Lab 7: Impulse and Momentum

Regular lab. You turn in the part you worked on. In a group of two, each of you turns in your own part plus CON.

Everyone turns in CON this week. The report template says it in capitals: every student submits the two-paragraph conclusion, whichever job they had. QPD and UDA are still turned in by the people who did them.

At a glance

• Questions. Two parts. Part 1: how does the change in a cart’s velocity depend on the impulse the wall gives it, and does it keep its kinetic energy through the bounce? Part 2: how does the momentum change of one cart compare with the momentum change of the other when they bounce, stick or fly apart, and what happens to the kinetic energy in each case?
• You measure. Part 1: the cart’s velocity before and after it hits the wall and the impulse from the force sensor, ten runs at different speeds, plus the cart’s mass on the balance. Part 2: the velocity of cart A before the interaction and the velocities of both carts after it, nine runs (three kinds of interaction, three ways of placing the mass bar), plus the masses of both carts and the bar.
• You calculate. Part 1: \(\Delta v\) and the kinetic energy before and after. Part 2: \(\Delta p\) and \(\Delta K\) for each cart. You type the formulas for \(\Delta v\) and both \(\Delta p\) columns yourself. The sheet builds \(M_A\) and \(M_B\) for every row from the three masses typed in above the Part 2 table.
• You graph. Four graphs. Part 1: \(\Delta v\) against impulse, slope \(1/M\), and \(K_F\) against \(K_I\), slope compared with 1. Part 2: \(\Delta p_B\) against \(\Delta p_A\) for all nine runs, slope compared with \(-1\), and \(\Delta K_B\) against \(\Delta K_A\) for the three bouncy runs only, slope compared with \(-1\).
• You conclude. Two paragraphs. Momentum first, both parts, with the cart mass and its percent difference from the balance. Kinetic energy second, both parts, with the wall bounce classified.

QPD

This lab has two parts, and the report template has a QPD slot for each. Each slot wants a board photo, an annotated photo of the setup, and a screenshot of the Graphical Analysis screen where the velocities were read off, with the impulse as well in Part 1. Split one board or use two.

The questions

Part 1: What is the relationship between the impulse the wall delivers to the cart (\(J\)) and the change in the cart’s velocity (\(\Delta v\))?

Part 2: What is the relationship between the change in momentum of cart A (\(\Delta p_A\)) and the change in momentum of cart B (\(\Delta p_B\)) when the two carts interact?

The template says “questions”, and each part has a second one about energy. Write it as a second line under the first. For Part 1 it is the relationship between the cart’s kinetic energy before the bounce (\(K_I\)) and after it (\(K_F\)). For Part 2 it is how the kinetic energy of the two carts after bouncing, sticking or flying apart compares with the kinetic energy before.

What the procedure must make clear

This page was written from the 2026 templates, not the handout, so the details below are the usual way this lab runs. Where the handout differs, the handout wins.

• Part 1: what the cart hit, what was on the front of the force sensor, how the cart was set moving (a push by hand, let go well before the wall), and that the ten runs were at different speeds so the impulses spread out along the graph.
• Part 1: where each number came from. \(v_i\) and \(v_f\) are read from the flat parts of the velocity–time graph just before and just after the collision. The impulse is the area under the force–time graph across the collision only, from the Integral tool with just the collision selected. Say that the sensor was zeroed first.
• Part 2: which cart was A (the one pushed, yellow on the sheet) and which was B (the one at rest, blue), which ends of the carts faced each other in each kind of run, and that in the explodey runs both carts started at rest touching, with the plunger of one pressed against the other, so \(v_{i,A}\) is zero for those three rows.
• Part 2: where the mass bar sat in each of the three runs of a kind, in the sheet’s order: on B, then on A, then on neither. All three velocities come from the velocity–time graphs of the two carts, read from the flat stretches on either side of the interaction.
• The units as recorded: m/s for every velocity, N·s for the impulse, kg for every mass.

Symbol	Quantity	How it is measured	Unit
\(M_\text{CART}\)	mass of the Part 1 cart with its sensor	electronic balance	kg
\(v_i\), \(v_f\)	cart velocity before and after it meets the wall	flat parts of the velocity–time graph beside the collision	m/s
\(J\)	impulse from the wall on the cart	area under the force–time graph across the collision	N·s
\(M_\text{YEL}\), \(M_\text{BLU}\), \(M_\text{bar}\)	the two carts and the mass bar	electronic balance	kg
\(v_{i,A}\)	velocity of cart A before the interaction	velocity–time graph of cart A	m/s
\(v_{f,A}\), \(v_{f,B}\)	velocities of A and B after the interaction	velocity–time graph of each cart	m/s

Cart B always starts at rest. That is why the sheet has no column for its initial velocity.

The diagram

Part 1: a photo of the cart, its force sensor and the wall it hits, labelled with the positive direction, \(v_i\) toward the wall, \(v_f\) away from it, the force from the wall on the cart, and \(M_\text{CART}\). Part 2: a photo of the two carts on the track, labelled A and B, with the mass bar, the ends that meet, the positive direction and the three velocities. The Graphical Analysis screenshots go in the same QPD slots, so mark on them where \(v_i\), \(v_f\) and the impulse were read.

UDA

In Excel

Part 1 has ten rows, Part 2 has nine, three for each kind of interaction. These are the calculated columns.

Column	Formula	Unit	Sample calculation on the board
\(\Delta v\) (Part 1)	\(v_f - v_i\)	m/s	yes
\(K_I\), \(K_F\) (Part 1)	\(\tfrac12 M_\text{CART}\,v_i^2\) and \(\tfrac12 M_\text{CART}\,v_f^2\)	J	yes, both from one row, with their difference
\(M_\text{EXP}\) (Part 1)	\(1/\text{slope}\) of the \(\Delta v\) against \(J\) graph	kg	yes
\(M_A\), \(M_B\) (Part 2)	built by the sheet from \(M_\text{YEL}\), \(M_\text{BLU}\) and \(M_\text{bar}\) for each row	kg	no
\(\Delta p_A\) (Part 2)	\(M_A\,(v_{f,A} - v_{i,A})\)	kg·m/s	yes
\(\Delta p_B\) (Part 2)	\(M_B\,v_{f,B}\)	kg·m/s	yes
\(K_{I,A}\), \(K_{F,A}\), \(K_{F,B}\) (Part 2)	\(\tfrac12 M v^2\) for each	J	inside the two \(\Delta K\) calculations
\(\Delta K_A\) (Part 2)	\(K_{F,A} - K_{I,A}\)	J	yes
\(\Delta K_B\) (Part 2)	\(K_{F,B}\), because B started with none	J	yes

Three calculations on the Part 1 board and four on the Part 2 board, which is what the report template counts.

Some things about this sheet that are not obvious from looking at it:

• What you type. The \(\Delta v\) column in Part 1 and the \(\Delta p_A\) and \(\Delta p_B\) columns in Part 2 are empty in the template. Enter those formulas yourself in the first row, then fill down. \(M_\text{CART}\) goes in the cell at the top of Part 1 and the three Part 2 masses in the row above its table, and every \(M_A\) and \(M_B\) is built from those.
• Signs. Positive is the direction the cart was pushed. In Part 1 the cart comes back, so \(v_f\), \(J\) and \(\Delta v\) are all negative, and the sheet says so under the table. In Part 2, \(v_{i,A}\) is positive (zero in the explodey rows), \(\Delta p_A\) is negative, \(\Delta p_B\) is positive, and \(v_{f,A}\) can go either way. The mean row already handles the signs, so \(\delta_r = \delta_a / \text{mean}\) works as it always does.
• What is filled in. Part 1 gives \(\delta_a = 0.01\ \text{m/s}\) for the velocities, \(0.001\ \text{N·s}\) for the impulse and \(0.001\ \text{kg}\) for the mass. Part 2 fills in the mass \(\delta_a\) and leaves the velocity cells blank. Use the same \(0.01\ \text{m/s}\), it is the same instrument.
• The two uncertainty rows. \(\Delta v\) is a difference, so its \(\delta_a\) is the larger of the two velocity values, \(0.01\ \text{m/s}\), which is what the green cell is waiting for. The \(\Delta p\) columns are products, so their \(\delta_r\) is the largest \(\delta_r\) among their inputs, and the sheet colours them for you: the green inputs (\(M_A\), \(v_{i,A}\), \(v_{f,A}\)) feed the green \(\Delta p_A\), the blue inputs (\(M_B\), \(v_{f,B}\)) feed the blue \(\Delta p_B\), then \(\delta_a = \delta_r \times \text{mean}\). The uncertainty guide walks through both rules.
• Kinetic energy. In Part 1 the \(K_I\) and \(K_F\) columns have their own mean, \(\delta_a\) and \(\delta_r\) cells, labelled in column N, with nothing filled in. Average them yourself, then the product rule. Part 2’s kinetic energy columns have no uncertainty rows. The template holds the kinetic energy formulas in first rows only: the first row of Part 1 for \(K_I\) and \(K_F\), the first row of each kind of run for \(K_{I,A}\) and \(K_{F,A}\), and the very first row for \(K_{F,B}\). Fill them all down before anything is graphed, or the \(\Delta K\) columns, which are already filled in, show zeros.
• Decimal places. A kinetic energy here is a few hundredths of a joule, so the cells show 0.02 or 0.00 until decimals are added, and the sheet warns about this at the top. Add places until three significant figures show.

The graphs. Four of them, axes labelled with units on all four.

1. Part 1, \(\Delta v\) in m/s on the vertical axis against \(J\) in N·s on the horizontal, one linear trendline with its equation. The slope is \(1/M_\text{CART}\), in units of one over kilograms, so \(M_\text{EXP} = 1/\text{slope}\). If you plotted it the other way round, impulse vertical, the slope is the mass itself.
2. Part 1, \(K_F\) against \(K_I\), both in joules, one linear trendline. The slope has no units and is compared with 1.00, and \(1 - \text{slope}\) is the fraction of the kinetic energy that the bounce cost.
3. Part 2, \(\Delta p_B\) against \(\Delta p_A\), both in kg·m/s, all nine runs in one series, one linear trendline. The slope is compared with \(-1.00\).
4. Part 2, \(\Delta K_B\) against \(\Delta K_A\) in joules for the three bouncy rows only. The template has this chart drawn already and keeps the sticky and explodey \(\Delta K_B\) values in a separate column so they stay out of it. Add the linear trendline. Its slope is compared with \(-1\).

On the board

Left half, derivation. For Part 1, start from Newton’s second law in the form it was first written: force is the rate at which momentum changes, \(F = \Delta p / \Delta t\). Multiply the time back in and add up the collision:

\[J = \int F\,dt = \Delta p = M\,\Delta v \quad\Longrightarrow\quad \Delta v = \frac{1}{M}\,J.\]

A line through the origin with slope \(1/M\), which is why the slope of the first graph gives the cart’s mass. Then \(K = \tfrac12 M v^2\), so \(K_F / K_I = (v_f / v_i)^2\), and the second graph has slope 1 only if the cart leaves the wall as fast as it arrived.

For Part 2, use Newton’s third law. While the carts interact, the force of A on B and the force of B on A are equal and opposite at every instant, and they last for the same time, so the two impulses are equal and opposite:

\[\Delta p_B = -\,\Delta p_A, \qquad M_A v_{i,A} = M_A v_{f,A} + M_B v_{f,B}.\]

That is a line of slope \(-1\) through the origin on the third graph, and the second form is momentum conservation written out. Kinetic energy has no such law: \(\Delta K_A + \Delta K_B\) can be zero, negative or positive, and the three kinds of run exist to show all three.

Right half, sample calculations. Three for Part 1 and four for Part 2, each from one real row, every factor with its unit. Invented numbers, with \(M_\text{CART} = 0.261\ \text{kg}\), \(v_i = 0.420\ \text{m/s}\) and \(v_f = -0.378\ \text{m/s}\):

\[\Delta v = -0.378\ \text{m/s} - 0.420\ \text{m/s} = -0.798\ \text{m/s},\]

\[K_I = \tfrac12 (0.261\ \text{kg})(0.420\ \text{m/s})^2 = 0.0230\ \text{J}, \qquad K_F = \tfrac12 (0.261\ \text{kg})(0.378\ \text{m/s})^2 = 0.0186\ \text{J}, \qquad \Delta K = -0.0044\ \text{J},\]

\[M_\text{EXP} = \frac{1}{\text{slope}} = \frac{1}{3.76\ \text{kg}^{-1}} = 0.266\ \text{kg}.\]

For Part 2, take a bouncy row with the bar on B: \(M_A = 0.257\ \text{kg}\), \(M_B = 0.513\ \text{kg}\), \(v_{i,A} = 0.502\ \text{m/s}\), \(v_{f,A} = -0.158\ \text{m/s}\), \(v_{f,B} = 0.325\ \text{m/s}\):

\[\Delta p_A = (0.257\ \text{kg})(-0.158\ \text{m/s} - 0.502\ \text{m/s}) = -0.170\ \text{kg·m/s}, \qquad \Delta p_B = (0.513\ \text{kg})(0.325\ \text{m/s}) = 0.167\ \text{kg·m/s},\]

\[\Delta K_A = \tfrac12 (0.257\ \text{kg})\left[(0.158\ \text{m/s})^2 - (0.502\ \text{m/s})^2\right] = -0.0292\ \text{J}, \qquad \Delta K_B = \tfrac12 (0.513\ \text{kg})(0.325\ \text{m/s})^2 = 0.0271\ \text{J}.\]

The two momentum changes nearly cancel and the two energy changes nearly cancel, which is what a bouncy row should do.

CON

Two paragraphs this week, and everyone writes them. The first is about momentum in both parts, the second about kinetic energy in both parts. The report template lists what each must contain, so each item below becomes one sentence, in order. Numbers invented.

Paragraph 1, momentum.

1. The relationships in words, \(\Delta v\) against impulse in Part 1 and \(\Delta p_B\) against \(\Delta p_A\) in Part 2. → “The relationship between the impulse on the cart and its change in velocity is linear, and the relationship between the momentum change of cart A and the momentum change of cart B is also linear.”
2. The two equations. → “The trendlines were \(\Delta v = (3.76\ \text{kg}^{-1})\,J\) for the wall bounce and \(\Delta p_B = (-0.96)\,\Delta p_A\) for the two carts.” The first slope has units, one over kilograms. The second has none.
3. The cart mass from Part 1, and how constant momentum was in Part 2, the slope against \(-1.00\). → “The slope gives a cart mass of \(M_\text{EXP} = 1/(3.76\ \text{kg}^{-1}) = 0.266\ \text{kg}\), and the slope of \(-0.96\) against the expected \(-1.00\) means cart B gained 96% of the momentum cart A lost.”
4. The two comparisons: \(M_\text{EXP}\) against the balance value \(M_\text{THEO}\) as a percent difference, and the momentum change of B against A as the percent change \(1 - |\text{slope}|\). → “The balance gave \(M_\text{THEO} = 0.261\ \text{kg}\), a percent difference of \(200\%\,(0.266 - 0.261)/(0.266 + 0.261) = 1.9\%\), and the momentum change of cart B fell short of cart A’s by \(1 - 0.96 = 4\%\).”

Paragraph 2, kinetic energy.

1. Part 1: the slope of \(K_F\) against \(K_I\) compared with 1.00, and what kind of interaction that makes the wall bounce. → “The slope of final against initial kinetic energy was 0.81, so the cart kept 81% of its kinetic energy through the wall collision, which makes the bounce a bouncy interaction with some loss rather than a sticky one.”
2. The percent difference the template asks for, \(1 - \text{slope}\). → “The kinetic energy change in the wall collision was \(1 - 0.81 = 19\%\).”
3. Part 2: how the kinetic energy change went with mass across the three kinds of interaction. The template says there are two different answers here, and the bouncy graph gives the first. This item can take two sentences. → “In the bouncy runs cart B gained what cart A lost whichever cart carried the bar, with a slope of \(-0.93\) against the expected \(-1\). In the sticky runs kinetic energy was lost, most of it when the heavier cart was the one at rest, and in the explodey runs it was gained, with the lighter cart taking the larger share.”

Common slips

• Reporting the slope of \(\Delta v\) against \(J\) as the mass. The slope is \(1/M\), in one over kilograms. One divided by it is the mass.
• A positive \(v_f\). The cart comes back the way it came, so \(v_f\), \(J\) and \(\Delta v\) are all negative. If \(J\) comes out positive while \(\Delta v\) is negative, the sensor’s sign is reversed, and the slope and the mass come out negative with it.
• Reading \(v_i\) and \(v_f\) far from the collision. Friction slows the cart on the way in and on the way out, so take the flat stretches right next to the collision, and integrate the force across the collision only, because baseline noise adds up over a long selection.
• Kinetic energies showing as 0.02 J or 0.00 J. Until the decimals are added, every point on the \(K_F\) against \(K_I\) graph lands in the same place.
• Kinetic energy formulas not filled down, or the \(\Delta v\) and \(\Delta p\) formulas never typed in. The template gives the energy formulas in the first row of each block only, and leaves \(\Delta v\), \(\Delta p_A\) and \(\Delta p_B\) empty.
• The bar on the wrong cart for the row. The sheet builds \(M_A\) and \(M_B\) in a fixed order within each kind of run, bar on B, bar on A, no bar. Run the trials in that order or move the formulas.
• Sticky and explodey points on the bouncy energy graph. That graph is three points, which is why the sheet keeps the other six \(\Delta K_B\) values in a separate column.
• Treating CON as someone else’s job. This week it is everyone’s.

Going deeper

Impulse and momentum are the same unit. A newton is a kilogram metre per second squared, so a newton second is a kilogram metre per second, which is a momentum. The impulse the wall delivers is the momentum it takes from the cart, which is why one column is in N·s and the next is in m/s and the slope between them is in one over kilograms. Read that slope as a price: how much velocity one newton second buys. A heavy cart buys less. That is what mass means.

Why the slope is minus one. Newton’s third law says the two carts push on each other with equal and opposite forces at every instant. They also push for exactly the same time, because the interaction starts and stops for both at once. Equal force for equal time is equal and opposite impulse, so \(\Delta p_B = -\Delta p_A\) whatever the carts are made of and whatever their ends do. Momentum conservation is the third law with the time multiplied in. Nothing in that argument mentions kinetic energy, which is why the bouncy, sticky and explodey runs all land on the same line of slope \(-1\) and still do three different things to \(K\).

What the three mass arrangements show. For a bouncy collision with B at rest, \(v_{f,A} = \dfrac{M_A - M_B}{M_A + M_B}\,v_{i,A}\) and \(v_{f,B} = \dfrac{2M_A}{M_A + M_B}\,v_{i,A}\). Light A on heavy B: A bounces back. Heavy A on light B: A keeps going, slower. Equal masses: A stops dead and B leaves with A’s speed, which is Newton’s cradle in one click. So the \(v_{f,A}\) column should change sign between the first two bouncy rows and sit near zero in the third. In an explodey run the two carts leave with equal and opposite momentum, and since \(K = p^2 / 2M\) the lighter cart carries more of the energy, with the same \(p\) and a smaller \(M\) under it. In a sticky run the fraction of kinetic energy lost is \(M_B / (M_A + M_B)\), so the heavier the cart at rest, the more is lost.

Where the wall fits. Part 1 is Part 2 with a cart B of enormous mass. The cart’s momentum is not conserved on its own, because the wall, and the Earth it is bolted to, takes the difference. Put \(0.2\ \text{kg·m/s}\) into \(6 \times 10^{24}\ \text{kg}\) and the Earth’s share of the velocity change is about \(3 \times 10^{-26}\ \text{m/s}\), which no instrument will ever see. Push \(M_B\) to infinity in the sticky fraction above and the loss goes to 100%, so a cart that stuck to the wall would stop dead and keep none of its energy. The wall bounce in Part 1 lives somewhere between that and a perfect rebound, and the slope of \(K_F\) against \(K_I\) says where.