Lesson 1 · 20 min
Why Impulse and Momentum?
A bat meets a ball for a millisecond. A car brakes for a few seconds. Two railcars couple, a ball bounces, an airbag inflates. The questions are about forces that act over a stretch of time, often a very short one, and about the velocities before and after. Newton's second law answers them once it is integrated in time. Do that integration once, in general, and you get the principle of impulse and momentum.
Learning objectives
- Explain how integrating \(\sum\Fvec = m\,d\vvec/dt\) over time relates the velocities at two instants to the forces acting between them.
- Recognize the questions that impulse and momentum answer directly, and the ones that call for work and energy or for \(\sum\Fvec = m\avec\).
- Read the module's color code: one color for momentum, and one for each kind of force by the impulse it delivers.
Tile or carpet?
Your phone slips out of your hand and lands flat, without bouncing. You drop it from the same height twice, once onto a tile floor and once onto thick carpet. The tile stops it in about \(2\ \text{ms}\); the carpet gives way and takes about \(20\ \text{ms}\). Before you look at Figure 1.1, make a prediction.
From \(1.2\ \text{m}\) the phone lands at \(4.85\ \text{m/s}\) with momentum \(mv = 0.970\ \text{kg·m/s}\), and leaves the floor with none. Whatever the floor, that momentum has to go, and the area under the force–time curve is what removes it. Spread the same area over ten times as long and the average force is ten times smaller: \(485\ \text{N}\) on tile, \(48.5\ \text{N}\) on carpet. That is the whole idea behind airbags, crumple zones, packing foam and bending your knees when you land.
Newton's second law, integrated in time
Newton's second law for a particle of constant mass is \(\sum\Fvec = m\avec = m\,d\vvec/dt\). Multiply by \(dt\) and integrate from time \(t_1\), when the velocity is \(\vvec_1\), to time \(t_2\), when it is \(\vvec_2\):
Newton's second law, integrated in time
\[ \int_{t_1}^{t_2} \textstyle\sum\Fvec\,dt = m\vvec_2 - m\vvec_1 \]Velocities at two instants, connected by the forces acting during the time between them. No position, no acceleration.
The left side is the impulse of the forces, the subject of Lesson 2. The right side is the change in the linear momentum \(\colL{\Lvec = m\vvec}\). Lesson 3 turns this into the working form \(m\vvec_1 + \sum\int\Fvec\,dt = m\vvec_2\).
Compare it with the principle of work and energy, which integrates the same law along the path. That integral keeps only the tangential components of force and gives one scalar equation in speed and position. This one integrates over time, keeps every force, and gives a vector equation in velocity and time: one scalar equation for each direction.
Example 1.1 — A stopping time, two ways
A \(0.17\ \text{kg}\) hockey puck slides across the ice at \(10\ \text{m/s}\). The coefficient of kinetic friction between the puck and the ice is \(0.1\). How long does it take to stop?
Show solution
The ice pushes up with \(N = mg\), so friction is \(\colF{F_f} = \mu_k mg\), against the motion. Take the direction of motion as positive.
With Newton's second law and kinematics. \(\sum F_x = ma\) gives \(-\mu_k mg = ma\), so \(a = -\mu_k g = -0.981\ \text{m/s}^2\), constant. Then \(v = v_1 + at\) with \(v = 0\):
\[ t = \frac{v_1}{\mu_k g} = \frac{10}{(0.1)(9.81)} = 10.2\ \text{s} \]With the integrated law. Along the ice the only force is friction, constant, so its impulse is \(-\mu_k mg\,t\):
\[ \colF{-\mu_k mg\,t} = 0 - m v_1 \quad\Rightarrow\quad t = \frac{v_1}{\mu_k g} = 10.2\ \text{s} \]Same answer, and the mass cancels again. The second route needs no acceleration and no constant-acceleration formula, so it still works when the force changes with time, where \(v = v_1 + at\) does not apply. Vertically, the impulses of \(N\) and the weight, \(1.67\ \text{N}\) times \(10.2\ \text{s}\) each, cancel: the puck never moves up or down.
What impulse and momentum answer
Impulse and momentum are the tool of choice whenever the question links a velocity to a time, or asks what happens across a collision, where the forces are huge, brief and unknown.
| You want | Use | Where |
|---|---|---|
| A velocity after a given time, or the time to reach a velocity | Impulse and momentum | Lessons 3–4 |
| The response to a force that changes with time | Impulse and momentum, with \(\int F\,dt\) | Lesson 4 |
| The average force during a contact of known duration | Impulse and momentum | Lessons 2–3 |
| Velocities after bodies push apart, couple or collide | Conservation of momentum, and the coefficient of restitution | Lessons 5–7 |
| A speed at a position, or the distance to stop | Work and energy | The Work and Energy module; Lesson 8 |
| An acceleration, or a force at one instant | Newton's second law | Earlier in the course |
Color tells you what changes the momentum
Every figure, readout and highlighted formula in this module uses the same colors. Momentum and velocity are green. Each force is drawn in the color of its impulse, so a ledger of impulses reads like the arrows in the figure.
- Momentum and velocityThe momentum \(\colL{m\vvec}\), its total for a system, and the velocity arrow \(\colL{\vvec}\)
- GravityThe weight \(\colVg{W}\) and its impulse \(\colVg{W\,\Delta t}\)
- SpringsThe spring force \(\colVe{F_s}\) and its impulse
- FrictionThe friction force \(\colF{F_f}\) and its impulse
- Applied forcesA push or pull \(\colP{P}\) and its impulse
- Impact forcesThe large, brief contact forces between colliding bodies, \(\colI{F}\), and their impulse
- Normal forcesThe push \(N\) of a supporting surface, drawn in black
The road ahead
- Lesson 2 defines linear momentum and the impulse of a force, and reads impulse as the area under a force–time graph.
- Lesson 3 states the principle of linear impulse and momentum with a procedure, and Lesson 4 applies it to forces that change with time.
- Lesson 5 adds up the momenta of several bodies and finds when the total is conserved.
- Lessons 6 and 7 analyze collisions, head-on and at an angle, with the coefficient of restitution.
- Lesson 8 chains impact, momentum and energy in problems with several stages.
Check your understanding
Key takeaways
- Newton's second law integrated in time gives \(\int\sum\Fvec\,dt = m\vvec_2 - m\vvec_1\): velocity and time, with no position.
- Every force acting during the interval contributes, including forces that do no work, like the normal force.
- The same change in momentum can come from a large force over a short time or a small force over a long time. Stretching the time lowers the force.
- Next, Lesson 2 defines both sides of the integrated law: linear momentum and impulse.