Conservation of Linear Momentum
One sentence that solves collisions you have no other way to touch, in one dimension and in two.
Describe the total linear momentum of a system, explain when and why that total does not change, and apply it in one dimension and in two.
Everything on this page serves that one sentence.
Topic 4.2 said that a change in momentum is an impulse from outside. Read that backward and you have this topic. If there is no external impulse, there is no change, and the total momentum of the system is the same after the event as before it. You do not need to know anything about what happened during the collision: not the force, not the contact time, not the shape of the objects.
1. The total momentum of a system
Pick a group of objects and call it the system. Its momentum is the vector sum of the momenta of its parts, which is where Topic 4.1 left off.
\(\vec{p}_{\text{sys}} = \sum_i (m_i \vec{v}_i)\)add the vectors
That whole collection can also be described as one object moving at one velocity, the velocity of its center of mass.
\(\vec{v}_{cm} = \frac{\sum_i \vec{p}_i}{\sum_i m_i} = \frac{\sum_i (m_i \vec{v}_i)}{\sum_i m_i}\)the center of mass velocity
This is worth having because of what comes next. With no net external force on the system, this velocity does not change at all, however violently the parts rearrange themselves inside. Two carts can collide, stick, bounce or explode, and the center of mass carries on exactly as it was going. Watching a problem from the frame that moves with the center of mass often turns it into an easier problem.
Topic questions
1 A 3.0 kg cart moving right at 4.0 m/s collides with a 2.0 kg cart at rest, and the two stick together. The velocity of the pair afterward is
2 Two carts of unequal mass move toward each other on a level track with a total momentum of zero. They collide. Immediately afterward, the velocity of the center of mass of the pair is
2. When the total cannot change
If the net external force on a system is zero, the total momentum of that system is constant.
The reason is Topic 4.2 and nothing more. The change in the system’s momentum equals the impulse delivered from outside it. No external force means no external impulse, and no external impulse means no change.
Forces between parts of the system do not count, and this is the point people lose. By Newton’s third law the push cart A gives cart B is equal and opposite to the push B gives A, and the two forces act over exactly the same interval. The two integrals are therefore equal and opposite too. Whatever momentum one cart gains, the other loses.
When the net external force is not zero, nothing is broken. The momentum of the system then changes by exactly the impulse delivered from outside:
\(\vec{J}_{\text{ext}} = \int \vec{F}_{\text{ext}}\,dt = \Delta \vec{p}_{\text{sys}}\)a transfer across the boundary
Topic questions
3 Which condition guarantees that the total momentum of a chosen system is constant?
4 A 5.0 kg cart rolling at 6.0 m/s slows to 4.0 m/s because of friction from the track. Taking the cart alone as the system, the impulse delivered to it is
3. The system is your choice, and the choice decides the answer
Nothing in the last section said which objects to put in the system. That is because nothing can. The system is a boundary you draw, and you may draw it wherever you like. What you cannot do is draw it and then ignore what crosses it.
A ball falling toward the ground is the cleanest case. Take the ball alone and its momentum grows steadily downward, because gravity from the Earth is an external force on it. Take the ball and the Earth together and the same event has a constant total momentum, because that force is now internal and the Earth moves up to meet the ball by an amount far too small to measure. One event, two descriptions, both correct.
Choose the system so that the forces you do not want to deal with end up inside it. That is the whole skill.
The same two carts, on the same track, with a line you draw around whatever you like. Drag either edge of the dashed box. Nothing about the physics changes when you move it. What changes is which forces count as external.
Test thisPut the line around cart A alone, then around both carts, then around both carts and the Earth. Read the verdict each time.
Report what happenedFor each of the three, say whether the total momentum inside the line is constant and name the force that decides it.
Now without the simulationA ball falls freely toward the ground. For which choice of system is the total momentum constant?
Topic questions
5 Two students argue about a collision between two carts on a track. One says momentum is conserved and the other says it is not. They are both describing the same event correctly. The best explanation is that
6 During a collision between cart A and cart B, the impulse A delivers to B is 6.0 N·s to the right. The impulse B delivers to A is
4. Two dimensions, solved
Momentum is a vector, so conservation is a vector statement, and a vector statement in a plane is two separate statements that do not mix.
\(\sum p_{x,\text{before}} = \sum p_{x,\text{after}}\)across
\(\sum p_{y,\text{before}} = \sum p_{y,\text{after}}\)and up
Resolve every velocity into components before you start. Keep the signs. Then you have two equations, and two equations will find two unknowns, which is usually a speed and an angle for one of the objects.
The check that saves the most time. If everything started along one line, the total momentum perpendicular to that line was zero. It is still zero afterward, so any component one object gains across the line is matched exactly by the other object gaining the same amount back the other way.
Puck A slides east across a frictionless table and strikes puck B, which is at rest. Drag the blue handle to choose where A goes afterward, in speed and in direction at once. B is then not free to do anything: the two component equations fix it completely, and the panel shows both of them balancing.
Test thisLeave the masses equal. Drag the handle so that A leaves at a steeper angle while keeping its speed about the same, and watch B. Then drag it so that A barely turns at all.
Report what happenedDescribe what B does as A’s angle steepens. Quote the up and down totals from the panel in two of your settings, and say why they come out as they do.
Now without the simulationA 4.0 kg puck slides east at 5.0 m/s and strikes a stationary 4.0 kg puck. Afterward the first puck moves at 3.0 m/s at 53° north of east. The velocity of the second puck is
One warning the simulation makes visible. The two component equations constrain the outcome, and they do not determine it: there are outcomes that balance both of them and still create kinetic energy, which no real collision does. Momentum alone never rules anything out on energy grounds. That is what Topic 4.4 is for.
Topic questions
7 A puck slides east and strikes a second puck at rest. After the collision the first puck moves northeast. Which statement about the second puck must be true?
8 In a two dimensional collision with two unknown quantities afterward, conservation of momentum supplies
Topic 4.4 asks the other question about a collision. The momentum is always conserved. The kinetic energy is not, and what separates the two cases has a name and supplies the second equation you will often need.