Momentum Calculator with Steps | Impulse, Collisions, 2D

A momentum calculator multiplies mass by velocity: p = mv, in kg·m/s.

A 0.160 kg cricket ball bowled at 35.0 m/s (126 km/h) has p = 0.160 kg × 35.0 m/s = 5.60 kg·m/s, pointing the way the ball moves.

Rule of thumb: momentum in kg·m/s equals the impulse in N·s needed to stop the object. Stopping that 5.60 kg·m/s ball in 0.100 s takes an average force of 56.0 N.

Momentum is the quantity of motion an object has: its mass multiplied by its velocity, p = mv. A momentum calculator does that multiplication for you in any units, and it can also run it backwards to find the mass or the velocity when you know the momentum.

Momentum is a vector. It has a size and a direction, and the direction is always the same as the velocity. In one dimension we show the direction with a sign: + for one way, − for the other. That sign is where most exam marks are lost.

A worked one-liner: a 0.160 kg cricket ball at 35.0 m/s has p = 0.160 × 35.0 = 5.60 kg·m/s. A 1500 kg car at 25 m/s has p = 37 500 kg·m/s, about 6700 times more.

What this calculator does: six calculators in one, chosen with the tabs in Section 2.

• Momentum p = mv: solve for momentum, mass or velocity.
• Impulse and change in momentum: Δp = m(v − u) = FΔt; solve for the average force, the final velocity, the contact time or the mass.
• Collisions (conservation of momentum): elastic, perfectly inelastic, any coefficient of restitution, or the second final velocity when you know the first.
• 2D momentum: add up to three momentum vectors by components to get the total momentum and its direction.
• Momentum without velocity: p = √(2mK) from mass and kinetic energy.
• Angular momentum: L = Iω for a spinning body, or L = mvr sin θ for a point mass.

Who uses it: CBSE and ICSE students in Class 9 (Force and Laws of Motion) and Class 11 (Laws of Motion, System of Particles), JEE and NEET aspirants, IGCSE, GCSE and A level students, AP Physics 1 and 2 students, and first-year university mechanics. Engineers use the same equations for crash design, sports scientists for bat and racket impacts, and forensic investigators for vehicle collision reconstruction.

A short history: Isaac Newton's second law in the Principia (1687) was written in terms of the change in the "quantity of motion", which is what we now call momentum. In 1918 Emmy Noether showed that conservation of momentum follows from a simple fact: the laws of physics are the same everywhere in space.

Choose a calculator tab, choose what to solve for, then type each value and pick its unit. The unit next to the unknown sets the unit of the answer. Nothing is calculated until you press Calculate.

Sections 4.1 to 4.4 fill in with your own numbers after you press Calculate. Sections 4.5 to 4.8 explain the physics behind every answer.

4.1 What your number means

___ Press Calculate to see what your momentum means in plain words.

4.2 Is it a sensible size?

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4.3 How it scales

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4.4 Direction and sign

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4.5 Momentum and kinetic energy are not the same thing

Momentum p = mv grows in step with speed. Kinetic energy K = ½mv² grows with the square of speed. They are linked by K = p² ÷ 2m. Compare a 1500 kg car at 2.0 m/s with a 0.010 kg bullet at 300 m/s. They have very different momenta: the car has 3000 kg·m/s and only 3000 J, the bullet has just 3.0 kg·m/s but 450 J. Momentum tells you how hard something is to stop; energy tells you how much damage it can do while stopping.

4.6 What the charts show

For p = mv, the first two charts are straight lines through the origin, because momentum is proportional to both velocity and mass. The kinetic energy chart curves upwards. For impulse, the area under the force-time graph is the impulse, and it equals Δp. For collisions, the momentum bars for the total are the same height before and after, while the kinetic energy bars usually are not.

4.7 Where the simple model stops working

• Very high speeds: above about 0.1c (30 000 km/s) use relativistic momentum γmv. At 0.5c the classical value is 13% too small. The calculator switches badges automatically.
• Outside forces during a collision: friction or a push during the impact changes the total momentum. For very short impacts the change is small, which is why conservation still works well for car crashes and ball games.
• Changing mass: rockets and leaking carts need the full form F = dp/dt, not F = ma with a fixed mass.

4.8 Where this leads next

Momentum connects straight to energy (the kinetic energy calculator), to motion (the velocity calculator and acceleration calculator), and to forces in contact (the friction force calculator). In a collision problem you usually need both momentum and energy to find two unknown speeds.

Five detailed, ready-to-copy write-ups of your answer: a full exam answer with every step, the final answer with all key values, a lab report results section, a teacher's mark scheme with common wrong answers, and a plain-text version for chat. Below they show the worked cricket ball (0.160 kg at 35.0 m/s). Press Calculate and they rewrite themselves with your own numbers, for whichever calculator tab you used.

Every formula this momentum calculator uses, with the SI unit of each symbol. The worked numbers are the cricket ball from Section 0.

This flowchart shows how to calculate momentum, step by step, for the same worked problem used in Sections 0, 9, 10, 11 and 14: a 0.160 kg cricket ball bowled at 126 km/h.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10, the Python script in Section 11 and the hand method in Section 14 all give 5.60 kg·m/s for this ball.

1. Choose the calculator tab. Pick p = mv, Impulse and Δp, Collision, 2D momentum, From kinetic energy, Angular momentum, or Batch.
2. Choose what to solve for. Tap a Solve for chip. That box turns into a dashed question mark and its unit sets the unit of the answer.
3. Pick a positive direction. Decide which way is positive and give velocities in the other direction a minus sign.
4. Type each value. Type numbers only, such as 0.160 or 35. Scientific notation like 9.109e-31 works.
5. Choose each unit. Pick g, kg, t or lb for mass, and m/s, km/h, mph or ft/s for velocity. The calculator converts to SI for you.
6. Set the significant figures. Leave Match the data on, or choose 2, 3, 4 or 6 significant figures.
7. Or load a sample problem. Ten samples cover p = mv, impulse, airbags, collisions, 2D, kinetic energy, angular momentum and a not-physical case.
8. Press Calculate. The answer appears with its unit, other units, a units check and a substitution check. Press Enter in any box to do the same.
9. Read the steps and charts. Follow the GRESS table and numbered steps, then look at the live diagram and the four charts.
10. Download or copy the working. Use Download Doc or Download PDF, or copy a ready-made answer from Section 5.

Excel is ideal for momentum because one row can hold one problem, with the units written in the header row. Fill one formula down and every row is solved. This section uses the same cricket ball as the rest of the page: 0.160 kg at 35.0 m/s gives 5.60 kg·m/s.

9.1 The functions you need

9.2 Step by step

9.3 The awkward cases

• Trig in degrees: =COS(90) gives −0.448, not 0, because Excel works in radians. Always write COS(RADIANS(90)). It returns about 6.1E−17 rather than exactly 0, which is fine.
• ATAN2 order: Excel's ATAN2(x, y) takes x first. Python and R take y first. Swapping them gives the wrong angle.
• CONVERT strings: they are case-sensitive and give #N/A when wrong. Test each one on a value you know before you trust a whole column.
• Display versus value: Excel shows 5.6 for 5.60. Use Format Cells, Number, 2 decimal places, or =TEXT(C2,"0.00"), to show the significant figures you mean.
• Precision: Excel keeps about 15 significant digits, which is far more than any physics data needs.

9.4 Why Excel might disagree with R or Python

For p = mv all three agree exactly, because no constant is involved. Differences come from rounding: Excel's ROUND rounds halves away from zero, while R's round() and Python's round() round halves to the nearest even digit, so 2.5 becomes 2 in R and Python and 3 in Excel. They also come from -3^2, which Excel reads as 9 and R and Python read as −9. Momentum does not use g, so the g value your syllabus uses does not matter here.

9.5 Charting it

Choose Insert, Scatter, Scatter with Straight Lines. Do not use the Line chart type, which spaces x values evenly whatever they are. Put velocity on the x axis and momentum on the y axis, label both with units, and add a linear trendline with its equation. For p against v the gradient is the mass.

9.6 Errors you will hit

R needs no package for momentum: it is one multiplication. The script below calculates momentum, kinetic energy, a change in momentum with its average force, and a collision, prints every answer with its unit, checks the answer by substitution, and saves one 300 dpi figure.

10.1 The complete script

R, base only, saves momentum_plot.png

# ============================================================
# Momentum calculator in R: p = mv, impulse and a collision
# StatsUnlock - https://statsunlock.com/momentum-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Inputs (SI units) -----------------------------------
m <- 0.160      # mass, kg (cricket ball)
v <- 35.0       # velocity, m/s (126 km/h)

# ---- 2. Momentum p = m v ------------------------------------
p <- m * v                     # kg*m/s
K <- 0.5 * m * v^2             # kinetic energy, J
cat(sprintf("Momentum p = %#.3g kg*m/s (= %#.3g N*s)\n", p, p))
cat(sprintf("Kinetic energy K = %#.3g J; check sqrt(2 m K) = %#.3g kg*m/s\n",
            K, sqrt(2 * m * K)))

# ---- 3. Change in momentum and average force ----------------
u_hit <- -35.0; v_hit <- 45.0; dt <- 0.00200   # m/s, m/s, s
dp <- m * (v_hit - u_hit)                       # kg*m/s
F  <- dp / dt                                   # N
cat(sprintf("Change in momentum = %#.3g kg*m/s; average force = %#.3g N\n", dp, F))

# ---- 4. Collision: conservation of momentum -----------------
m1 <- 2.00; u1 <- 3.00; m2 <- 1.00; u2 <- 0
v_stick <- (m1 * u1 + m2 * u2) / (m1 + m2)
cat(sprintf("Trolleys stick together: v = %#.3g m/s\n", v_stick))

# ---- 5. Substitution check ----------------------------------
stopifnot(abs(p / m - v) < 1e-12)
cat("Units: kg x m/s = kg*m/s = N*s (check passed)\n")

# ---- 6. One 300 dpi figure: p against v for this mass -------
vv <- seq(0, 2 * v, length.out = 100)
png("momentum_plot.png", width = 6, height = 4, units = "in", res = 300)
plot(vv, m * vv, type = "l", lwd = 2, col = "#2563eb",
     xlab = "velocity v (m/s)", ylab = "momentum p (kg m/s)",
     main = "Momentum of a 0.160 kg cricket ball")
points(v, p, pch = 19, cex = 1.6, col = "#9333ea")
grid()
legend("topleft", bty = "n", col = c("#2563eb", "#9333ea"),
       lwd = c(2, NA), pch = c(NA, 19),
       legend = c("p = mv, m = 0.160 kg", "v = 35.0 m/s, p = 5.60 kg m/s"))
dev.off()
cat("Saved momentum_plot.png\n")

Printed output:

Momentum p = 5.60 kg*m/s (= 5.60 N*s)
Kinetic energy K = 98.0 J; check sqrt(2 m K) = 5.60 kg*m/s
Change in momentum = 12.8 kg*m/s; average force = 6.40e+03 N
Trolleys stick together: v = 2.00 m/s
Units: kg x m/s = kg*m/s = N*s (check passed)
Saved momentum_plot.png

10.2 What each part does

• Section 1 holds the inputs in SI units, with the unit in a comment. Convert before this point (grams ÷ 1000, km/h ÷ 3.6).
• Section 2 is the whole of p = mv. sprintf("%#.3g", p) prints 3 significant figures and keeps trailing zeros, so 5.6 shows as 5.60.
• Section 3 uses signed velocities: the ball arrives at −35.0 m/s and leaves at +45.0 m/s, so Δp = 12.8 kg·m/s.
• Section 4 applies conservation of momentum for two objects that stick.
• Section 5 substitutes the answer back. stopifnot stops the script if the check fails.
• Section 6 draws p against v and saves it with png(..., res = 300). Always finish with dev.off() or the file stays empty.

R traps: sin() and cos() take radians (use cos(theta * pi / 180)); atan2(y, x) takes y first; round(2.5) gives 2 because R rounds halves to even; -3^2 is −9; and log() is the natural log.

10.3 What the figure shows

A straight line through the origin: momentum is proportional to velocity. Its gradient is the mass, 0.160 kg. The purple point is the bowled ball at 35.0 m/s and 5.60 kg·m/s.

10.4 Common problems and fixes

• The PNG is blank: dev.off() was not run. Run the whole block, not line by line.
• Answer 1000 times too big: mass typed in grams. Divide by 1000 first.
• Prints 5.6 instead of 5.60: use sprintf("%#.3g", p), not print(p).

10.5 Useful one-liners

m * v                                  # momentum, kg*m/s
m * (v - u)                            # change in momentum
m * (v - u) / dt                       # average force, N
sqrt(2 * m * K)                        # momentum from kinetic energy
(m1*u1 + m2*u2) / (m1 + m2)            # stick-together velocity
sqrt(sum(px)^2 + sum(py)^2)            # 2D total momentum
atan2(sum(py), sum(px)) * 180 / pi     # its direction, degrees

The Python version does exactly the same job as the R script, with NumPy for the arrays and Matplotlib for the figure. It prints the same numbers: 5.60 kg·m/s, 98.0 J, 12.8 kg·m/s, 6.40 × 10³ N and 2.00 m/s.

Install once: pip install numpy matplotlib. Runs in a terminal (python momentum.py), Jupyter or Google Colab.

11.1 The complete script

Python 3, NumPy and Matplotlib, saves momentum_plot.png

# ============================================================
# Momentum calculator in Python: p = mv, impulse and a collision
# StatsUnlock - https://statsunlock.com/momentum-calculator/
# Needs: pip install numpy matplotlib
# ============================================================
import numpy as np
import matplotlib.pyplot as plt

# ---- 1. Inputs (SI units) -----------------------------------
m = 0.160      # mass, kg (cricket ball)
v = 35.0       # velocity, m/s (126 km/h)

# ---- 2. Momentum p = m v ------------------------------------
p = m * v                      # kg*m/s
K = 0.5 * m * v**2             # kinetic energy, J
print(f"Momentum p = {p:#.3g} kg*m/s (= {p:#.3g} N*s)")
print(f"Kinetic energy K = {K:#.3g} J; check sqrt(2 m K) = {np.sqrt(2*m*K):#.3g} kg*m/s")

# ---- 3. Change in momentum and average force ----------------
u_hit, v_hit, dt = -35.0, 45.0, 0.00200   # m/s, m/s, s
dp = m * (v_hit - u_hit)                  # kg*m/s
F = dp / dt                               # N
print(f"Change in momentum = {dp:#.3g} kg*m/s; average force = {F:#.3g} N")

# ---- 4. Collision: conservation of momentum -----------------
m1, u1, m2, u2 = 2.00, 3.00, 1.00, 0.0
v_stick = (m1*u1 + m2*u2) / (m1 + m2)
print(f"Trolleys stick together: v = {v_stick:#.3g} m/s")

# ---- 5. Substitution check ----------------------------------
assert abs(p / m - v) < 1e-12, "substitution check failed"
print("Units: kg x m/s = kg*m/s = N*s (check passed)")

# ---- 6. One 300 dpi figure: p against v for this mass -------
vv = np.linspace(0, 2*v, 100)
fig, ax = plt.subplots(figsize=(6, 4))
ax.plot(vv, m*vv, color="#2563eb", lw=2, label="p = mv, m = 0.160 kg")
ax.plot(v, p, "o", color="#9333ea", ms=9, label="v = 35.0 m/s, p = 5.60 kg·m/s")
ax.set_xlabel("velocity v (m/s)")
ax.set_ylabel("momentum p (kg·m/s)")
ax.set_title("Momentum of a 0.160 kg cricket ball")
ax.grid(alpha=0.3)
ax.legend()
fig.savefig("momentum_plot.png", dpi=300, bbox_inches="tight")
print("Saved momentum_plot.png")

Printed output (tested):

Momentum p = 5.60 kg*m/s (= 5.60 N*s)
Kinetic energy K = 98.0 J; check sqrt(2 m K) = 5.60 kg*m/s
Change in momentum = 12.8 kg*m/s; average force = 6.40e+03 N
Trolleys stick together: v = 2.00 m/s
Units: kg x m/s = kg*m/s = N*s (check passed)
Saved momentum_plot.png

11.2 What each part does

• Inputs are plain floats in SI units.
• f"{p:#.3g}" prints 3 significant figures and keeps the trailing zero. Without the #, Python prints 5.6.
• v**2 is v squared. In Python ^ means bitwise XOR and will raise an error or give nonsense.
• assert is the substitution check; the script stops with a message if it fails.
• fig.savefig(..., dpi=300, bbox_inches="tight") writes a print-quality figure.

Python traps: math.sin and numpy.cos take radians, so use np.radians(theta); np.arctan2(y, x) takes y first; round(2.5) gives 2 (half to even); and float precision is about 15 to 16 digits. Momentum needs no physical constant, so scipy.constants is only needed for the relativistic case (from scipy.constants import c).

11.3 What the figure shows

The same straight line as the R figure: p = 0.160v, with the bowled ball marked at 35.0 m/s and 5.60 kg·m/s.

11.4 Common problems and fixes

• ModuleNotFoundError: numpy: run pip install numpy matplotlib in the same environment you run the script in.
• No window appears: the script saves a file instead. Add plt.show() at the end to see it on screen.
• The angle is wrong in 2D: you used degrees in np.cos, or swapped the arguments of np.arctan2.

11.5 Useful one-liners

p = m * v                                   # momentum
dp = m * (v - u); F = dp / dt               # impulse and average force
p = np.sqrt(2 * m * K)                      # from kinetic energy
v = (m1*u1 + m2*u2) / (m1 + m2)             # stick together
P = np.hypot(px.sum(), py.sum())            # 2D magnitude
theta = np.degrees(np.arctan2(py.sum(), px.sum()))

12.1 Constants used by this calculator

Source: NIST CODATA 2022 recommended values (Mohr et al., 2025). Momentum itself needs no constant; c is used only for the relativistic check.

12.2 Momentum unit conversions

12.3 Typical momenta

Each value is mass × speed for the stated numbers, which are round illustrative figures, not measurements.

12.4 Typical contact times

Contact times vary widely with speed and materials. Measure them where you can, for example with a phone accelerometer (Vogt & Kuhn, 2014).

12.5 Coefficient of restitution e

12.6 Unit factors built into the calculator

Eight worked problems, from the textbook case to the traps that cost marks. Each one is different from the cricket ball used in the rest of the page.

This is the written method examiners expect, worked on the same example as Sections 7, 9, 10 and 11: a 160 g cricket ball is bowled at 126 km/h. Find its momentum.

14.1 The written method (GRESS)

1. Given. m = 160 g = 0.160 kg; v = 126 km/h = 126 ÷ 3.6 = 35.0 m/s, towards the batter (+).
2. Required. p, the momentum, in kg·m/s.
3. Equation. p = mv.
4. Substitute, with units. p = (0.160 kg)(35.0 m/s).
5. Solve. p = 5.60 kg·m/s.
6. Check the units and the size. kg × m/s = kg·m/s ✓. A ball should have a few kg·m/s, far less than a walking person (about 100), so 5.60 is sensible.
7. Write the answer with units, direction and significant figures. p = 5.60 kg·m/s towards the batter (3 s.f., matching 0.160 and 35.0).

14.2 On a scientific calculator

The display shows 5.6; you write 5.60 because the data have 3 significant figures. Use the (−) negative key, not the subtract key, for a negative velocity. For any cos or sin, check the screen shows D (degrees). Menu keystrokes for stored constants differ between models, so check your calculator's manual.

14.3 Measure it yourself: a school lab practical

Paste any row as "mass, speed" into the Batch tab (for example 0.500, 0.84 and 1.00, 0.42) to get p before and p after. Main errors: friction on the runway (about 2 to 3% loss in row 3), a card not square to the beam, and a push that continues after release.

14.4 Practice problems with answers

14.5 Mistakes that lose marks

• Using grams or km/h without converting to kg and m/s.
• Forgetting the minus sign on a rebound, which makes Δp far too small (1.6 instead of 12.8 kg·m/s for the bat).
• Adding momenta at angles as plain numbers instead of as vectors.
• Assuming kinetic energy is conserved in every collision. Only momentum always is.
• Writing kg/m/s or kgm/s². The unit is kg·m/s or N·s.
• Leaving out the direction of a momentum or an impulse.
• Quoting 5.6 when the data justify 5.60, or 5.6000 when they do not.

Use p = mv when:

• ✓ You know the mass and velocity of one object and want its momentum.
• ✓ The speed is below about 0.1c (30 000 km/s), which covers everything outside particle physics.
• ✓ The mass is constant.

Use something else when:

• ✗ The velocity changes: use Δp = m(v − u) and the impulse FΔt.
• ✗ Two or more objects interact: use conservation of momentum.
• ✗ The objects move at angles: add momentum vectors by components.
• ✗ The object spins: use angular momentum L = Iω.
• ✗ The speed is close to c: use p = γmv.
• ✗ The mass changes (rockets): use F = dp/dt with the mass flow.

Real examples:

1. Sport: a bat or racket hit is an impulse problem; contact time sets the force.
2. Road safety: crumple zones and airbags lengthen Δt so the same Δp needs a smaller force.
3. Accident reconstruction: skid marks and final positions give speeds after, and conservation of momentum gives speeds before.
4. Space: a rocket pushes exhaust backwards and gains equal momentum forwards.
5. Particle physics: momenta of tracks in a detector, using γmv.

Decision tree: One object, constant velocity → p = mv. Velocity changes because of a force → Δp = FΔt. Objects collide or push apart with no outside force → conservation of momentum (in a line: one equation; at angles: one for x and one for y). They also bounce perfectly → add conservation of kinetic energy, or use e = 1. Only mass and energy known → p = √(2mK). Spinning → L = Iω.

Assumptions

1. Constant mass. Check: nothing is added or thrown off. If it fails (a rocket, a leaking cart), F = ma underestimates the thrust needed; use F = dp/dt.
2. Speed well below c. Check: v under 3 × 10⁷ m/s. If it fails, p = mv underestimates the momentum (13.4% low at 0.5c).
3. Isolated system in a collision. Check: friction and outside pushes are small during the impact. If it fails, the calculated final speeds are too high when friction acts against the motion.
4. Average force in the impulse tab. Check: you want the mean, not the peak. For a smooth, half-sine-shaped impact the peak is about 1.6 times the average (π/2), so the average underestimates the peak.
5. Point objects in one line (1D tab). Check: the hit is head-on. Off-centre hits add spin and sideways motion, so 1D results overestimate the speed along the line.
6. Rigid body about a fixed axis (angular tab). Check: I is about the actual axis. Using I about the centre for an off-centre axis underestimates L (parallel axis theorem).

Limitations

• The collision tab is one-dimensional. For angled collisions use the 2D tab to sum momenta, and solve x and y separately.
• Three or more bodies colliding at once are not modelled; treat them as a series of two-body collisions.
• Rolling and spinning balls also carry angular momentum, which can change the bounce (Cross, 2002, 2005).
• No air resistance is included; over long distances drag removes momentum.
• Relativistic momentum is reported for the p = mv tab only; relativistic collisions need four-momentum.
• The definition of mass itself has subtleties at relativistic speeds (Okun, 1989; Hecht, 2006). This calculator uses invariant (rest) mass throughout.

What the answer is

A momentum calculator turns two measurements, mass and velocity, into one of the most useful quantities in mechanics: p = mv, in kg·m/s or N·s. For the cricket ball on this page that is 5.60 kg·m/s.

___ Press Calculate to add your own result here.

What it means physically

Momentum measures how hard something is to stop. Its unit is the same as the unit of impulse, so the momentum of an object is the impulse needed to bring it to rest. Spread that impulse over a longer time and the force falls, which is the idea behind airbags, gloves and bent-knee landings. In every collision without an outside force, the total momentum stays the same, even when kinetic energy is lost.

Watch out for

Convert to kg and m/s before multiplying. Give velocities a sign. Add momenta at angles as vectors. Do not assume kinetic energy is conserved unless the collision is elastic. Above 0.1c, switch to γmv.

Next steps

Go from momentum to energy with the kinetic energy calculator, find speeds with the velocity calculator, or work out stopping forces with the friction force calculator. Whenever a problem asks how hard something hits, how fast things move after a crash, or how a skater spins faster, come back to this momentum calculator.

Peer-reviewed journal articles supporting the physics, constants, laboratory methods and teaching points of this momentum calculator. Every DOI was checked in October 2026.

1. Clement, J. (1982). Students' preconceptions in introductory mechanics. American Journal of Physics, 50(1), 66–71. https://doi.org/10.1119/1.12989
2. Crouch, C. H., & Mazur, E. (2001). Peer instruction: Ten years of experience and results. American Journal of Physics, 69(9), 970–977. https://doi.org/10.1119/1.1374249
3. Cross, R. (1999). The bounce of a ball. American Journal of Physics, 67(3), 222–227. https://doi.org/10.1119/1.19229
4. Cross, R. (2002). Grip-slip behavior of a bouncing ball. American Journal of Physics, 70(11), 1093–1102. https://doi.org/10.1119/1.1507792
5. Cross, R. (2005). Bounce of a spinning ball near normal incidence. American Journal of Physics, 73(10), 914–920. https://doi.org/10.1119/1.2008299
6. Frohlich, C. (1979). Do springboard divers violate angular momentum conservation? American Journal of Physics, 47(7), 583–592. https://doi.org/10.1119/1.11759
7. Hake, R. R. (1998). Interactive-engagement versus traditional methods: A six-thousand-student survey of mechanics test data for introductory physics courses. American Journal of Physics, 66(1), 64–74. https://doi.org/10.1119/1.18809
8. Halloun, I. A., & Hestenes, D. (1985). The initial knowledge state of college physics students. American Journal of Physics, 53(11), 1043–1055. https://doi.org/10.1119/1.14030
9. Hecht, E. (2006). There is no really good definition of mass. The Physics Teacher, 44(1), 40–45. https://doi.org/10.1119/1.2150758
10. Hestenes, D., Wells, M., & Swackhamer, G. (1992). Force concept inventory. The Physics Teacher, 30(3), 141–158. https://doi.org/10.1119/1.2343497
11. Lawson, R. A., & McDermott, L. C. (1987). Student understanding of the work-energy and impulse-momentum theorems. American Journal of Physics, 55(9), 811–817. https://doi.org/10.1119/1.14994
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