 kg cart moving left at 1 m/s. The carts stick together. Which direction will the joined carts move, and what will their speed be? Students must first commit individually by writing a direction and estimated speed, then compare with a partner.
AskWhat made you choose that direction? Do not resolve the answer yet. Collect two or three predictions, especially any claim that the heavier cart must win or that the carts must stop.
Direct instruction
DemonstrateRoll the carts together so students observe that the carts can move as one object after the collision. State the model clearly: for this lesson, the two carts stick together, so they have one shared final velocity. The system is both carts. Take right as positive.
ExplainMomentum is mass times velocity, p = mv. Momentum has direction because velocity has direction. For a closed system during a short collision, total momentum before equals total momentum after. The heavier object does not automatically determine the result. The direction depends on the total signed momentum.
Worked example
1 kg cart: p1 = (1 kg)(+4 m/s) = +4 kg·m/s 3 kg cart: p2 = (3 kg)(-1 m/s) = -3 kg·m/s Total initial momentum: +4 + (-3) = +1 kg·m/s Total mass after sticking: 1 kg + 3 kg = 4 kg Final momentum: pf = (4 kg)vf
Think aloudI keep the negative sign because the heavier cart moves left. I add the signed momenta before dividing. The common error is to add 4 and 3 as positive numbers, which would give the wrong total of 7 kg·m/s. Now solve:
(4 kg)vf = +1 kg·m/s vf = +0.25 m/s
The joined carts move right at 0.25 m/s. The lighter cart did not win because it was stronger. Its larger speed gave it slightly more initial momentum.
ConfrontIf equal 1 kg carts move at +2 m/s and -2 m/s, their total initial momentum is +2 + (-2) = 0. When they stick, vf = 0/(2 kg) = 0 m/s. Momentum did not disappear. The system had zero total momentum before and after because the two opposite momenta canceled. If a system that starts with nonzero total momentum later stops, momentum was transferred to something outside the chosen system, such as the floor or Earth.
Guided practice
ScaffoldStudents use this organizer for every problem: 1. Choose positive direction. 2. Write each initial momentum with its sign. 3. Add the initial momenta. 4. Add the masses. 5. Set total initial momentum equal to total final momentum and solve for vf. 6. Interpret the sign and include units.
Worked exampleSolve the opening prediction together, then have students explain why the answer is not based only on comparing 3 kg with 1 kg.
Check for understandingPresent 1 kg moving right at 5 m/s and 4 kg moving left at 1 m/s. Students show only the signed total momentum and predicted direction on a mini-whiteboard or paper. Expected: +5 + (-4) = +1 kg·m/s, so the joined objects move right. Ask students to identify the exact step that would fail if they ignored direction.
Turn and talkIn the equal-and-opposite example, was each cart's individual momentum zero before the collision? Expected: No. Each had nonzero momentum, but the system total was zero.
Independent application
Have students solve the following collisions. In every case, the objects stick together. They must show signed momentum, total mass, an equation, a final velocity, and a one-sentence interpretation.
- A 2 kg cart moves right at 3 m/s. A 4 kg cart moves right at 1 m/s. Find the shared final velocity.
- A 0.50 kg cart moves right at 6 m/s. A 1.50 kg cart is stationary. Find the shared final velocity.
- A 2 kg cart moves left at 2 m/s. A 1 kg cart moves right at 5 m/s. Find the shared final velocity.
Expected1. Total momentum is 6 + 4 = 10 kg·m/s, total mass is 6 kg, so vf = +1.67 m/s. 2. Total momentum is 3 + 0 = 3 kg·m/s, total mass is 2 kg, so vf = +1.5 m/s. 3. Total momentum is -4 + 5 = +1 kg·m/s, total mass is 3 kg, so vf = +0.33 m/s.
Look forStudents who use the heavier object's direction automatically, omit negative signs, or divide by only one mass. Ask them to point to the system, the signed total momentum, and the total mass.
ExtensionCompare a 1 kg cart moving right at 5 m/s with a 4 kg cart moving left at 1 m/s. Then reverse the speeds and directions. Predict before calculating which case changes direction. Expected results are +0.20 m/s for the first case and -0.20 m/s for the reversed case. Explain that reversing every velocity reverses the final velocity, while mass stays the same.
Closing check
Exit ticketA 2 kg cart moves right at 3 m/s and collides with a 4 kg cart moving left at 1 m/s. They stick together. Calculate the final velocity and write a two-sentence claim-evidence-reasoning response: state the direction and speed, give the signed momentum calculation as evidence, and explain why the heavier cart does not automatically determine the result.
ExpectedInitial momentum is (2)(+3) + (4)(-1) = +6 - 4 = +2 kg·m/s. Total mass is 6 kg, so vf = +2/6 = +0.33 m/s. The joined carts move right at about 0.33 m/s because the system's total initial momentum is positive. A heavier object does not always determine the outcome; both mass and signed velocity contribute to total momentum.