Linear momentum and Newton's second law
Momentum is mass times velocity, p = mv. Force equals the rate of change of momentum, which gives F = ma. The SI unit of force is the newton.
Linear momentum
Linear momentum (p) of a body is the product of its mass and velocity: p = m × v. It is a vector: it points in the direction of the velocity. Its SI unit is kg m s⁻¹. A loaded lorry and a bicycle moving at the same speed have very different momentum. The lorry has far more momentum, so it is much harder to stop.
Deriving F = ma from the second law
**Newton's second law**: the force acting on a body is **directly proportional to the rate of change of its linear momentum**. The change in momentum takes place in the direction of the force.
- 1A body of mass m moves with initial velocity u. A force F acts on it for time t, and its velocity becomes v.
- 2Initial momentum = mu. Final momentum = mv. Change in momentum = mv − mu = m(v − u).
- 3Rate of change of momentum = m(v − u)/t.
- 4Acceleration a = (v − u)/t, so the rate of change of momentum = ma.
- 5★By the second law, F ∝ ma, so F = k ma. The unit of force is chosen so that k = 1. So F = ma.
★ key step
Units of force
| System | Unit | Definition | Value |
|---|---|---|---|
| SI | newton (N) | Force that gives 1 kg an acceleration of 1 m s⁻² | 1 N = 1 kg m s⁻² |
| CGS | dyne | Force that gives 1 g an acceleration of 1 cm s⁻² | 1 N = 10⁵ dyne |
| Gravitational (SI) | kilogram force (kgf) | Force of gravity on a 1 kg mass | 1 kgf = 9.8 N |
| Gravitational (CGS) | gram force (gf) | Force of gravity on a 1 g mass | 1 gf = 980 dyne |
Impulse
An impulsive force is a large force that acts for a very short time. Impulse (J) is force × time, and it equals the change in momentum: J = F × t = Δp. Its unit is N s (the same as kg m s⁻¹). For the same change in momentum, a longer time means a smaller force. A cricket fielder pulls the hands back while catching. This increases the time of impact and reduces the force on the hands. Sand in a long-jump pit and the shock absorbers of vehicles work the same way.
Force on a fielder's hands
A cricket ball of mass 0.15 kg moving at 20 m s⁻¹ is stopped by a fielder in 0.1 s. Find (i) its initial momentum (ii) the force on the hands.
- Given: m = 0.15 kg, u = 20 m s⁻¹, v = 0, t = 0.1 s.
- (i) Formula: p = mu = 0.15 kg × 20 m s⁻¹ = 3 kg m s⁻¹.
- (ii) Change in momentum = mv − mu = 0 − 3 = −3 kg m s⁻¹.
- Formula: F = (mv − mu)/t = −3 kg m s⁻¹ ÷ 0.1 s = −30 N.
- The minus sign shows that the force acts opposite to the motion of the ball.
Momentum = 3 kg m s⁻¹; force = 30 N, opposite to the ball's motion. If the fielder pulls the hands back and takes 0.3 s, the force drops to 3 ÷ 0.3 = 10 N.
Catching a fast ball with the hands moving back, or dropping an egg on a pillow, gives a longer stopping time and so a smaller force. Helmets, air bags and autorickshaw shock absorbers use the same idea.
2 marks: “Define 1 newton.” 1 newton is the force that produces an acceleration of 1 m s⁻² in a body of mass 1 kg. 1 N = 1 kg m s⁻².
2 marks: “A force of 20 N acts on a body of mass 4 kg. Find its acceleration.” a = F/m = 20 N ÷ 4 kg = 5 m s⁻².
4 marks: “State Newton's second law of motion and derive F = ma.” Law in one line; momentum changes from mu to mv in time t; rate of change = m(v − u)/t = ma; F ∝ ma, F = kma, k = 1, so F = ma; unit newton.
Push changes p: force is the rate of change of momentum, F = (mv − mu)/t = ma.