Section 1 of 9Newton's second law in terms of momentum
Newton originally stated his second law in terms of momentum: the resultant force on an object is equal to the rate of change of its momentum. For a force acting for a time Δt that changes the momentum by Δp:
F = Δp / Δt
F is the resultant force in N, Δp the change in momentum in kg m/s and Δt the time in s. The direction of the force is the direction of the change in momentum.
The equation shows that force and rate of change of momentum are the same thing. If momentum changes quickly the force is large; if it changes slowly the force is small. A constant force gives a steady change of momentum, represented by a straight line on a momentum–time graph whose gradient equals the force.
Section 2 of 9Link to F = ma
If the mass is constant, Δp = m(v − u), so F = m(v − u)/Δt = ma, because acceleration a = (v − u)/Δt. The momentum form is more general, since it also works when mass changes (for example a rocket burning fuel) and it is convenient for impacts, where the time of contact is short and unknown.
In practice the momentum form is used when the time of contact is known from a measurement, such as high-speed video of a ball hitting a wall, or a light gate timing a collision.
Units help as a checkkg m/s divided by s gives kg m/s2, which is the newton.
Section 3 of 9Impulse (Supplement)
Rearranging gives F Δt = Δp.
A force–time graph for an impact, such as a ball hitting a bat, is normally a curve: the force rises to a peak and falls back to zero over a few milliseconds. The area under the curve gives the impulse, which also equals the change in momentum. The average force is the impulse divided by the contact time.
Section 4 of 9Change in momentum
Change in momentum = final momentum − initial momentum, and direction matters.
- which is greater than if the ball simply stopped,
- because the direction reverses.
Section 5 of 9Everyday applications
Fluids give another example. When a hose directs 5 kg of water per second at a wall at 10 m/s and the water stops, momentum lost per second = 5 × 10 = 50 kg m/s per s, so the force is 50 N. This is how a water jet exerts a force.
Section 6 of 9Method for calculations
Always check that the direction of the force agrees with the direction of Δp. This links directly to the safety features in a later lesson.
Section 7 of 9Worked examples
A 0.057 kg tennis ball is served and leaves the racket at 50 m/s from rest. The contact time is 0.005 s. Find the average force.
Δp = 0.057 × 50 − 0 = 2.85 kg m/s.
F = Δp/Δt = 2.85 / 0.005 = 570 N.
A 1200 kg car slows from 15 m/s to rest in 3.0 s. Find the average braking force.
Δp = 0 − 1200 × 15 = −18 000 kg m/s.
F = −18 000 / 3.0 = −6000 N, so 6000 N backwards.
A 0.50 kg ball hits a wall at 8.0 m/s and rebounds at 6.0 m/s. The contact lasts 0.020 s. Find the average force on the ball.
Take the initial direction as positive. Δp = 0.50 × (−6.0) − 0.50 × 8.0 = −3.0 − 4.0 = −7.0 kg m/s.
F = −7.0/0.020 = −350 N; the force on the ball has size 350 N, directed away from the wall.
Section 8 of 9Key terms
| Term | Meaning |
|---|---|
| Impulse | Force multiplied by time, equal to change in momentum. |
| Rate of change of momentum | Change in momentum divided by time taken. |
| Contact time | Time during which two objects interact. |
| Average force | Total change in momentum divided by time. |