Change in Momentum and Impulse
A derivative, an integral, and the one case where force is not mass times acceleration.
Describe the change in the linear momentum of a system, relate it to the time integral of the net force, and handle a system whose mass is changing.
Everything on this page serves that one sentence.
Topic 4.1 said what a momentum is. This one says what changes it. The answer is not a force on its own and it is not a length of time on its own. It is the force integrated over the time it acted, that integral has its own name, and once you have the name two graphs you have been drawing since Unit 1 start telling you things they were not telling you before.
It also fixes something you may not have noticed was broken. Newton’s second law in the form you have been using assumes the mass is constant. Most of the time it is. When it is not, the momentum statement still works and the familiar one does not.
1. The impulse is an integral
Push a cart gently for a long time and push it hard for an instant, and you can arrive at the same final speed. What the cart responds to is the accumulated effect of the force over time.
\(\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\,dt\)newton seconds
That is the impulse. Its units are newton seconds, and a newton second is the same thing as a kilogram meter per second: \(\mathrm{N\cdot s} = (\mathrm{kg\cdot m/s^{2}})(\mathrm{s}) = \mathrm{kg\cdot m/s}\). Impulse and momentum share a unit because they are the same kind of quantity, which is the first hint of what is coming.
\(\vec{J} = \int \vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}\)the impulse momentum theorem
When the net force happens to be constant the integral collapses to a product, \(\vec{J} = \vec{F}\Delta t\), and that is the only case AP Physics 1 is given. Treat it as the easy special case rather than as the definition, because a collision force is never constant.
Impulse is a vector, and it points along the net force that delivered it. A force to the left delivers an impulse to the left and makes the momentum more negative. Signs are not bookkeeping here, they are the answer.
One push on a cart that starts at rest, drawn twice. The top graph is the force against time. The bottom graph is its integral, the momentum. Drag the gold handle: sideways changes how long the push lasts, upwards changes how hard it is. The shaded area above and the rise below are the same number.
Test thisTick Hold the impulse fixed, then drag the handle sideways until the push lasts about twice as long. Read the peak force before and after, and watch the steepest slope of the lower curve.
Report what happenedGive both peak forces and both durations. Say what happened to the steepest slope of the momentum curve and why that had to happen.
Now without the simulationA net force on a cart rises linearly from zero, \(F(t) = kt\), and is switched off at time \(T\). The impulse delivered is
Topic questions
1 A 4.0 kg cart is at rest on a level track. A constant net force of 8.0 N acts on it for 2.5 s. The speed of the cart at the end is
2 The net force on an object is \(F(t) = 60t\) newtons, with \(t\) in seconds, acting from \(t = 0\) to \(t = 0.40\) s. The impulse delivered is
2. A change is the end minus the start
\(\Delta\vec{p} = \vec{p} – \vec{p}_0\)after, then before
Written out that looks too obvious to need a line of its own. It gets one because the single most expensive mistake in this unit is subtracting in a hurry when the two momenta point in opposite directions.
A ball of mass \(m\) arrives at a wall at speed \(v\) and stops. Take the direction the ball is traveling as positive, so \(p_0 = +mv\) and \(p = 0\), which gives \(\Delta p = -mv\). Now let the same ball bounce straight back at the same speed. Then \(p = -mv\) and \(\Delta p = -2mv\). Bouncing costs the wall twice what stopping costs it, and in both cases the minus sign says the impulse points back the way the ball came.
This is why things that bounce hit harder. A ball that rebounds does not only have its momentum removed, it has the same amount supplied again the other way. Two deliveries, not one.
Topic questions
3 A 0.25 kg ball travels at 8.0 m/s toward a wall and rebounds along the same line at 6.0 m/s. The magnitude of the change in its momentum is
4 The same ball instead hits the wall and stops. Compared with the rebound, the magnitude of the impulse the wall delivers is
3. The area, and the slope
Two graphs now carry the whole topic between them.
On a force against time graph the impulse is the area. Area above the axis counts as positive, area below it counts as negative, and the two can cancel.
This is worth more than it looks. It means you never need the detailed shape of a collision force to find what the collision did. Integrate it, or break the shape into a triangle and a rectangle, and you have the change in momentum without knowing anything about what happened inside.
\(\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}\)the net external force
On a momentum against time graph the net external force is the slope. A steeper curve means a larger force. A horizontal line means no net force at all.
The two statements are the same statement read in opposite directions, which is why the lower panel of the first simulation is the upper one integrated, and why its steepest point sits directly under the peak of the triangle.
Topic questions
5 The momentum of a cart is plotted against time and the graph is a straight line rising from 4.0 kg·m/s to 16 kg·m/s over 3.0 s. The net force on the cart is
6 A car is designed so that its front end crumples during a crash. Compared with a rigid front end, the crumple zone
4. When the mass is changing
Everything above holds whether the mass is constant or not, because it is written in terms of momentum. The familiar form of Newton’s second law does not.
\(\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}\)the general statement
Hold the mass constant and the derivative gives \(m\,d\vec{v}/dt = m\vec{a}\), and you are back to Unit 2. Newton’s second law is therefore not an independent law at all. It is what this statement becomes in the common case.
Now hold the velocity constant instead and let the mass change. The framework asks you to be able to handle exactly this case.
\(\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{dm}{dt}\vec{v}\)constant velocity, changing mass
Nothing is accelerating, so \(m\vec{a}\) is zero, and yet a force is required. The momentum of the system rises because its mass rises, and a rising momentum needs a force whatever the reason for the rise.
A hopper drops sand onto a conveyor belt that is held at a constant speed. Drag the sliders to set how fast the sand arrives and how fast the belt runs. Nothing on this screen is accelerating.
Test thisLeave the belt speed at 2.0 m/s and double the rate the sand arrives. Then put the rate back and double the belt speed instead. Read the force both times.
Report what happenedGive the force in all three settings. The force is proportional to one of the two sliders and to the other as well. Say which change doubled it and which quadrupled the power, and why those are not the same.
Now without the simulationSand falls onto a belt at a steady 5.0 kg/s. The belt is held at a constant 2.0 m/s. The horizontal force needed to keep the belt at that speed is
The belt is the standard example and it is worth sitting with. A rocket is the other one, and it is the case where both the mass and the velocity change at once. This course asks you to handle the constant velocity case quantitatively and no more than that.
Topic questions
7 Sand falls vertically onto a horizontal conveyor belt at a steady rate \(dm/dt\). The belt moves horizontally at a constant speed \(v\). The horizontal force that must be applied to the belt is
8 A student says that because the belt is not accelerating, the net horizontal force on it must be zero. The error is that
Topic 4.3 takes the same idea and points it inside a system. If nothing pushes from outside, the total momentum cannot change at all, and that single sentence solves collisions in one dimension and in two.