Momentum & Collision Calculator (Final Velocity)

Enter masses and velocities and the calculator finds momentum (p = mv), impulse, the perfectly inelastic final velocity, or both 1D elastic final velocities. Every mode is built on conservation of momentum; the elastic mode also conserves kinetic energy.

Mass m (kg)Velocity v (m/s)Negative means moving left.

momentum (p)

6 kg m/s

Momentum keeps the sign of the velocity: negative means moving left.

Steps

1.p = m v
2.p = 2 kg x 3 m/s = 6 kg m/s

AP Physics: Unit 4 (topics 4.1 Linear Momentum, 4.2 Change in Momentum and Impulse, 4.3 Conservation of Linear Momentum, 4.4 Elastic and Inelastic Collisions). Built for AP Physics 1 Unit 4, Linear Momentum, worth 10 to 15% of the multiple-choice section. The same momentum and collision analysis appears in AP Physics C: Mechanics Unit 4 for the calculus-based course.

What the calculator above solves

The calculator above handles the four core Unit 4 computations: momentum of a single object with $p = mv$ , impulse from force and contact time or from a change in momentum, the shared final velocity after a perfectly inelastic collision, and both final velocities after a one-dimensional elastic collision. Enter masses in kilograms and velocities in meters per second, and it applies the matching conservation equation.

One rule carries across every mode: momentum is a vector. Pick a positive direction before you enter anything, then type any velocity pointing the opposite way as a negative number. Most wrong collision answers trace back to a dropped sign, not bad algebra. The sections below walk through the conservation setup behind each mode, so you can reproduce every result by hand on Unit 4 free-response problems.

Momentum: p = mv

Momentum is mass times velocity: $p = mv$ , with units of $\text{kg} \cdot \text{m/s}$ . It is a vector, so in one dimension its sign tells you which way the object is moving. A 0.145 kg baseball thrown at 40.0 m/s carries $p = (0.145\ \text{kg})(40.0\ \text{m/s}) = 5.80\ \text{kg} \cdot \text{m/s}$ , roughly the momentum of a 1200 kg car creeping along at 5 mm per second. Mass and speed trade off directly.

Momentum matters because the total momentum of an isolated system never changes. However messy the interaction, if no external force acts along the direction you care about, the sum of every $mv$ before the collision equals the sum after. That single idea generates every collision equation the calculator uses, as the conservation of momentum guide develops in detail.

Impulse: how force changes momentum

Impulse links force to momentum change: $J = F_{avg} \Delta t = \Delta p$ , exactly as printed on the AP equation sheet. In impulse mode the calculator accepts either pair of inputs. Give it an average force and a contact time and it multiplies them; give it a mass with initial and final velocities and it computes $\Delta p = m(v_f - v_i)$ , signs included.

The payoff on exam problems is usually the trade between force and time. The momentum change in a collision is fixed by the velocities, so stretching the contact time (airbags, crumple zones, bending your knees on a landing) cuts the average force in the same proportion. The impulse-momentum theorem guide covers the graphical version too: impulse is the area under a force versus time graph, which shows up constantly in Topic 4.2 questions.

Perfectly inelastic collisions: the objects stick

When two objects stick together, they share one final velocity, and conservation of momentum gives it directly. Set total momentum before equal to total momentum after:

m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f

Then solve for the final velocity:

v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}

That is the entire derivation, and it is what the calculator computes in inelastic mode. Note what is not conserved: kinetic energy. Some of it becomes thermal energy and deformation during impact, and a perfectly inelastic collision loses the maximum amount consistent with momentum conservation. If a problem asks how much, compute $K = \frac{1}{2} m v^2$ before and after (the kinetic energy calculator speeds this up) and subtract. Never set kinetic energy before equal to kinetic energy after in a sticking collision.

1D elastic collisions: two conservation laws

Elastic collisions conserve both momentum and kinetic energy, which gives two equations in the two unknown final velocities:

m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2

Solving the system produces the formulas the calculator uses in elastic mode:

v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} + \frac{2 m_2}{m_1 + m_2} v_{2i}

v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} + \frac{m_2 - m_1}{m_1 + m_2} v_{2i}

Three limiting cases are worth memorizing. Equal masses swap velocities. A light object bouncing off a much heavier stationary one reverses direction at nearly the same speed. A heavy object hitting a much lighter stationary one barely slows, while the light one flies off at almost twice the heavy object's speed.

How to check any collision answer

Run three quick checks before trusting a result, whether it came from the calculator or your own algebra.

Total momentum before must equal total momentum after, signs included. Add the $mv$ terms both ways.
Kinetic energy after can never exceed kinetic energy before in a collision. Equal means elastic, smaller means inelastic, larger means a sign or arithmetic error.
For elastic collisions only, the relative speed reverses: the objects separate exactly as fast as they approached.

The third check is the fastest elastic test on the exam. If two carts approach at a combined 6.0 m/s, they must separate at 6.0 m/s after an elastic collision, no algebra needed. Since collision energy accounting overlaps with conservation of energy problems, keep the bookkeeping straight: momentum conservation needs an isolated system, while kinetic energy conservation additionally requires the collision to be elastic.

Perfectly inelastic: car rear-ends a stopped SUV

A 1200 kg car moving at 20.0 m/s rear-ends a stationary 1800 kg SUV. The bumpers lock and the vehicles move together. Find their common final velocity and the kinetic energy lost.

Choose the car's direction of motion as positive. Initial momenta: $p_{car} = (1200\ \text{kg})(20.0\ \text{m/s}) = 24000\ \text{kg} \cdot \text{m/s}$ and $p_{SUV} = 0$ .
Apply conservation of momentum with one shared final velocity: $24000\ \text{kg} \cdot \text{m/s} = (1200\ \text{kg} + 1800\ \text{kg}) v_f = (3000\ \text{kg}) v_f$ .
Solve: $v_f = 24000 / 3000 = 8.00\ \text{m/s}$ in the car's original direction.
Kinetic energy before: $K_i = \frac{1}{2}(1200\ \text{kg})(20.0\ \text{m/s})^2 = 2.40 \times 10^5\ \text{J}$ . After: $K_f = \frac{1}{2}(3000\ \text{kg})(8.00\ \text{m/s})^2 = 9.60 \times 10^4\ \text{J}$ .
Energy lost: $\Delta K = 2.40 \times 10^5\ \text{J} - 9.60 \times 10^4\ \text{J} = 1.44 \times 10^5\ \text{J}$ , which is 60.0% of the initial kinetic energy.

$v_f = 8.00\ \text{m/s}$ forward, and $1.44 \times 10^5\ \text{J}$ of kinetic energy is converted to thermal energy and deformation.

Elastic collision on an air track

A 0.500 kg glider moving at +4.00 m/s collides elastically with a 1.50 kg glider at rest on a frictionless air track. Find both final velocities and verify that kinetic energy is conserved.

The mass sum is $m_1 + m_2 = 0.500\ \text{kg} + 1.50\ \text{kg} = 2.00\ \text{kg}$ . With $v_{2i} = 0$ , the elastic formulas simplify.
First glider: $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} = \frac{0.500 - 1.50}{2.00}(4.00\ \text{m/s}) = (-0.500)(4.00\ \text{m/s}) = -2.00\ \text{m/s}$ . It rebounds.
Second glider: $v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} = \frac{2(0.500)}{2.00}(4.00\ \text{m/s}) = (0.500)(4.00\ \text{m/s}) = +2.00\ \text{m/s}$ .
Momentum check: before, $(0.500\ \text{kg})(4.00\ \text{m/s}) = 2.00\ \text{kg} \cdot \text{m/s}$ . After, $(0.500)(-2.00) + (1.50)(2.00) = -1.00 + 3.00 = 2.00\ \text{kg} \cdot \text{m/s}$ . Conserved.
Kinetic energy check: before, $\frac{1}{2}(0.500)(4.00)^2 = 4.00\ \text{J}$ . After, $\frac{1}{2}(0.500)(2.00)^2 + \frac{1}{2}(1.50)(2.00)^2 = 1.00\ \text{J} + 3.00\ \text{J} = 4.00\ \text{J}$ . Conserved, so the collision is elastic.

The light glider rebounds at 2.00 m/s and the heavy glider moves forward at 2.00 m/s. Kinetic energy stays at 4.00 J, confirming an elastic collision.

Impulse: ball rebounding off a wall

A 0.600 kg ball hits a wall horizontally at 8.00 m/s and rebounds at 6.00 m/s. Contact lasts 0.0500 s. Find the impulse on the ball and the average force the wall exerts.

Take the direction toward the wall as positive: $v_i = +8.00\ \text{m/s}$ and $v_f = -6.00\ \text{m/s}$ , because the ball rebounds.
Change in momentum: $\Delta p = m(v_f - v_i) = (0.600\ \text{kg})(-6.00\ \text{m/s} - 8.00\ \text{m/s}) = (0.600\ \text{kg})(-14.0\ \text{m/s}) = -8.40\ \text{kg} \cdot \text{m/s}$ .
Impulse equals the momentum change: $J = \Delta p = -8.40\ \text{kg} \cdot \text{m/s}$ , magnitude $8.40\ \text{N} \cdot \text{s}$ directed away from the wall.
Average force: $F_{avg} = J / \Delta t = -8.40 / 0.0500 = -168\ \text{N}$ , magnitude 168 N pointing away from the wall.
Common error check: entering $v_f = +6.00\ \text{m/s}$ (forgetting the rebound) gives $\Delta p = -1.20\ \text{kg} \cdot \text{m/s}$ , seven times too small. The sign flip is most of the momentum change.

Impulse is 8.40 N s away from the wall, and the average force is 168 N away from the wall.

Frequently asked questions

What is the difference between an elastic and a perfectly inelastic collision?

Both conserve total momentum. An elastic collision also conserves kinetic energy, so the objects bounce apart carrying the same total kinetic energy they brought in. In a perfectly inelastic collision the objects stick together and move with one shared velocity, losing the maximum kinetic energy that momentum conservation allows. Most real collisions fall between these two extremes.

How do I enter velocity directions in the calculator?

Pick one direction as positive and stay consistent. If the two objects approach each other, one velocity must be entered as a negative number. The sign of each output then tells you its direction: a negative final velocity means that object moves in the direction you labeled negative.

Is momentum conserved even when kinetic energy is lost?

Yes. Conservation of momentum only requires that no net external force act on the system during the collision. Kinetic energy can transform into thermal energy, sound, and deformation without changing the total momentum, which is why the inelastic formula works even though kinetic energy drops.

Do the elastic collision formulas work when both objects are moving?

Yes. The general 1D elastic formulas the calculator uses include both initial velocities, so they handle head-on approaches, rear-end catches, and moving targets. Encode directions with signs and the same two formulas apply. Special cases like equal masses swapping velocities fall out of the general result.

How much of the AP Physics 1 exam covers momentum?

Unit 4, Linear Momentum, carries 10 to 15% of the multiple-choice section weight. Its topics are 4.1 Linear Momentum, 4.2 Change in Momentum and Impulse, 4.3 Conservation of Linear Momentum, and 4.4 Elastic and Inelastic Collisions, all of which this calculator practices.