There’s a deeper way to state Newton’s second law than F = ma. Force is really the rate of change of momentum — how quickly an object’s momentum shifts. This version is more powerful: it still gives F = ma for ordinary objects, but it also handles situations where the mass itself changes, like a rocket burning fuel or sand piling onto a moving belt.
📘 What you need to know
The resultant force on a body equals the rate of change of its momentum
As an equation: F = Δp ÷ Δt, where Δp = pf − pi
The change in momentum is in the same direction as the resultant force
This form works even when the mass is not constant
When the mass is constant, it reduces to the familiar F = ma
Forces come in equal and opposite pairs (Newton’s third law), so a change of momentum in one body means an equal and opposite change in another
Force as rate of change of momentum
The most general statement of Newton’s second law is:
Newton’s second law (momentum form)
The resultant force on a body equals the rate of change of its momentum
The change in momentum is just the final momentum minus the initial:
Change in momentum
Δp = pf − pi
and the force is that change divided by the time it takes:
Force and momentumF = Δp ÷ Δt
Here F is the resultant force (N) and Δt is the time over which the momentum changes (s). The big advantage of this version is that it can be used in situations where the mass of the body is not constant — something F = ma alone can’t handle.
Why it gives F = ma
For an ordinary object with constant mass, this momentum form collapses straight back into the equation you already know. Starting from force as the rate of change of momentum, and using the fact that momentum is mv and that the rate of change of velocity is acceleration:
F = Δp/Δt
→
F = Δ(mv)/Δt
→
F = m(Δv/Δt)
→
F = ma
The middle step relies on the mass being constant, so it can come outside the change. The last step uses Δv/Δt = a. So F = ma is just the constant-mass special case of the deeper momentum law — use F = ma when mass is fixed, and the momentum form when it isn’t.
Think of F = ma as the everyday tool and F = Δp/Δt as the master version behind it. A rocket loses mass every second as it burns fuel, so m isn’t constant and you can’t just use ma. But momentum still changes at a definite rate, so the force is still Δp/Δt. Whenever a problem mentions changing mass — fuel, falling sand, a leaking tank — reach for the momentum form.
WE 1
A resultant force acts on a body, changing its momentum by 3000 kg m s−1 over a time of 6.0 s. Calculate the resultant force.
Step 1 — use the momentum form
F = Δp ÷ Δt
Step 2 — substituteF = 3000 ÷ 6.0F = 500 NThe force points in the same direction as the change in momentum.
Direction of forces
Force and momentum are both vectors, so direction matters. Take the initial direction of motion as positive; then a force that opposes that motion comes out negative. When two objects interact, the force one exerts on the other is matched by an equal and opposite force back — Newton’s third law — so the momentum gained by one is exactly the momentum lost by the other.
Picture a car driving into a wall. The car pushes on the wall, and the wall pushes back on the car with an equal and opposite force:
The car exerts a force on the wall (red), and the wall exerts an equal and opposite force back on the car (blue). The force on the car is minus the force on the wall — the momentum the car loses, the wall (and Earth) gains.
WE 2
A car of mass 1200 kg travelling at 15 m s−1 hits a wall and rebounds at 5.0 m s−1. It is in contact with the wall for 0.20 s. Calculate the average force the wall exerts on the car.
Step 1 — list values (initial direction = positive)
m = 1200 kg, u = +15 m/s, v = −5.0 m/s (rebounds)
Δt = 0.20 s
Step 2 — find the change in momentumΔp = m(v − u) = 1200(−5.0 − 15) = −24000 kg m s⁻¹Step 3 — apply F = Δp ÷ ΔtF = −24000 ÷ 0.20F = −1.2 × 10⁵ N (120 kN)The minus sign shows the force acts backward, opposing the car’s initial motion — exactly what a wall does.
💡 Top tips
Use F = Δp/Δt when mass changes; use F = ma when it’s constant. They agree whenever both apply.
Mind the sign of a rebound — the return velocity is negative, making Δp larger.
The force is in the direction of Δp — a negative answer just means it points backward.
Force pairs are equal and opposite — one body’s momentum gain is the other’s loss.
Watch the units — Δp in kg m s−1, time in s, force in N.
Quick recap: force is the rate of change of momentum, F = Δp/Δt. This is the general form of Newton’s second law — it reduces to F = ma when mass is constant, but also works when mass changes. The force points in the direction of the momentum change, and interacting bodies exert equal and opposite forces on each other.
⚠ Common mistakes
Using F = ma when the mass changes — you need the momentum form there
Forgetting the sign of a rebound velocity, which makes Δp too small
Dropping the minus sign in the answer — it tells you the force’s direction
Mixing up Δp (change) with p (the momentum itself)
Confusing the force on one body with the force on the other — they’re equal and opposite
You now have both faces of Newton’s second law and the full toolkit for momentum in a straight line. Next the topic turns to what happens to energy in a collision — whether kinetic energy survives or is lost — starting with Collisions & Explosions in One Dimension and the elastic–inelastic distinction.
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