Linear Momentum
Mass in motion, and why the direction is half the answer.
Describe the linear momentum of an object or system, and predict how it changes when the mass or the velocity is scaled.
Everything on this page serves that one sentence.
You already have one way to describe how hard something is to stop. Kinetic energy, from Unit 3, is a scalar and it grows as the square of the speed. Momentum is the other way, it is a vector, and it grows as the first power. A loaded truck rolling slowly and a bullet are both hard to stop, and they are hard to stop differently.
There is a second reason this quantity is about to matter more than kinetic energy did. In the next topic you will meet it inside a derivative, and the law you have been calling Newton’s second law will turn out to be a special case of a statement about momentum. Keep that in view while you read this page.
1. Momentum is mass times velocity
\(\vec{p} = m\vec{v}\)kilogram meters per second
There is no separate name for the unit. It is written \(\mathrm{kg\cdot m/s}\) and read as it is written. If a quantity you have calculated does not come out in kilogram meters per second, it is not a momentum.
Momentum is a vector. It points in exactly the same direction as the velocity, because mass is a positive scalar and multiplying a vector by a positive scalar does not turn it. On a straight track a momentum to the right is positive and a momentum to the left is negative, and that sign is part of the answer rather than decoration on it.
Because the relationship is linear in both inputs, scaling questions are easy and the exam asks them often. Double the mass and you double the momentum. Triple the speed and you triple it. Do both and you multiply by six. Nothing is squared here, which is exactly where momentum parts company with kinetic energy.
Two carts on a level track. Drag a cart to move it. Drag the tip of its velocity arrow to change how fast it goes and which way. Drag the mass sliders underneath. The momentum arrow below each cart is drawn to scale, and the bottom arrow is the two of them added.
Test thisSet cart A to 1.0 kg and cart B to 4.0 kg. Drag cart A’s velocity arrow until it reads about \(+4.0\) m/s, and read the system total. Now drag cart B’s arrow until that total disappears.
Report what happenedWhat velocity did cart B need? Write the system momentum as a function of cart B’s velocity, and say what that function’s zero is.
Now without the simulationA 4.0 kg cart moves right at 3.0 m/s. A 6.0 kg cart moves left at speed \(v\). The system momentum is zero when \(v\) equals
Topic questions
1 A 1.5 kg cart moves west at 6.0 m/s. Its momentum is
2 An object’s mass is tripled while its speed is halved. Its momentum is multiplied by a factor of
2. How momentum and kinetic energy scale differently
Momentum answers a question speed alone cannot: how much motion there is to get rid of. A loaded shopping cart and an empty one rolling at the same speed are not equally easy to stop, and the arithmetic says so before your arms do.
\(p = mv \qquad K = \tfrac{1}{2}mv^{2} = \frac{p^{2}}{2m}\)the same motion, two descriptions
The second form of the kinetic energy is worth having. It says that at a fixed momentum, a heavier object carries less kinetic energy. That is why a heavy slow object and a light fast one with the same momentum are not equally dangerous, and it will come back in Topic 4.4.
Worth keeping straight. Kinetic energy is a scalar and goes as \(v^{2}\). Momentum is a vector and goes as \(v\). A pair of objects can have a total momentum of zero while carrying a great deal of kinetic energy between them. The reverse cannot happen.
Topic questions
3 A cart’s speed is tripled and its mass is unchanged. Its momentum and its kinetic energy are multiplied by factors of
4 Object X has mass \(3m\) and speed \(v\). Object Y has mass \(m\) and speed \(3v\). Which statement is correct?
3. A system has one momentum, and it is a sum
Pick any group of objects and call it a system. The momentum of that system is the vector sum of the momenta of its parts. That is all, and it is why the bottom arrow in the first simulation is simply the two arrows above it laid head to tail.
\(\vec{p}_{\text{sys}} = \sum_i \vec{p}_i = \sum_i m_i \vec{v}_i\)add the vectors, not the sizes
Adding the sizes is the most common way to get this wrong. Two carts of momentum \(6\ \mathrm{kg\cdot m/s}\) moving toward each other have a total of zero, not twelve. The sizes do add to twelve, and the sizes are not what the physics asked for.
For a system whose mass is spread out rather than concentrated in a few objects, the sum becomes an integral, \(\vec{p} = \int \vec{v}\,dm\). You will not need to evaluate one in this unit, but the idea is the same and the notation should not surprise you when it appears.
Topic questions
5 Cart A has mass 4.0 kg and moves right at 2.5 m/s. Cart B has mass 2.0 kg and moves left at 5.0 m/s. The momentum of the system is
6 Which statement about the momentum of a system of two objects is always correct?
4. Two models: the collision and the explosion
Momentum earns its keep in two situations the course framework names as models.
A collision is an interaction in which the forces the objects exert on each other are much larger than any net external force on them. For the fraction of a second two carts are touching, the push between them dwarfs friction from the track and everything else acting from outside. That is what lets you ignore the outside world for the length of the interaction.
Because only the state before and the state after are examined, each object can be treated as a single point carrying a mass. Its shape, its spin and where it was struck never enter the analysis. The framework calls that the object model, and it is allowed here precisely because the inside of the interaction is never looked at.
An explosion is the same idea pointed the other way. Forces inside the system move its parts apart: a spring released between two carts, a firework, a person stepping off a skateboard. Nothing is added from outside, so whatever momentum the system had before, it still has after.
A cart with a compressed spring between two halves, at rest on a level track. Drag the divider to decide how the mass is shared, then release the spring. Watch the two momentum arrows.
Test thisDrag the divider until one piece is about three times the mass of the other, release the spring, and read both speeds. Then raise the spring energy and release again.
Report what happenedGive the ratio of the two speeds in both runs, and the ratio of the two momenta. Which ratio depended on the spring energy and which did not?
Now without the simulationA stationary object at rest breaks into two pieces, one of mass \(m\) and one of mass \(4m\). The ratio of the speed of the lighter piece to the speed of the heavier piece is
Notice what the second simulation refuses to do. It never lets the two arrows come out unequal. Change the split, change how hard the spring pushes, and the two momenta still match in size and oppose in direction. That is not the simulation being polite. It is the only outcome the physics allows, and the next two topics are about why.
Topic questions
7 Two gliders collide on an air track. Which condition makes it reasonable to model the interaction as a collision?
8 A student models two colliding carts as single points with mass, ignoring their shape and any rotation. This is justified because
Topic 4.2 asks what changes a momentum. The answer is the time integral of the net force, and for a system whose mass is changing it is the only way to get the right answer at all.