HS-PS2-2: Using Math to Show That Momentum Is Conserved
In plain English: HS-PS2-2 is the NGSS high school performance expectation that asks students to use mathematical representations to support the claim that the total momentum of a system is conserved when no net force acts on it. Students compute p = mv for two objects before and after a collision or push-apart along a line and compare the totals. It is usually taught in Physics or Physical Science.
Use mathematical representations to support the claim that the total momentum of a system of objects is conserved when there is no net force on the system.
Clarification Statement: Emphasis is on the quantitative conservation of momentum in interactions and the qualitative meaning of this principle.
Assessment Boundary: Assessment is limited to systems of two macroscopic bodies moving in one dimension.
Next Generation Science Standards · Disciplinary Core Idea: Motion and Stability: Forces and Interactions · Official standard
Students use equations, data tables, position-time graphs and momentum bar charts to support the claim that the total momentum of a two-object system stays the same when no net outside force acts on it. They compute each object's momentum, p = mv, with a sign for direction, add the two values before and after a collision or a push-apart, and judge whether the totals agree within measurement uncertainty. They also explain the meaning of the principle: during an interaction the two objects push on each other with equal and opposite forces for the same time, so whatever momentum one object gains, the other loses.
The NGSS Clarification Statement says: "Emphasis is on the quantitative conservation of momentum in interactions and the qualitative meaning of this principle." The Assessment Boundary says: "Assessment is limited to systems of two macroscopic bodies moving in one dimension." Every example on this page has exactly two objects moving along one line, with right (or forward) taken as positive, and every data set is invented for teaching. Students do the practice of using mathematical representations: they build tables and bar charts, write and solve conservation equations, and use the results as evidence for or against the claim.
Learning Objectives
By the end of this lesson, students will be able to:
Calculate the momentum of an object moving along a line, p = mv, using a sign to show direction and the unit kg·m/s
Write and solve a conservation equation, m1v1 + m2v2 = m1v1' + m2v2', for sticking collisions, bouncing collisions and push-apart interactions of two objects
Use data tables, position-time graphs and momentum bar charts to compare total momentum before and after an interaction and judge agreement within measurement uncertainty
Decide whether a net outside force acts on a chosen two-object system and explain when momentum is and is not conserved
Explain, using Newton's third law, why the momentum one object gains equals the momentum the other object loses
Prior Knowledge Required
Students should already be comfortable with:
Newton's third law in collisions between two objects MS-PS2-1
Newton's second law and the idea of net force HS-PS2-1
Velocity as the slope of a position-time graph, with a sign for direction
Solving a linear equation for one unknown HSA.CED.A.4
Set a stationary cart on the track and roll a second, identical cart into it with a Velcro bumper so the two stick. Before anyone measures, ask:
Warm-Up Prompt
"A moving cart runs into an identical cart at rest and the two stick together. Predict their speed right after the collision compared with the first cart's speed before it. What quantity do you think stays the same, and what would you measure to check?"
Record predictions (same speed, half the speed, zero). Many students predict half the speed without a reason; ask them to name what is shared between the two carts. Keep the list: the lesson tests it with numbers.
Direct Instruction15-20 minutes
Momentum as a mathematical representation. The momentum of an object is p = mv, measured in kg·m/s. Along one line, velocity has a sign, so momentum does too: pick one direction as positive and give motion the other way a negative value. The system is the two objects. If no net outside force acts on the system, the claim is that ptotal, before = ptotal, after. Why: during the interaction, object 1 pushes on object 2 and object 2 pushes back with a force of equal size in the opposite direction (Newton's third law) for exactly the same time. The two changes in momentum are equal in size and opposite in sign, so they add to zero. Show three representations for each case: an equation, a before-and-after table, and a momentum bar chart (Diagram 2).
Sticking collision
A 0.50 kg cart moving at +1.20 m/s hits a 0.30 kg cart at rest, and the Velcro bumpers make them stick.
Result: Before: (0.50)(1.20) + (0.30)(0) = 0.60 kg·m/s. After: (0.80)v' = 0.60, so v' = +0.75 m/s. The pair moves more slowly than the first cart because the same momentum is shared by more mass.
Push-apart from rest
A 0.40 kg cart and a 0.60 kg cart sit together at rest. A spring plunger pushes them apart, and the 0.60 kg cart moves off at +0.50 m/s.
Result: Before: total p = 0. After: (0.40)v + (0.60)(0.50) = 0, so v = -0.30 ÷ 0.40 = -0.75 m/s. The lighter cart moves faster, in the opposite direction.
Head-on bounce
A 1.0 kg cart moving at +0.80 m/s meets a 0.50 kg cart moving at -0.40 m/s. Magnetic bumpers make them bounce, and the 1.0 kg cart stops.
Result: Before: (1.0)(0.80) + (0.50)(-0.40) = +0.60 kg·m/s. After: 0 + (0.50)v = 0.60, so the 0.50 kg cart moves at +1.20 m/s.
Testing the claim with lab data
A 0.60 kg cart at +0.80 m/s hits and sticks to a 0.40 kg cart at rest. Phone video shows the pair moving at +0.47 m/s.
Result: Before: 0.48 kg·m/s. After: (1.00)(0.47) = 0.47 kg·m/s. The difference is 0.01 kg·m/s, about 2.1% of 0.48, which is within the uncertainty of reading video frames: the data support conservation.
Choosing the system
A 0.050 kg lump of clay falls at 4.0 m/s and stops on the floor. Is momentum conserved?
Result: For the clay alone, no: the floor exerts a net outside force and p drops from 0.20 kg·m/s to 0. For the clay-Earth system, yes: Earth (6.0 × 10²⁴ kg) gains 0.20 kg·m/s, which changes its velocity by only about 3.3 × 10⁻²⁶ m/s.
After each example, ask: "What is the system? Is there a net outside force on it during the interaction? What does the equation predict, and what would you measure to test it?" Stress that friction from the track is an outside force, but in a quick collision it acts for so short a time that it changes the total very little.
Guided Practice15-20 minutes
Pairs analyze invented phone-video data of a sticking collision. Cart A (0.75 kg) rolls toward cart B (0.25 kg), which is at rest with its Velcro bumper at x = 0.58 m. The table gives cart A's position; after t = 0.6 s the carts move together. Diagram 1 shows the same data.
Guided practice data: position of cart A, then of the stuck pair
t (s)
0
0.2
0.4
0.6
0.8
1.0
1.2
x (m)
0.10
0.26
0.42
0.58
0.70
0.82
0.94
Find cart A's velocity before the collision as the slope of the first four points (+0.80 m/s).
Find the velocity of the stuck pair from the last four points (+0.60 m/s).
Complete a before-and-after table: pA, pB and the total, then draw a momentum bar chart.
Compare the totals (0.60 kg·m/s before and after) and write one sentence that uses them as evidence for the claim. Then discuss: real video data rarely agree this exactly. How big a difference would still count as support?
Independent Practice10-15 minutes
Each student analyzes a push-apart trial. Two carts sit together at rest on a level track: cart P (1.50 kg) and cart Q (0.50 kg). A spring plunger pushes them apart. In the first 1.00 s after release, cart P moves 0.30 m to the left and cart Q moves 0.87 m to the right (invented data; friction is small over this time).
Students find each velocity, each momentum and the total after the push, then decide whether the data support the claim. (Cart P: -0.45 kg·m/s; cart Q: +0.435 kg·m/s; the total, -0.015 kg·m/s, is about 3% of either momentum, so it is close to the zero total before the push.) They finish with a claim, evidence and reasoning paragraph that explains, with Newton's third law, why the two momenta must be equal and opposite.
Closure5 minutes
Exit ticket: "A 0.20 kg cart moving at +0.90 m/s hits and sticks to a 0.40 kg cart at rest. Find the velocity of the pair and draw the momentum bar chart." (+0.30 m/s; the total stays 0.18 kg·m/s.) Return to the warm-up predictions and ask which one the math supports for identical carts, and why.
Differentiation Strategies
For Struggling Students
Give a before-and-after table with columns m, v and p = mv already drawn, and have students fill it in before writing any equation
Use a number line with arrows to decide the sign of each velocity before multiplying
Start with sticking collisions and push-apart from rest, where only one unknown appears, before head-on bounces
For Advanced Students
Ask students to estimate the uncertainty in each measured velocity from video (for example, one frame at 30 frames per second) and decide whether a measured difference in total momentum is larger than that uncertainty
Have students prove algebraically that for any push-apart from rest, the ratio of speeds equals the inverse ratio of the masses
Ask students to explain why a longer, gentler friction force over a whole trip changes the total momentum far more than friction during a 0.05 s collision
Assessment Guidance
What to Look For
Check that students give velocities a sign and carry it into p = mv, especially in head-on and push-apart cases. Look for claims that compare two numbers (total before, total after) and say whether the difference is within measurement uncertainty. Listen for students who say momentum is "used up" in a sticking collision, who add speeds instead of momenta, or who apply conservation to a single object that is pushed by something outside the system.
02
Classroom Activities
3 Activities
1
Stick or Bounce: Collision Lab
25 minGroups of 3-4
Groups run collisions between two carts on a level track, once with Velcro bumpers (the carts stick) and once with magnetic bumpers (the carts bounce). They measure velocities from phone video or photogates, build a before-and-after momentum table for each trial, and decide whether the totals agree.
Procedure
Measure each cart's mass on a balance; add slotted masses to one cart for trial 2.
Tape a meter stick along the track and film from the side at 30 or 60 frames per second, or set two photogates on each side of the collision point.
Give cart A a gentle push toward cart B, which starts at rest. Read velocities just before and just after the collision (use a short stretch of track so friction has little time to act).
Safety: put end stops on the track, keep fingers out from between the bumpers, and do not launch carts off the table.
Sample Data (invented)
Trial 1 (Velcro): A 0.50 kg at +0.62 m/s, B 0.50 kg at rest; after, together at +0.30 m/s. Trial 2 (Velcro): A 0.50 kg at +0.70 m/s, B 1.00 kg at rest; after, together at +0.23 m/s. Trial 3 (magnets): A 0.50 kg at +0.60 m/s, B 0.50 kg at rest; after, A at +0.02 m/s and B at +0.57 m/s. Students compute the totals: 0.310 and 0.300 kg·m/s, 0.350 and 0.345 kg·m/s, 0.300 and 0.295 kg·m/s.
Discussion Questions
In all three trials the total after is slightly smaller than the total before. What outside force could explain this, and why is its effect small?
Trial 1 has the largest percent difference, about 3.2%. Is that small enough to count as support for the claim? What would you change to reduce it?
In trial 3, cart A almost stops. Use the momentum table to explain where its momentum went.
Modification for Distance Learning
Students use a free online collision simulation with two carts on a line, record three trials of their own, and submit their momentum tables and bar charts to the class discussion board.
2
Explosion Carts: Where Do They Land at the Same Time?
20 minPairs
Two carts sit together at rest with a spring plunger between them. Pairs place an end stop in front of each cart and slide the stops until both carts hit at the same moment (the two clicks sound as one). Because both carts travel for the same time, the distances are in the same ratio as the speeds, so a conservation claim predicts m1d1 = m2d2.
Procedure
Mark the starting point of the carts on the track with masking tape.
Choose a mass pair, predict where each end stop should go, then release the plunger with a pencil tap from above (not with fingers between the carts).
Adjust the stops until the clicks sound as one, and record both distances.
Sample Data (invented)
Masses 0.50 kg and 0.50 kg: distances 0.60 m and 0.60 m. Masses 1.00 kg and 0.50 kg: 0.40 m and 0.80 m. Masses 1.50 kg and 0.50 kg: 0.30 m and 0.90 m.
Discussion Questions
For every pair, the product of mass and distance is the same for both carts (0.30, 0.40 and 0.45 kg·m). Why does this show that the two momenta are equal in size?
What is the total momentum before release, and why must the carts move in opposite directions afterward?
Why is it important that the two carts reach their stops at the same time?
Challenge Variation
With a 2.00 kg cart and a 0.50 kg cart and only 1.20 m of track between the two stops, predict where each stop should go, then test it.
3
Is Momentum Conserved? System Stations
25 minGroups of 3-4
Four stations each describe one two-object interaction with invented data. At each station, groups name the system, decide whether a net outside force acts during the interaction, and use a momentum table to support or reject the claim that the total is conserved. Groups rotate every 6 minutes.
Station Cards
Station 1, ice skaters: a 60 kg skater and a 40 kg skater stand together at rest and push apart. The 40 kg skater glides off at 1.5 m/s. Predict the other skater's velocity.
Station 2, carts on carpet: a cart rolls 2 m across carpet and then sticks to a second cart. The total momentum measured at the start of the roll is 0.40 kg·m/s, and right after the collision it is 0.28 kg·m/s.
Station 3, rail cars: a 20,000 kg rail car at 2.0 m/s couples with a 30,000 kg car at rest on a level track. Predict their speed together.
Station 4, dropped basketball: a student says momentum is not conserved when a basketball bounces, because it changes direction. Decide which system makes the claim true.
Discussion Questions
At Station 2, the total dropped by 30%. Which part of the motion (the roll or the collision) is responsible, and what does that tell you about choosing the time interval?
At Stations 1 and 3, what outside forces act, and why can we ignore them during the push or the coupling?
At Station 4, what is the second object in a system whose momentum is conserved?
Modification for Distance Learning
Post each station as a slide. Groups work in breakout rooms and submit one momentum table and one sentence about the system for each station on a shared document.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Position-Time Data for a Sticking Collision
The guided-practice data drawn to scale. Before t = 0.6 s, cart A's slope is +0.80 m/s and cart B stays at 0.58 m. After the collision the stuck pair has a smaller slope, +0.60 m/s. The total momentum is (0.75)(0.80) = 0.60 kg·m/s before and (1.00)(0.60) = 0.60 kg·m/s after.
Diagram 2: Momentum Bar Chart for a Head-On Bounce
Bars drawn to scale for the head-on worked example. Before the collision the carts have +0.80 and -0.20 kg·m/s; after it, 0 and +0.60 kg·m/s. The individual bars change, but the total bar is +0.60 kg·m/s both times, which is what conservation of momentum claims.
04
Homework Assignment
~30 min
HS-PS2-2 Homework: Momentum Before and After
Directions: Every problem involves two objects moving along one line. Take right or forward as positive. All data are invented for practice. Show each momentum with its sign and the unit kg·m/s, and end each problem with a sentence that states what your numbers show about the claim.
Part 1: Using the Conservation Equation (Problems 1-2)
On an icy road, a 1,500 kg car moving at 12 m/s hits a 1,000 kg car stopped at a light, and the two lock together. (a) Find the total momentum before the collision. (b) Find the velocity of the two cars just after the collision. (c) Friction from the road acts on the cars. Explain why you may still use conservation of momentum for the moment of the collision.
A 52 kg student stands at rest on a 4.0 kg skateboard and throws a 4.0 kg medicine ball forward at 7.0 m/s. (a) Write the conservation equation for the system of student-plus-board and ball. (b) Find the velocity of the student and board. (c) Explain in words why the student moves backward.
Part 2: Analyzing Data and Representations (Problems 3-4)
A lab group ran three sticking collisions. Trial 1: a 0.40 kg cart at 0.90 m/s hits a 0.40 kg cart at rest; the pair moves at 0.44 m/s. Trial 2: a 0.40 kg cart at 0.90 m/s hits a 0.80 kg cart at rest; the pair moves at 0.30 m/s. Trial 3: a 0.80 kg cart at 0.50 m/s hits a 0.40 kg cart at rest; the pair moves at 0.27 m/s. (a) Make a table of total momentum before and after for each trial. (b) Find each percent difference. (c) Which trials support conservation within 5%? (d) Suggest one specific cause for the trial that does not.
A 2.0 kg cart moving at +1.5 m/s meets a 1.0 kg cart moving at -0.60 m/s. After they bounce, the 2.0 kg cart moves at +0.10 m/s. (a) Find the velocity of the 1.0 kg cart after the collision. (b) Draw a momentum bar chart with a bar for each cart and for the total, before and after. (c) Use the chart to explain what changes and what stays the same.
Part 3: Explaining and Critiquing (Problems 5-6)
A 3,000 kg truck moving at 10 m/s hits a 1,000 kg parked car. Just after the collision the truck moves at 6.0 m/s in the same direction. (a) Find the car's velocity. (b) Find the change in momentum of the truck and of the car. (c) A classmate says the truck's momentum changed more because it is bigger. Use your numbers and Newton's third law to respond.
A 0.25 kg cart and a 0.50 kg cart are pushed apart from rest. Video gives -1.20 m/s for the 0.25 kg cart and +0.62 m/s for the 0.50 kg cart. (a) Find the total momentum after the push. (b) Is this result consistent with a total of zero, given that each velocity may be off by about 0.02 m/s? (c) Explain qualitatively why the two momenta should be equal and opposite.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
Momentum Calculations
Correct p = mv for every object, with signs and units, and a correctly solved conservation equation
Correct method with one sign, unit or arithmetic error
Representations
Tables and bar charts are complete and consistent with the equations
Representation present but incomplete or inconsistent with the numbers
Evidence and Reasoning
Claim compares totals before and after, judges agreement against uncertainty, and names the system and any outside force
Claim stated but the comparison or the system is missing
Qualitative Meaning
Explains conservation with equal and opposite forces acting for the same time
Mentions Newton's third law without connecting it to the change in momentum
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Every question involves two objects moving along one line, with right or forward positive. Data are invented. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which unit is a correct unit for momentum?
Answer: C
Momentum is mass times velocity, so its unit is kg·m/s (which also equals N·s). Choice A is the newton, a unit of force: it has one extra "per second". Choice D is the joule, a unit of energy, which comes from mass times velocity squared.
Question 2 of 20 · Multiple Choice
A 0.17 kg hockey puck slides across the ice at 30 m/s. What is its momentum?
Answer: A
p = mv = 0.17 × 30 = 5.1 kg·m/s. Choice B divides the speed by the mass instead of multiplying. Choice C divides the mass by the speed. Choice D doubles the speed and ignores the mass.
Question 3 of 20 · Multiple Choice
A 3.0 kg cart moving at 2.8 m/s hits a 1.0 kg cart at rest, and they stick together. What is their velocity just after the collision?
Answer: B
The total momentum is 3.0 × 2.8 = 8.4 kg·m/s before, and it is shared by 4.0 kg after, so v = 8.4 ÷ 4.0 = 2.1 m/s. Choice A ignores the added mass. Choice D is the momentum, not the velocity. Choice C divides the mass by the momentum.
Question 4 of 20 · Multiple Choice
Two carts sit together at rest on a level track. A spring pushes them apart. What is the total momentum of the two-cart system just after the push?
Answer: D
The spring force is internal to the system, and no net outside force acts, so the total stays at its starting value of zero: the two momenta are equal in size and opposite in sign. Choice B adds speeds without signs, a frequent error when both objects are moving. Choice A forgets the other cart's momentum.
Question 5 of 20 · Multiple Choice
A 0.20 kg cart and a 0.80 kg cart are pushed apart from rest. The 0.80 kg cart moves at +0.25 m/s. What is the velocity of the 0.20 kg cart?
Answer: A
0 = (0.20)v + (0.80)(0.25), so v = -0.20 ÷ 0.20 = -1.0 m/s. Choice C has the right size but the wrong direction: the carts must move opposite ways for the total to stay zero. Choice B gives both carts the same speed, which only works for equal masses. Choice D multiplies by the mass ratio the wrong way.
Question 6 of 20 · Multiple Choice
For which two-object system is the net outside force closest to zero during the interaction?
Answer: C
On a level air table, gravity is balanced by the air cushion and friction is tiny, so the only strong forces are the ones the pucks exert on each other. In choice B the road pushes on the car through friction, an outside force. In choice D the batter's hands exert a large outside force on the bat. In choice A the hand belongs to a person who pushes against the floor, so the floor's friction is an outside force on the hand-cart system.
Question 7 of 20 · Multiple Choice
A 2.0 kg cart moves at +3.0 m/s toward a 1.0 kg cart moving at -2.0 m/s. What is the total momentum of the system?
Answer: B
Total p = (2.0)(3.0) + (1.0)(-2.0) = 6.0 - 2.0 = +4.0 kg·m/s. Choice A ignores the minus sign of the second velocity. Choice C adds the two velocities and ignores the masses. Choice D assumes carts moving toward each other always cancel.
Question 8 of 20 · Multiple Choice
A group measures a total momentum of 0.84 kg·m/s before a collision and 0.81 kg·m/s after it. Which conclusion is best supported?
Answer: D
The difference is 0.03 ÷ 0.84, about 3.6%, which is within typical cart-lab uncertainty and in the direction friction would cause. Choice A demands perfect agreement from real measurements. Choice B confuses momentum with energy: energy can become thermal energy, but momentum is not converted into other forms. Choice C overstates what data can show.
Question 9 of 20 · Multiple Choice
During a collision, cart 1 pushes on cart 2 and cart 2 pushes back on cart 1. Why does the total momentum stay the same?
Answer: A
By Newton's third law the forces are equal and opposite at every instant, and they act for the same contact time, so one cart gains exactly the momentum the other loses. Choice B contradicts the third law: the forces are equal even when the masses are not. Choice D is false: the lab data in this lesson include unequal masses.
Question 10 of 20 · Multiple Choice
A 0.50 kg cart moving at +2.0 m/s hits a 1.5 kg cart at rest. After the collision the 0.50 kg cart bounces back at -0.55 m/s. What is the velocity of the 1.5 kg cart?
Answer: C
Before: (0.50)(2.0) = 1.0 kg·m/s. After: (0.50)(-0.55) + 1.5v = 1.0, so 1.5v = 1.275 and v = +0.85 m/s. Choice A treats the rebound momentum as positive (+0.275 instead of -0.275). Choice B forgets the rebounding cart altogether. Choice D divides by the wrong mass (0.75 kg).
Question 11 of 20 · Multiple Choice
Video shows a 0.45 kg cart moving at steady speed, covering 0.90 m in 1.5 s. What is its momentum?
Answer: B
The velocity is the slope, 0.90 ÷ 1.5 = 0.60 m/s, so p = 0.45 × 0.60 = 0.27 kg·m/s. Choice A multiplies the mass by the distance, not the velocity. Choice C is the velocity alone. Choice D multiplies distance by velocity.
Question 12 of 20 · Multiple Choice
A tennis ball is dropped, hits the floor and bounces up. For which system is momentum conserved during the bounce?
Answer: D
The floor (attached to Earth) exerts an outside force on the ball, so the ball alone does not keep its momentum (choice A). If Earth is part of the system, that force is internal: Earth gains the momentum the ball loses, although Earth's huge mass means its velocity change is far too small to notice. Choice C mixes up a change in direction with a change in the system's total.
Question 13 of 20 · Multiple Choice
Two carts are pushed apart from rest. Afterward, cart A moves three times as fast as cart B. How does the mass of cart A compare with the mass of cart B?
Answer: A
The momenta must be equal in size: mA(3v) = mBv, so mA = mB/3. Choice B reverses the relationship: the heavier cart moves more slowly. Choice D squares the ratio, as if momentum depended on v².
Question 14 of 20 · Multiple Choice
A heavy moving truck hits a small parked car. Compared with the change in the car's momentum, the change in the truck's momentum is:
Answer: C
The truck and car exert equal and opposite forces on each other for the same time, so their momentum changes are equal in size and opposite in sign, which is why the total stays the same. Choice A confuses change in momentum with change in velocity: the truck's velocity changes less because its mass is larger. Choice D ignores that the truck slows down.
Question 15 of 20 · Short Answer
A 12,000 kg rail car moving at 3.0 m/s couples with an 18,000 kg rail car moving at 1.0 m/s in the same direction on a level track. Find their velocity together.
Total momentum = (12,000)(3.0) + (18,000)(1.0) = 36,000 + 18,000 = 54,000 kg·m/s. The coupled mass is 30,000 kg, so v = 54,000 ÷ 30,000 = 1.8 m/s in the original direction.
Question 16 of 20 · Short Answer
An 80 kg astronaut (including suit) floats at rest next to the space station and throws a 2.0 kg tool bag away at 4.0 m/s. Find the astronaut's velocity afterward and explain the direction.
Total momentum stays 0: (80)v + (2.0)(4.0) = 0, so v = -8.0 ÷ 80 = -0.10 m/s. The astronaut drifts at 0.10 m/s in the direction opposite to the throw, because the bag pushes back on the astronaut with a force equal and opposite to the throw.
Question 17 of 20 · Short Answer
A 1.20 kg cart moving at +0.50 m/s hits a 0.60 kg cart moving at -0.40 m/s, and they stick. Video gives the pair a velocity of +0.19 m/s. Do the data support conservation of momentum? Show the comparison.
Before: (1.20)(0.50) + (0.60)(-0.40) = 0.60 - 0.24 = 0.36 kg·m/s. After: (1.80)(0.19) = 0.342 kg·m/s. The prediction for the pair is 0.36 ÷ 1.80 = 0.20 m/s. The difference is 0.018 kg·m/s, 5% of 0.36, so the data support the claim within the uncertainty of video measurements, with a small loss that friction could explain.
Question 18 of 20 · Short Answer
A cart rolls slowly across a rough carpet for 3 s and then hits a second cart. Explain why the total momentum of the two carts is not conserved over the whole 3 s but is very nearly conserved during the 0.05 s collision.
Friction from the carpet is an outside force on the two-cart system. Over 3 s it has time to change the total momentum a lot (the cart slows down before the collision). During the 0.05 s collision, the forces between the carts are much larger than friction, and friction acts for only a short time, so its effect on the total is very small and the before-and-after totals nearly match.
Question 19 of 20 · Short Answer
A 0.30 kg cart moving at 1.4 m/s hits a cart of unknown mass that is at rest. They stick and move at 0.35 m/s. Find the unknown mass.
Before: 0.30 × 1.4 = 0.42 kg·m/s. After: (0.30 + M)(0.35) = 0.42, so 0.30 + M = 1.20 and M = 0.90 kg.
Question 20 of 20 · Short Answer
A student claims: "When a heavy cart hits a light cart at rest, the total momentum goes up, because the light cart shoots off fast." In a test, a 1.0 kg cart at 0.80 m/s hits a 0.25 kg cart at rest. Afterward the 1.0 kg cart moves at 0.48 m/s and the 0.25 kg cart at 1.28 m/s. Use the data to respond.
Before: (1.0)(0.80) = 0.80 kg·m/s. After: (1.0)(0.48) + (0.25)(1.28) = 0.48 + 0.32 = 0.80 kg·m/s. The total does not go up: the light cart's high speed comes with a small mass, so its momentum (0.32 kg·m/s) equals exactly what the heavy cart lost. The claim is not supported.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HS-PS2-2 mean?
HS-PS2-2 asks students to use math to show that the total momentum of a group of objects stays the same when no net outside force acts on them. Students calculate p = mv for each object, add the values before and after an interaction, and use the comparison as evidence. The code means High School, Physical Science, core idea 2 (Motion and Stability: Forces and Interactions), performance expectation 2.
What is the assessment boundary for HS-PS2-2?
Assessment is limited to systems of two macroscopic objects moving in one dimension. Two carts on a track, two skaters pushing apart or two rail cars coupling are in scope. Collisions at an angle (two-dimensional momentum), systems of three or more objects, and particle collisions are beyond this standard.
Is HS-PS2-2 taught in physics or physical science?
It is usually taught in a high school Physics course, often right after Newton's laws, and it also appears in Physical Science and integrated science courses. The math is multiplication with signed numbers and solving a one-step or two-step linear equation.
What is momentum, and what are its units?
Momentum is mass times velocity, p = mv, and its unit is kg·m/s. Because velocity has a direction, so does momentum: along a line, one direction is positive and the other is negative. A slow, heavy truck and a fast, light ball can have the same momentum.
Why is momentum conserved in a collision?
Because the two objects push on each other with equal and opposite forces for the same amount of time. By Newton's third law the forces match at every instant, so the momentum one object gains is exactly the momentum the other loses, and the total does not change. This is the qualitative meaning the Clarification Statement asks students to explain.
Is kinetic energy also conserved when momentum is conserved?
Not always. Momentum is conserved in both sticking and bouncing collisions when no net outside force acts, but kinetic energy usually decreases, especially when the objects stick. Energy accounting is part of the HS-PS3 energy standards and is beyond this standard, so this lesson keeps the focus on momentum.
What counts as "no net force on the system" in a real lab?
It means the outside forces on the two objects cancel or are too small to matter during the interaction. On a level track, gravity and the track's support force cancel. Friction is still there, but in a quick collision the forces between the carts are far larger and act only briefly, so friction changes the total very little. Over a long roll, friction does matter, so measure velocities just before and just after the collision.
Why doesn't lab data show perfect conservation of momentum?
Because every measurement has uncertainty and friction acts for a short time. Reading positions from video frames, timing photogates and measuring mass all add small errors. A difference of a few percent between the totals, especially when the total after is a little smaller, is normal and still counts as support for the claim.
Can students do HS-PS2-2 without carts or sensors?
Yes. Phone video next to a meter stick gives good velocities from the slope of position against time. Two identical balls or pucks on a smooth floor, a free online collision simulation, or the invented data sets on this page also work, as long as students build the tables, bar charts and equations themselves.
What comes after HS-PS2-2?
HS-PS2-3 comes next: students use the idea that force equals change in momentum divided by time to design and refine a device, such as a helmet or a parachute, that reduces the force on an object in a collision. The HS-PS3 standards then track energy in the same interactions.
07
Related Standards
5 standards
These standards connect to HS-PS2-2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
MS-PS2-1Prerequisite
Apply Newton's third law to design a solution for two colliding objects
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