Linear Momentum & Systems

The momentum of a particle is p = m v, so it points along the velocity and carries the unit kg m s⁻¹, which is the same as N s. Work in components: a 0.50 kg ball moving with v = (6.0, −8.0) m s⁻¹ has p = (3.0, −4.0) kg m s⁻¹, of magnitude 5.0 kg m s⁻¹. Because p is a vector, a change of direction is a change of momentum. A 0.20 kg ball that meets a wall at 15 m s⁻¹ and leaves at 15 m s⁻¹ goes from +3.0 to −3.0 kg m s⁻¹, so Δp = −6.0 kg m s⁻¹, not zero. Nor is momentum kinetic energy in disguise. K = p²/(2m) weights the same momentum very differently: a 0.145 kg baseball at 40 m s⁻¹ has p = 5.8 kg m s⁻¹ and K = 116 J, while a 1500 kg car creeping at 3.9 mm s⁻¹ has the same momentum and only 0.011 J of kinetic energy.

Write the second law as Fₙₑₜ = dp/dt. With constant mass, dp/dt = m dv/dt = m a, so nothing new is claimed; what changes is the question you ask. The force says how fast, and in which direction, p is changing, not which way p points. A 0.40 kg ball thrown with v = (12, 9.0) m s⁻¹ has p = (4.8, 3.6) kg m s⁻¹. In flight, ignoring air, the only force is its weight, (0, −3.92) N. So pₓ stays at 4.8 kg m s⁻¹ for the whole flight, while p_y falls by 3.92 kg m s⁻¹ every second: zero after 0.92 s at the top, and −3.6 kg m s⁻¹ when the ball is back at launch height. The momentum follows the path; the force points straight down throughout. The same reading turns data into force: if a sensor gives p(t) = 2.0 + 0.30t² kg m s⁻¹ along a line, the net force is F = dp/dt = 0.60t N, which is 1.8 N at t = 3.0 s.

A system is whatever collection of particles you decide to treat together, and its total momentum is the vector sum P = Σ mᵢ vᵢ. Three pucks on an air table move at (5.0, 0) m s⁻¹ with mass 0.20 kg, at (−2.0, 3.0) m s⁻¹ with mass 0.30 kg, and at (0, −1.2) m s⁻¹ with mass 0.50 kg. Their momenta are (1.0, 0), (−0.60, 0.90) and (0, −0.60) kg m s⁻¹, so P = (0.40, 0.30) kg m s⁻¹, of magnitude 0.50 kg m s⁻¹. Adding the three magnitudes instead gives 1.0 + 1.08 + 0.60 = 2.68 kg m s⁻¹, more than five times too large and not a property of anything. A system can have P = 0 while every part moves fast: two carts racing apart, or a spinning wheel. So P says nothing about how much motion is going on inside a system; it describes only how the system as a whole is carried along.

Write the second law for each particle i. The force on it splits into the external force from outside the system, Fᵢ(ext), and internal forces Fⱼᵢ, each exerted on i by another particle j of the system: dpᵢ/dt = Fᵢ(ext) + Σⱼ Fⱼᵢ. Now add these equations for all N particles. The left side becomes dP/dt. On the right, every internal force Fⱼᵢ appears alongside its third-law partner Fᵢⱼ = −Fⱼᵢ, so the internal terms cancel pair by pair, leaving dP/dt = Σ Fₑₓₜ. Internal forces are not small; they are often the largest forces present. They only move momentum from one part to another. Two skaters of 60 kg and 45 kg on smooth ice, pushing apart with 150 N: one's momentum changes at 150 kg m s⁻¹ each second in one direction, the other's at 150 kg m s⁻¹ each second in the opposite direction, and the total changes at 0. Their accelerations, 2.5 and 3.3 m s⁻², differ; their momentum rates cannot.

Internal and external are not properties of a force; they depend on the system you chose. When a skater hauls a sled on a rope, the tension is external to the sled alone and internal to skater, rope and sled together. Choose the system so that the forces you cannot measure are internal and the external ones are known or negligible. A falling 0.50 kg ball, taken alone, has its 4.9 N weight acting from outside, so its momentum grows downward at 4.9 kg m s⁻¹ each second. Put Earth inside the system and gravity becomes internal: Earth's momentum grows upward at exactly the same rate, which for 5.97 × 10²⁴ kg is an acceleration of 8.2 × 10⁻²⁵ m s⁻², real but far beyond measurement. A system is effectively isolated when its external forces are small beside its internal ones; judging whether that holds is the first step in every problem in this unit.

Velocity is measured relative to a frame, so momentum is too. An observer moving at constant velocity u measures v′ = v − u for every particle, and therefore P′ = P − M u, where M is the total mass. On a straight road, a 1200 kg car at 25 m s⁻¹ and a 9000 kg bus at 20 m s⁻¹ have P = 2.10 × 10⁵ kg m s⁻¹ in the road frame. A cyclist riding the same way at 5.0 m s⁻¹ measures 1.59 × 10⁵ kg m s⁻¹, and a passenger on the bus measures only 6.0 × 10³ kg m s⁻¹. Every observer gets a different number, but because M u is constant, all of them see P change by the same amount at the same rate and infer the same external force. That is why dP/dt = Σ Fₑₓₜ holds in every inertial frame. In an accelerating frame u changes with time, and an extra −M du/dt appears on the right-hand side: a force that no object exerts.