Momentum & Impulse Calculator

Momentum and Impulse Guide

What should I know about Momentum Fundamentals?

Momentum (p) = mass (m) × velocity (v), measured in kg⋅m/s. It is a vector quantity — direction matters. A 1,000kg car at 20 m/s has the same momentum as a 2,000kg car at 10 m/s (20,000 kg⋅m/s each). Newton's second law in terms of momentum: F = dp/dt (force equals rate of change of momentum). This is more fundamental than F = ma, which assumes constant mass. Conservation of momentum: in any isolated system, total momentum before equals total momentum after a collision.

What's the key thing to understand about Impulse?

Impulse (J) = Force × time = change in momentum. This is why airbags save lives — they increase the time of impact, reducing the force for the same change in momentum. A head hitting the steering wheel in 5ms vs the airbag in 50ms: same impulse, 10× longer time, so 10× less force. Same principle: catching a cricket ball by drawing your hands back, bending your knees when landing, crumple zones in cars, sports padding and helmets.

What do I need to know about Conservation of Momentum?

In elastic collisions, both momentum and kinetic energy are conserved (billiard balls, atomic collisions). In perfectly inelastic collisions, only momentum is conserved and objects stick together (clay, car crashes). Total momentum before = total momentum after, regardless of collision type. For a 70kg person at 5 m/s colliding with a stationary 50kg person in a perfectly inelastic collision: final velocity = (70×5 + 50×0)/(70+50) = 350/120 = 2.92 m/s.

What do I need to know about Everyday Momentum?

Sports: a cricket ball (0.16kg) at 90mph (40 m/s) has p = 6.4 kg⋅m/s. A rugby tackle: a 100kg player at 6 m/s has p = 600 kg⋅m/s — much more momentum despite lower speed. Rockets: Newton's third law means the momentum of exhaust gases equals the momentum gained by the rocket. Recoil: when a gun fires, the bullet and gun gain equal and opposite momenta. The gun's recoil velocity = (bullet mass × bullet velocity) / gun mass.

What's the key thing to understand about Elastic and Inelastic Collisions?

Collisions are classified by whether kinetic energy is conserved alongside momentum, which is always conserved in any collision (assuming no external forces). In a perfectly elastic collision, both momentum and total kinetic energy are conserved — no energy is lost to heat, sound, or permanent deformation — a close real-world approximation is two billiard balls or hard steel spheres colliding. In an inelastic collision, momentum is still conserved, but some kinetic energy converts into other forms (heat, sound, deformation of the colliding objects), which is why the objects move slower, in total, after the collision than a purely elastic collision would predict — a car crash is a clear real-world example, where crumpling metal absorbs kinetic energy. A perfectly inelastic collision is the extreme case where the colliding objects stick together and move afterward with a single shared velocity — the maximum possible amount of kinetic energy is lost (while momentum conservation still determines that final shared velocity exactly), such as a dart embedding itself in a dartboard, or two train carriages coupling together on impact.

What should I know about Momentum as a Vector Quantity?

Momentum, defined as p = mv, is a vector, meaning its direction matters just as much as its magnitude — a critical fact when analysing collisions along a straight line, where moving objects are typically assigned a positive or negative sign depending on their direction of travel relative to a chosen reference direction. For a head-on collision between two objects moving toward each other, one object's momentum is positive and the other's is negative (even though both have positive mass and positive speed), and the law of conservation of momentum still requires the vector sum — including these signs — to remain constant before and after the collision. This is why two identical objects of equal mass approaching each other at equal speed can collide and both come to a complete stop (their equal-and-opposite momenta sum to zero both before and after), despite each object clearly having kinetic energy and speed right up until the moment of impact. Forgetting to assign the correct sign to momentum moving in different directions is one of the most common sources of error in solving collision problems, since it's easy to add speeds as if they were simply positive numbers.

What should I know about Momentum Conservation in Two Dimensions?

When a collision isn't head-on, momentum conservation must be applied separately along two perpendicular axes: the total x-component of momentum before a collision equals the total x-component after, and likewise for the y-component, independently. This is solved by resolving every velocity into its x and y components before applying conservation, then recombining the resulting components back into a magnitude and direction for each object afterward. Billiards, car-crash reconstruction, and particle-physics collision analysis all rely on exactly this component-wise approach whenever the interaction isn't constrained to a single line.

Worked Example: A Perfectly Inelastic Collision

Cart A (mass 3 kg) moves at +4 m/s and collides head-on with Cart B (mass 2 kg) moving at −1 m/s (opposite direction). After the collision they stick together. Total momentum is conserved: p_initial = (3 × 4) + (2 × −1) = 12 − 2 = 10 kg·m/s. Since the carts move together afterward with combined mass 5 kg, the final velocity is v_final = p_initial / m_total = 10 / 5 = 2 m/s, in the direction Cart A was originally moving. Kinetic energy, unlike momentum, is not conserved in this collision: KE_initial = ½(3)(4²) + ½(2)(1²) = 24 + 1 = 25 J, while KE_final = ½(5)(2²) = 10 J. The missing 15 J was converted to heat and deformation — exactly what defines a collision as perfectly inelastic, even though momentum was conserved throughout.

Reference Table: Momentum of Everyday Moving Objects

Comparing the momentum of familiar objects helps build intuition for why momentum depends on both mass and speed, not just one or the other:

Notice the bullet, despite its very high speed, has far less momentum than a walking person — its tiny mass dominates the calculation. This is why momentum and kinetic energy (which depends on v², not v) can give very different rankings for the same set of objects, and why a bullet's danger comes mainly from its concentrated energy and penetrating force rather than from its overall momentum.

Worked Example: Average Force from an Impulse

Impulse-momentum theorem states that impulse J equals the change in momentum, J = Δp = mΔv, and also equals average force multiplied by contact time, J = F_avg × Δt. Combining these gives a way to estimate forces during very short collisions. A 0.50 kg ball arrives at a bat moving at −5 m/s (toward the batter is taken as negative) and leaves moving at +8 m/s after being struck, with the bat in contact for Δt = 0.02 s. The change in momentum is Δp = m(v_f − v_i) = 0.50 × (8 − (−5)) = 0.50 × 13 = 6.5 kg·m/s. The average force the bat exerts on the ball is F_avg = Δp / Δt = 6.5 / 0.02 = 325 N. This is why bats, gloves, and airbags all aim to extend the contact time Δt: for a fixed required change in momentum, a longer Δt directly produces a smaller average force, which is the whole physical principle behind "following through" in sports and behind crumple zones in cars.

Why Momentum Conservation Only Holds for the Total System

Momentum is conserved for a system only when no net external force acts on it; internal forces between objects in the system (like the force a bat exerts on a ball) always come in equal-and-opposite pairs by Newton's third law, so they cancel out of the total. But a force from outside the system — gravity, friction with the ground, or air resistance — is not automatically canceled and can change total momentum. This is why momentum problems always specify "two carts on a frictionless track" or "ignoring air resistance": those conditions are what make the system's momentum conserved. A ball dropped and bouncing off the floor does not conserve its own momentum, because the floor (and the Earth behind it) is exerting an external force during the bounce; momentum is still conserved for the ball-plus-Earth system, but the Earth's mass is so enormous that its own velocity change is immeasurably small.