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Laws of Motion

Momentum, the second law and impulse

Apply F = ma and impulse = Δp to solve problems.

The momentum of a body is p = mv: mass × velocity. It is a vector pointing the same way as the velocity, measured in kg m s⁻¹. A body at rest has zero momentum. A slow truck and a fast cricket ball can have the same momentum.

The second law: the rate of change of momentum equals the net external force, F = dp/dt. When the mass stays the same, this becomes F = ma. (A rocket burning fuel loses mass, so for it you must use F = dp/dt.)

The first law hides inside the second: if F = 0, then a = 0, and the velocity stays constant.

F = ma works for each direction separately: Fₓ = maₓ and F_y = ma_y. 1 newton (N) = 1 kg m s⁻². Two examples:

  • A force F = 5î + 12ĵ N gives a body an acceleration of 2.6 m s⁻². The force's size is √(25 + 144) = 13 N, so m = 13/2.6 = 5 kg.
  • A 2 kg body has velocity 10î m/s, and a steady force F = –4î + 2ĵ N acts on it. Then a = –2î + ĵ m s⁻², so vₓ = 10 – 2t. At t = 5 s vₓ = 0, and the body is moving straight along y at 5 m/s.

The force here and now fixes the acceleration here and now. It does not remember the past motion. A stone dropped from a speeding train feels only gravity once it has left your hand.

Impulse = force × time = change in momentum (F Δt = Δp). Very large forces that act for a very short time, like a bat on a ball, a hammer on a nail, or a fall onto a hard floor, are easiest to handle through their impulse.

Momentum is a vector, so its change must be found direction by direction. A ball with momentum 6 kg m s⁻¹ strikes a wall at 60° to the normal and bounces off at 60°. The part along the wall (6 sin 60°) does not change; the part normal to the wall (6 cos 60° = 3) reverses. So Δp = 2 × 3 = 6 kg m s⁻¹, straight out from the wall.

Why does a fielder move the hands back when catching? The ball's change in momentum is fixed. Making the stopping time longer makes the force smaller. A 0.2 kg ball at 20 m s⁻¹ stopped in 0.01 s needs 400 N; stopped in 0.1 s, only 40 N. Shock absorbers in cars and sand pits for long jumpers work the same way.

Net force = mass × acceleration; impulse = change in momentum.

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