Momentum problem studio

Momentum is conserved for a system over an interval, never for an event in general. Choose the system so the large forces are internal, then estimate the external impulse over the interval you mean to use. Two carts of total mass 2.0 kg carrying 3.0 kg m s⁻¹ across a floor with μₖ = 0.30 feel at most 0.30 × 2.0 × 9.81 = 5.9 N of friction. Over a 12 ms contact that is 0.071 N s, 2.4% of their momentum, so conservation across the contact is sound; over the 0.50 s skid afterwards the same force delivers 2.9 N s, nearly all of it. Brevity only defeats forces that stay finite. When a 5.0 kg sandbag falling at 4.0 m s⁻¹ lands in a 20 kg cart rolling at 2.0 m s⁻¹, the track's normal force spikes as hard as the landing, so vertical momentum is not conserved. No horizontal force spikes, so horizontal momentum is: (20)(2.0) = (25)V gives V = 1.6 m s⁻¹.

Before writing an equation, list the stages and what each conserves. A stage ends whenever a force switches on or off: contact begins, a spring lets go, a surface turns rough. Two gliders of 0.50 kg and 0.25 kg are pushed apart from rest by a spring storing 0.75 J. Stage 1, the release, conserves momentum and mechanical energy: 0.50v₁ = 0.25v₂ and 0.25v₁² + 0.125v₂² = 0.75 give 1.0 and 2.0 m s⁻¹. Stage 2, the lighter glider coupling to a 0.75 kg glider, conserves momentum only: (0.25)(2.0) = (1.00)V, so V = 0.50 m s⁻¹ and the kinetic energy drops from 0.50 J to 0.125 J. Stage 3, the pair coasting up a ramp, conserves energy only: h = V²/2g = 1.3 cm. Carry energy straight through stage 2 and you predict 5.1 cm, four times too high — the factor (m + M)/m that the coupling threw away.

A one-dimensional collision has two unknown final velocities and momentum supplies one equation, so something else must close it: the bodies stick (e = 0), the collision is elastic (e = 1), e is measured, or one final velocity is. Prefer the restitution equation to the energy equation even when the collision is elastic, because energy is quadratic and brings a spurious root. For a 2.0 kg cart at 3.0 m s⁻¹ striking a 1.0 kg cart at rest elastically, 6.0 = 2.0v₁ + 1.0v₂ with v₂ − v₁ = 3.0 gives v₁ = 1.0 m s⁻¹ and v₂ = 4.0 m s⁻¹ in two lines; the energy route also returns v₁ = 3.0, v₂ = 0, the carts passing through each other untouched. In two dimensions the count is four unknown components against two momentum equations and at most one energy equation. One measured angle or speed is still missing, and without it the correct answer is a family of collisions.

Write each velocity as v_cm plus a velocity relative to the centre of mass. The cross terms vanish because the relative momenta sum to zero, leaving K = ½Mv_cm² + ½μuᵣₑₗ², with M the total mass and μ = m₁m₂/M. Internal forces cannot change v_cm, so the first term is untouchable; a collision acts only on the second, turning uᵣₑₗ into −e uᵣₑₗ and removing ½μ(1 − e²)uᵣₑₗ². The payoff is that damage tracks closing speed. A 1200 kg car at 15 m s⁻¹ runs into the back of an 1800 kg car doing 10 m s⁻¹ and the bumpers lock: μ = 720 kg and uᵣₑₗ = 5.0 m s⁻¹, so 9.0 kJ is dissipated out of 225 kJ carried. The same pair meeting at 5.0 m s⁻¹ with the front car parked loses the same 9.0 kJ, but out of only 15 kJ.

When a system starts at rest and no external horizontal force acts, its centre of mass stays where it began however the parts rearrange, so m₁Δx₁ + m₂Δx₂ = 0 with both displacements measured from the ground. The trap is the distance you are given, which is almost always relative. A 75 kg angler walks 5.0 m from stern to bow of a 175 kg boat on still water. Her displacement relative to the boat is Δx₁ − Δx₂ = 5.0 m; combining the two equations gives Δx₂ = −(75)(5.0)/250 = −1.5 m and Δx₁ = +3.5 m. Check: 75 × 3.5 = 262.5 = 175 × 1.5. Enter 5.0 m as her ground displacement instead and the boat moves 2.1 m, which would have her walk 7.1 m along a 5.0 m deck. The same trap sits in every launcher problem: a projectile's speed relative to the gun must be converted to the ground frame before it enters the momentum sum.

A result that conserves momentum can still be impossible, so every collision answer gets two more tests and a limit. Is the final kinetic energy no larger than the initial, unless a spring or explosive released energy? Is the body that was behind still no faster than the one ahead? A 1.5 kg cart at 4.0 m s⁻¹ hits a 0.50 kg cart at rest, carrying 6.0 kg m s⁻¹ and 12 J. The claim v₁ = −1.0 m s⁻¹, v₂ = 15 m s⁻¹ conserves momentum and makes 57 J out of 12. The claim v₁ = 3.5 m s⁻¹, v₂ = 1.5 m s⁻¹ conserves momentum and loses energy plausibly, but drives the rear cart through the front one. The allowed outcomes run from 3.0 m s⁻¹ each (stuck) to 2.0 and 6.0 m s⁻¹ (elastic). Then test a limit: a very heavy target must return v₁ = −e u₁, and equal masses in an elastic collision must swap velocities.