Short on time? This page is the recap. For explanations, graphs and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- The change in momentum is final minus initial momentum. With signs, a rebound gives a bigger change than a stop.
- The net external force on a system equals the rate of change of its momentum. Internal forces cannot change a system’s total momentum.
- Impulse is average force times the time it acts. It is a vector in the direction of the net force.
- Impulse–momentum theorem: the impulse of the net external force equals the change in momentum.
- For constant mass, the theorem becomes Newton’s second law, F_net = ma.
- Systems with changing mass (rockets, leaking carts) are described in words only in this course.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| Δp = p − p₀ = m(v − v₀) | kg·m/s, signed | find a change in momentum |
| J = F_avg Δt | N·s | find impulse from an average force |
| J = Δp | N·s = kg·m/s | link force and time to momentum change |
| F_net = Δp / Δt | N | find average net force from a momentum change |
| area under F_net–t graph | N·s | find impulse when force varies |
| slope of p–t graph | N | find the net external force |
| F_net = ma | constant mass only | recover Newton’s second law |
Assumptions behind the numbers
- The mass of the system stays constant during the interaction.
- In J = Δp, the impulse must come from the net external force. When a surface pushes and gravity also acts, the surface force = net force + weight (for an upward net force).
- During short, hard impacts, small external forces such as gravity and friction are often negligible; say so when you ignore them.
- Areas below the time axis on a force–time graph count as negative impulse.
Mistakes to avoid
- Unsigned velocities in Δp. A rebound from +4 m/s to −3 m/s is a change of 7 m/s, not 1 m/s.
- Peak force × time for a force that rises and falls. Use the area.
- “Soft landings reduce impulse.” They lengthen the time; the impulse stays the same.
- Forgetting the weight when asked for the force from the floor, not the net force.
- Treating impulse and momentum as different units. N·s = kg·m/s.
- Reading a flat p–t graph as “at rest”. It means constant momentum.
Quick self-check
- A constant 3.0 N net force acts for 0.50 s on a 1.5 kg cart at rest. What is its final speed? (J = 1.5 N·s, so v = 1.0 m/s.)
- A momentum–time graph rises in a straight line from 2.0 to 8.0 kg·m/s over 4.0 s. What is the net force? (1.5 N)
- Take +x towards a wall. A 0.10 kg ball arrives at +5.0 m/s and leaves at −3.0 m/s. What is Δp? (−0.80 kg·m/s)
Next: practice questions.