Momentum Calculator
Calculate linear momentum as mass times velocity, with results in kg m/s and the direction of travel preserved by the sign.
Momentum Calculator
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Formula: p = m × v
Worked example — Final velocity = 2 m/s (momentum conserved: 12 kg*m/s before and after)
Formula
p = m × v
Momentum equals mass times velocity.
Worked Examples
Example 1: Collision of Two Carts
Problem:A 4 kg cart moving at 3 m/s collides with a stationary 2 kg cart and they stick together (perfectly inelastic collision). What is the final velocity?
Solution:Initial momentum = m1*v1 + m2*v2 = (4)(3) + (2)(0) = 12 kg*m/s After sticking together, combined mass = 6 kg Final velocity = total momentum / combined mass = 12 / 6 = 2 m/s
Result:Final velocity = 2 m/s (momentum conserved: 12 kg*m/s before and after)
Example 2: Rifle Recoil
Problem:A 4 kg rifle fires a 0.01 kg bullet at 400 m/s. What is the rifle's recoil velocity?
Solution:Bullet momentum = 0.01 kg x 400 m/s = 4 kg*m/s (forward) By conservation of momentum, rifle momentum must be -4 kg*m/s Recoil velocity = -4 / 4 = -1 m/s
Result:Recoil velocity = 1 m/s backward
Frequently Asked Questions
What is momentum and why does it matter?
Momentum (p) is the product of an object's mass and velocity, p = mv, measured in kg·m/s. It matters because in a closed system with no external forces, total momentum is always conserved — this single rule lets you solve collision and recoil problems without ever knowing the details of the forces involved during impact.
How do you calculate momentum for a moving object?
Multiply mass in kilograms by velocity in meters per second. A 10 kg object moving at 5 m/s has momentum p = 10 × 5 = 50 kg·m/s. Because velocity is a vector, momentum also has direction — two objects with equal mass and speed but opposite directions have equal and opposite momentum.
How is momentum different from kinetic energy?
Momentum (p = mv) scales linearly with velocity, while kinetic energy (KE = ½mv²) scales with velocity squared. Doubling an object's speed doubles its momentum but quadruples its kinetic energy. This is why a car crash at twice the speed is far more than twice as destructive — the energy that must be dissipated rises much faster than the momentum.
What is conservation of momentum used for?
Conservation of momentum lets you find unknown velocities after a collision or explosion without needing to know the forces involved. In an elastic collision both momentum and kinetic energy are conserved; in an inelastic collision (like a car crash where vehicles crumple together) momentum is still conserved but kinetic energy is lost as heat and deformation.
How does momentum explain recoil in a gun or rocket?
Before firing, total momentum is zero. When a bullet is fired forward with momentum p_bullet, the gun must recoil backward with equal and opposite momentum p_gun = -p_bullet so the total stays zero. Because the gun's mass is much larger than the bullet's, its recoil velocity is proportionally much smaller — the same principle that propels rockets forward by ejecting exhaust mass backward.
What is impulse and how does it relate to momentum?
Impulse equals the change in momentum, J = Δp = FΔt, where F is the average force and Δt is the time over which it acts. This is why airbags and crumple zones reduce injury: stretching the stopping time (Δt) for the same change in momentum reduces the average force experienced by occupants.
Why can kinetic energy be lost in a collision when momentum never is?
Momentum conservation follows from Newton's third law: the internal impact forces are equal and opposite, so they cancel in the total and the sum of mv is untouched. Kinetic energy has no such protection — it can convert into deformation, heat, and sound. In the two-cart example, the 4 kg cart at 3 m/s carries 18 J before impact but the combined 6 kg mass at 2 m/s carries only 12 J, so 6 J went into crumpling and heating, while momentum stayed at 12 kg·m/s throughout.
What is the coefficient of restitution and what does it tell you?
The coefficient of restitution e is the ratio of the relative separation speed after a collision to the relative approach speed before it. e = 1 describes a perfectly elastic collision where kinetic energy is fully recovered, e = 0 describes a perfectly inelastic collision where the bodies move off together, and real impacts sit in between. It is a convenient second equation to pair with momentum conservation, since two unknown final velocities need two equations to solve.
How do you apply momentum conservation to a two-dimensional collision?
Momentum is a vector, so it is conserved independently along each axis. Resolve every velocity into x and y components, write one conservation equation for the sum of m*vx and a separate one for the sum of m*vy, then solve them together. A glancing collision between billiard balls is the classic case: the total x-momentum before equals the total x-momentum after, and likewise for y, even though neither ball travels along the original line of motion.
How does p = mv connect to Newton's second law?
Newton originally stated the second law as force equals the rate of change of momentum, F = dp/dt. Expanding the derivative of mv gives m*dv/dt plus v*dm/dt, and when mass is constant the second term vanishes and it collapses to the familiar F = ma. For a body that gains or sheds mass you cannot simply differentiate mv, because v*dm/dt uses the body's own velocity and the result is not frame-independent. Instead apply dp/dt to the whole system including the ejected or accreted mass, which gives F_ext + v_rel*dm/dt = m*dv/dt, where v_rel is the velocity of that mass relative to the body — the relation behind the rocket equation.
References
Background & Theory
Momentum in Everyday Collisions
Linear momentum p = mv is one of the most reliable tools in mechanics because, unlike force, it's conserved in any closed system regardless of how complicated the internal collision forces are. This makes it possible to solve for post-collision speeds without ever measuring the actual (often very brief and hard-to-measure) forces involved in an impact.
| Object | Typical mass | Typical speed | Approx. momentum |
|---|---|---|---|
| Thrown baseball | 0.145 kg | 40 m/s (fastball) | ~5.8 kg*m/s |
| Rifle bullet (.308) | 0.0098 kg | 850 m/s | ~8.3 kg*m/s |
| Sedan car | 1,500 kg | 27.8 m/s (100 km/h) | ~41,700 kg*m/s |
| Loaded cargo ship | 50,000,000 kg | 7 m/s (~13.6 knots) | ~350,000,000 kg*m/s |
The huge momentum figure for a cargo ship explains why large vessels take miles to stop even at modest speed — bringing that momentum to zero requires either a very large force or a very long time (impulse J = FΔt = Δp), and neither is available in an emergency stop at sea.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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