Newton’s second law has a deeper form than F = ma. The resultant force on a body equals the rate of change of its momentum. That version works even when the mass changes, and it’s the key to handling collisions and rebounds where the velocity flips sign.
📘 What you need to know
Newton’s second law (general form):F = Δp ⁄ Δt.
Δp = pf − pi = m(v − u).
The resultant force acts in the same direction as the change in momentum.
For constant mass this reduces to F = ma; the momentum form is needed when mass changes.
By Newton’s third law, in a collision each object feels an equal-and-opposite force from the other.
Force as the rate of change of momentum
Newton actually phrased his second law in terms of momentum:
Newton’s second law (momentum form)F = Δp ⁄ Δt
That is, the resultant force on a body is how quickly its momentum is changing. Because momentum is a vector, the force points in the same direction as Δp — which isn’t always the direction the object is moving.
Why this reduces to F = ma
When the mass is constant, only the velocity changes, so Δp = m(v − u) and:
Constant massF = m(v − u) ⁄ Δt = m × a
So F = ma is just a special case. The momentum form is the one to reach for whenever the velocity changes by a known amount in a known time — especially in collisions, where a brief force changes the momentum sharply.
Direction in a collision
The resultant force acts in the direction of Δp — not the velocity. When something rebounds, the velocity flips sign, the change in momentum is large and opposite to the initial motion, and the force on the object points backwards. By Newton’s third law, the object pushes the wall (or floor, or other object) forwards with the same size of force.
The wall pushes back on the car; by Newton’s third law the car pushes the wall the other way with the same size of force.
Quick reference:F = Δp ⁄ Δt = m(v − u) ⁄ Δt • points along Δp • reduces to F = ma for constant mass • use signed velocities for rebounds.
Worked examples
WE 1
A car rebounds from a wall
A 1500 kg car hits a wall at 15 m s⁻¹ and bounces back at 5.0 m s⁻¹. The collision lasts 3.0 s. Find the average force on the car during the impact.
Take the initial motion as positive: u = +15, v = −5Initial momentum: p_i = 1500 × 15= 22 500 kg m s⁻¹Final momentum: p_f = 1500 × (−5)= −7 500 kg m s⁻¹Δp = p_f − p_i = −30 000 kg m s⁻¹F = Δp / Δt = −30 000 ÷ 3.0F = −10 000 N (opposing motion)by Newton’s 3rd law the car pushes the wall with +10 000 N
WE 2
A skateboarder being pushed
A girl on a skateboard speeds up from 1.0 m s⁻¹ to 4.0 m s⁻¹ in 2.5 s while a 72 N force pushes her forwards. Calculate her combined mass (girl + skateboard).
Use F = m(v − u) / Δt → m = F·Δt / (v − u)m = (72 × 2.5) ÷ (4.0 − 1.0)m = 180 ÷ 3.0m = 60 kg
💡 Top tips
Pick a positive direction first so every velocity (and the resultant force) has a clear sign.
Use Δp with signs for any rebound — speed alone isn’t enough, you need velocity.
The force is along Δp, which is often not along the initial velocity.
Pair it with Newton’s third law: the force on the wall is equal in size to the force on the car, in the opposite direction.
Stretching the time reduces the average force — the impulse–force link from the previous topic.
⚠ Common mistakes
Calculating Δp as m(v + u) instead of m(v − u) on a rebound.
Dropping the sign of the force and then saying it acts in the wrong direction.
Using F = ma when the mass changes — for a rocket losing fuel, only the momentum form is right.
Confusing the forces in a third-law pair with two equal forces on the same object.
Notice the difference from the impulse page: there we asked “how much momentum changed?”, here we ask “how fast did it change?”. The first answers J = Δp; the second answers F = Δp ⁄ Δt. Up next: Collisions & Explosions in 1-D, where we put conservation of momentum together with kinetic energy to classify elastic and inelastic events.
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