Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which uses no calculus). For explanations, graphs and worked examples, use the full study guide.
Recap
- The net external force on a system equals the rate of change of its momentum. Internal forces cancel in pairs.
- Impulse is the integral of a force over a time interval. It is a vector, in N·s, pointing the way of the net force.
- Change in momentum is final minus initial: Δp = p_f − p_i.
- Impulse–momentum theorem: the impulse of the net external force equals Δp.
- Graphs: area under F–t = impulse; slope of p–t = net force.
- Special cases: constant mass gives F_net = ma; constant velocity with mass being added gives F_net = v dm/dt.
Key relationships
| Relationship | Units | Graph or condition |
|---|---|---|
| F_net = dp/dt | N | slope of p–t graph |
| J = ∫ F dt | N·s | signed area under F–t graph |
| Δp = p_f − p_i | kg·m/s | use signed values |
| J_net = Δp | N·s = kg·m/s | net external force only |
| F_avg = J/Δt | N | less than the peak for a smooth pulse |
| F_net = m dv/dt = ma | N | constant mass |
| F_net = v dm/dt | N | constant v; added mass has no momentum along v |
Assumptions behind the numbers
- The system is chosen first; only forces from outside it count in F_net.
- During short impacts, weight and friction are often negligible (the collision model, Topic 4.1). Check this with numbers.
- In two dimensions, apply J = Δp to each component separately.
Mistakes to avoid
- Speeds instead of velocities in Δp. Reversals double the change.
- Impulse in the direction of motion. It points along the net force.
- Peak force × time instead of the integral or the area.
- Forgetting p_i when building p(t) from F(t).
- One force’s impulse = Δp when other forces are not negligible.
- F = ma for a system gaining mass. Use dp/dt.
Quick self-check
- Take +x towards a wall. A 0.30 kg ball hits it at 6.0 m/s and rebounds at 4.0 m/s. What is the impulse on the ball? (Δp = 0.30 × (−4.0 − 6.0) = −3.0 N·s, i.e. 3.0 N·s away from the wall)
- F_x = 2.0t (N, t in s) acts from 0 to 3.0 s. What is the impulse? (∫ 2.0t dt = t² from 0 to 3.0 = 9.0 N·s)
- Sand falls onto a belt moving at a constant 3.0 m/s, at 0.50 kg/s. What extra horizontal force keeps the belt at that speed? (v dm/dt = 3.0 × 0.50 = 1.5 N)
Next: practice questions.