Momentum: definition and formula

Momentum, by definition and formula, is the mass of an object times its velocity: p = m × v, in kg·m/s. When a force acts for a short time, it changes that momentum by F × Δt, which we call the impulse. This relation is the key to propeller theory: a propeller pushes because it accelerates matter.

In this page:

1: Impulse and momentum: definition and formula

Let us take a simple example. A man pushes a cabinet mounted on rollers, and we neglect friction. He exerts a force F on it for a time interval Δt, and the cabinet's speed goes from v1 to v2.

A man pushes a cabinet on rollers; free-body diagram of the cabinet: applied force F, weight and normal force

On the left, the man and the cabinet; on the right, the forces on the cabinet. Weight and normal force cancel: the resultant force is F.

The average acceleration is the change in speed divided by the duration: a = (v2 − v1) / Δt. Newton's second law, F = m × a, then gives:

F = m × (v2 − v1) / Δt   that is   F × Δt = m × v2 − m × v1

The product F × Δt (in newton-seconds) is called the impulse. The product m × v is the momentum p, whose SI unit is kg·m/s. The impulse therefore equals the change in momentum:

F × Δt = p' − p   with p = m × v and p' = m × v'

This form lets us recover the average force of a violent event without knowing its details.

Example: a 0.15 kg baseball is at rest, then the batter hits it. Just after the hit, it leaves at 40 m/s, and the impact lasts 0.001 s. Its initial momentum is zero, and the final one is 0.15 × 40 = 6 kg·m/s. The average force is therefore:

F = (0.15 kg × 40 m/s) / 0.001 s = 6,000 N

Six thousand newtons is the weight of about six hundred kilos: the ball has every right to complain.

2: Conservation of momentum in a collision

Momentum is most useful for studying the interaction of two or more objects. The diagram below shows a collision between two objects sliding without friction. Each one is subject to its weight and a normal force, which cancel.

Two objects m1 and m2 sliding toward each other: before, during and after the collision, with forces F12 and F21 during the impact

Before, during and after the collision. During the impact, the two objects exert equal and opposite forces, F12 and F21, on each other.

These two forces are an action and a reaction: they have the same magnitude and opposite directions, so F12 + F21 = 0. If the collision lasts Δt, the change in momentum of each object is:

Adding the two equations, and since F12 + F21 = 0, we obtain:

p'1 − p1 + p'2 − p2 = 0   that is   p1 + p2 = p'1 + p'2

The total momentum of the system is therefore conserved. This result is valuable: we relate the speeds before and after the collision without knowing anything about the forces during the impact. It is enough that they are equal and opposite.

3: Momentum calculation example: the cannon on its railway car

A cannon is fixed on a railway car, initially at rest, that can move without friction. The cannon fires a ball of mass m = 5 kg at a horizontal speed v = 15 m/s relative to the ground, toward the opposite wall of the car. The total mass of the cannon and the car is M = 15,000 kg, and we neglect the mass of the explosion gases. Two questions arise: what is the car's speed V while the ball is in flight, and after the ball hits the wall?

A cannon fixed to the floor of a railway car fires a ball of mass m at speed v to the right; the car of mass M recoils to the left at speed V

The ball leaves to the right at speed v; the car carrying the cannon recoils to the left at speed V.

(a) Ball in flight. When the cannon fires, it pushes the ball to the right, and the ball pushes the cannon with an equal force to the left. No external horizontal force acts on the system, so its momentum stays zero. The ball's momentum, to the right, equals in magnitude that of the cannon and car, to the left:

m × v = M × V   that is   V = m × v / M = (5 kg × 15 m/s) / 15,000 kg = 0.005 m/s

The recoil is tiny, because the car is very heavy compared with the ball.

(b) Ball embedded in the wall. The ball pushes the wall to the right, the wall pushes it to the left, and everything stops. The total momentum is zero again. In the meantime, the car has moved to the left and the ball to the right.

Keep the idea for the propeller: it throws fluid backward, like the cannon its ball, and the fluid pushes it forward in return. Unlike the cannon, it does not fire one ball once, but a mass of fluid every second.

To go further, we can apply this balance to a propeller disc with Froude's theory for propulsive propellers, or see what it gives for the propulsive efficiency of a propeller.