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Study Notes

Unit 4 — Linear Momentum

How motion is transferred during interactions: momentum, impulse, conservation of momentum, and collisions. Each section ends with a short practice set — click "Show answer" to check yourself.

4.1

Linear Momentum

Linear momentum measures how hard it is to stop a moving object. It depends on both mass and velocity, and it points in the direction of the velocity.

p = m · v                     units: kg·m/s
p_system = p₁ + p₂ + …        (vector sum)
K = p² / (2m)                  (links momentum and kinetic energy)
  • Momentum is a vector. In one dimension, use signs: pick a positive direction and treat the opposite as negative.
  • The total momentum of a system is the vector sum of each object's momentum, so opposite directions can cancel.
  • Two objects can have the same kinetic energy but different momentum. Momentum has direction, energy does not.
  • An object at rest has zero momentum. Momentum depends on the reference frame.
Common mix-up: a heavy slow object and a light fast object can have the same momentum. Compare mv, not just mass or speed.
Practice

1. A 1500 kg car travels at 20 m/s. What is its momentum?

Show answer
p = mv = 1500 × 20 = 30,000 kg·m/s in the direction of motion.

2. A 2.0 kg cart moves at +3.0 m/s and a 3.0 kg cart moves at −2.0 m/s. What is the total momentum of the system?

Show answer
ptotal = (2.0)(+3.0) + (3.0)(−2.0) = 6.0 − 6.0 = 0 kg·m/s.
4.2

Change in Momentum and Impulse

Impulse is the product of the net force on an object and the time over which it acts. It equals the object's change in momentum (the impulse–momentum theorem).

J = F_net · Δt = Δp = m·v_f − m·v_i
F_net = Δp / Δt        (Newton's second law in momentum form)
  • Impulse is a vector with units of N·s, which are the same as kg·m/s.
  • On a graph of force vs. time, the area under the curve equals the impulse.
  • For the same Δp, a longer collision time means a smaller average force. This is why airbags, crumple zones, and padded landings work.
  • Bouncing gives a bigger Δp than sticking, because the velocity reverses: Δp = m(vf − vi) with opposite signs.
Tip: always compute final − initial with signs. If a ball reverses direction, its speeds add: 6 m/s in and 4 m/s out means |Δv| = 10 m/s, not 2 m/s.
Practice

1. A 0.50 kg ball hits a wall at +6.0 m/s and rebounds at −4.0 m/s. It is in contact for 0.020 s. Find the impulse and the average force on the ball.

Show answer
Δp = 0.50(−4.0 − 6.0) = −5.0 kg·m/s. Favg = Δp/Δt = −5.0/0.020 = −250 N (directed away from the wall).

2. Why is it safer to bend your knees when landing from a jump?

Show answer
Your momentum must drop to zero either way (same Δp). Bending your knees increases the stopping time, so the average force on your body is smaller.
4.3

Conservation of Linear Momentum

If the net external force on a system is zero (an isolated system), the total momentum of the system stays constant.

p_initial = p_final
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
  • The objects exert equal and opposite forces on each other for the same time (Newton's third law), so their impulses cancel and momentum is only exchanged.
  • Internal forces cannot change the total momentum; only external impulses can.
  • Momentum is conserved separately in each direction: x and y components are handled independently.
  • In explosions and recoil, a system starting at rest has zero total momentum, so the pieces move in opposite directions with equal and opposite momentum.
  • The velocity of the system's center of mass stays constant when the system is isolated.

Steps: define the system, confirm the external force is negligible during the interaction, choose a positive direction, write total p before and after, and solve.

Key idea: conservation of momentum works even when you don't know the forces during a collision or explosion.
Practice

1. A 60 kg skater and a 40 kg skater stand at rest on ice and push off each other. The 60 kg skater moves at +2.0 m/s. What is the 40 kg skater's velocity?

Show answer
0 = (60)(2.0) + (40)v → v = −3.0 m/s (3.0 m/s in the opposite direction).

2. A 5.0 kg rifle fires a 0.020 kg bullet at 500 m/s. What is the rifle's recoil speed?

Show answer
0 = (0.020)(500) + (5.0)v → v = −2.0 m/s. The rifle recoils at 2.0 m/s, opposite to the bullet.
4.4

Elastic and Inelastic Collisions

In an isolated system, momentum is always conserved in a collision. Whether kinetic energy is conserved depends on the type of collision.

TypeMomentumKinetic energyAfter the collision
ElasticConservedConservedObjects separate; no energy lost
InelasticConservedDecreasesObjects separate; some energy lost
Perfectly inelasticConservedMaximum lossObjects stick and move together
Perfectly inelastic:  m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)·v_f
  • "Lost" kinetic energy is transformed into thermal energy, sound, and deformation of the objects.
  • Most real collisions are inelastic. Elastic collisions are an ideal model (e.g. between hard steel balls or very springy objects).
  • In a 1D elastic collision between equal masses, the objects simply exchange velocities.
  • To classify a collision, compare Ktotal before and after: equal → elastic, smaller → inelastic.
Common mix-up: inelastic does not mean momentum is lost. Total momentum is conserved in every isolated collision.
Practice

1. A 2.0 kg cart at 3.0 m/s collides with a 1.0 kg cart at rest, and they stick together. Find the final speed and the kinetic energy lost.

Show answer
vf = (2.0)(3.0)/(3.0) = 2.0 m/s. Ki = ½(2.0)(3.0)² = 9.0 J; Kf = ½(3.0)(2.0)² = 6.0 J. Energy lost = 3.0 J.

2. A 1.0 kg block moving at 4.0 m/s hits a 1.0 kg block at rest. Afterward, the first moves at 1.0 m/s and the second at 3.0 m/s, both in the original direction. Is momentum conserved? What kind of collision is it?

Show answer
Momentum: before = 4.0, after = 1.0 + 3.0 = 4.0 → conserved. Kinetic energy: before = ½(1)(4)² = 8.0 J; after = ½(1)(1)² + ½(1)(3)² = 5.0 J. K decreased, so it is an inelastic collision (not perfectly inelastic, since the blocks don't stick).