Conservation of Linear Momentum
One sentence that solves collisions you have no other way to touch.
Describe the total linear momentum of a system, and explain when and why that total does not change.
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 impulse from outside, 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. You do not need the force, the contact time or the shape of the objects. You need the state before and the state after, and one equation connects them.
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.
\(\vec{p}_{\text{sys}} = \sum m_i \vec{v}_i\)add the vectors
That whole collection can also be described as one object moving at one velocity, the velocity of its center of mass.
\(\vec{v}_{cm} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}\)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.
Two carts on a level track, set up by hand and then released. Drag the sliders to choose the masses and the velocities, tick the box to decide whether they stick, then press Release. The two bars underneath are the total momentum before and the total momentum after.
Test thisTry to make the two bars disagree. Change both masses, make one velocity negative and then both, untick the box so the carts bounce apart, and release after each change.
Report what happenedGive three settings you tried and the two totals each one produced. Say what you were unable to do, and name the rule that stopped you.
Now without the simulationA 3.0 kg cart moving right at 2.0 m/s collides with a 1.0 kg cart moving left at 4.0 m/s, and the two stick together. Their common velocity afterward is
Topic questions
1 A 2.0 kg cart moving right at 3.0 m/s collides with a 1.0 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 for exactly the same length of time. So the two impulses are equal and opposite too. Whatever momentum one cart gains, the other loses, and the total does not move.
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, which is the same equation you already have:
\(\vec{J}_{\text{ext}} = \Delta \vec{p}_{\text{sys}}\)a transfer across the boundary
Topic questions
3 Which condition guarantees that the total momentum of a chosen system is constant?
4 A 4.0 kg cart rolling at 5.0 m/s slows to 3.0 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
Momentum is a vector, so conservation is a vector statement, and a vector statement in a plane is two separate statements.
\(\sum p_{x,\text{before}} = \sum p_{x,\text{after}}\)across
\(\sum p_{y,\text{before}} = \sum p_{y,\text{after}}\)and up
The two do not mix. A collision that sends one object away at an angle has not changed the total momentum across, and it has not changed the total momentum up either. Set the two equations down separately, resolve every velocity into components before you start, and keep the signs.
For AP Physics 1 the work here is setting those equations up correctly and reasoning about them. If the first object leaves at a steeper angle, the second must carry more momentum the other way to keep the up and down total at zero, and you can say that without solving anything.
A useful check. 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 by the other object gaining the same amount back the other way.
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 the same collision, the first puck instead leaves at a steeper angle north of east while its speed is unchanged. Compared with before, the southward momentum of the second puck is
9 AP Physics 2. A 2.0 kg puck slides east at 6.0 m/s and strikes a 2.0 kg puck at rest. Afterward the first puck moves at 3.0 m/s at 60° north of east. The velocity of the second puck is
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.