# Mechanics 7: Conservation of Momentum and Collisions

This chapter, we introduce a new quantity, called *momentum.* It is defined as Since velocity is a vector, so is momentum — it is a scalar () times a vector, which is a vector. Momentum is a way of quantifying the force and impact of a moving object in a collision. Think of it this way: If you were standing on a baseball field, would you rather be hit by a baseball moving 100 mph (45 m/s) or a car? Even though these objects would be moving at the same speed, the car would have 5000 times the momentum of the baseball due to having 5000 times more mass. That’s momentum!

### Momentum and Newton’s Second Law

We learned previously that Newton’s second law tells us that the sum of the forces on an object equals it mass times acceleration, or

If we do a little math, we can re-write this another way. If we take the derivative of the momentum equation, we get

since mass is a constant, we can re-write this as

and since we know that the time-derivative of velocity is acceleration, or we get

and setting this equal to the first equation, we have another version of Newton’s second law,

To summarize, the first version tells us that the sum of the forces on a object are proportional to its acceleration. This second form tells us that the sum of the forces on a object tell us how its momentum changes with time. Same thing, just two different ways of seeing/thinking about it.

### Collisions and Conservation of Momentum

Conservation of momentum means that

*total*momentum does not change throughout an interaction (). This is usually some type of collision. The two types of collisions are

*elastic*and

*inelastic.*For both types of collision, momentum is conserved. For an

*elastic*collision, energy is also conserved. Energy is not conserved for the

*inelastic*collision because work is done by one particle on the other, like crumpling or deformation, and probably heating.

For an *elastic* collision, where the masses do not change and energy and momentum are conserved, we can write (subscript for initial, for final)

and the energy (kinetic) is also conserved, written as

Some algebra with these two equations gives

#### Examples

- Say we have mass 1 of kg traveling to the right at m/s, and a mass 2 of kg traveling left at m/s. What are their final velocities?
Using the above equations, we get

now insert the values for initial velocities

So what are these numbers telling us? Mass 1 enters the collision headed right (positive velocity) and leaves headed left (negative velocity). Vice versa for mass 2 — enters headed left and leaves headed right. Also, is half the mass of , and it enters with the velocity of , and leaves only as fast.

Finally, we check the conservation of momentum. The initial momentum is kg-m/s and the final momentum is kg-m/s, so momentum is conserved.

- What if the masses and velocities were equal? Let and and so we get
So the final velocities are equal and opposite from the initial velocities. In other words, the masses leave with the same velocity they started with, but headed in the opposite direction. Intuitively, this makes sense.

- Now, let the masses and velocities of one be the other, like so: This gives us
which simplifies to

So this tells us that the first mass leaves at a velocity 3x larger than it started with, and headed in the opposite direction. The second mass comes to a dead stop!

- Now, let’s do an example of a 2-dimensional elastic collision. That means we will need to work out the and coordinates separately, and this will be an opportunity to use the vectors that we talked about previously. Let mass one be kg with an initial velocity of m/s, and mass two is kg with initial velocity of m/s. Now, we do the same calculation as before, but we need one for and one for
coordinate

coordinate

So in summary, we get (in units of meters-per-second)

Let’s check conservation of momentum. The initial & final momentum are

direction

direction

which shows that momentum is conserved.

Finally, we’ll finish with a diagram. Remember, vectors have a magnitude and a direction, which we can calculate. The magnitude formula is just the distance formula from geometry, and the direction (angle) is from trig, or So, we get

Using these magnitudes (lengths) and direction (angles), we get this diagram:

Which shows us that mass comes in at a angle and leaves slightly slower and at a little sharper angle, but mass comes in headed straight “south” and leaves moving faster and headed “northeast.” You can use this technique to calculate the vectors (magnitude and direction) for any elastic collision. I’ll let you try a few in the homework.

### Inelastic Collisions

The second type of collision is

*inelastic,*where the masses stick together and there is a single final velocity; here momentum is conserved but energy is not conserved

which can be re-written as

### Examples

**One dimensional:**Say we have two masses that collide inelastically. Mass 1 is kg with an initial velocity of m/s to the right. Mass 2 is kg and is sitting still, so m/s. Find the final velocity of the combined masses after the collision.Use the formula above to get

So the final velocity is 1/3 less that the initial, also moving to the right.

**Two dimensional:**Suppose we have two masses moving toward each other and they collide inelastically. Again, two dimensions means that we will have to use vectors. Mass 1 is kg with an initial velocity of m/s. Mass 2 is kg with velocity m/s. Find the final velocity of the combined masses after the collision.We can do this just as before, but we have to do each direction ( and ) separately.

direction

direction

So the final velocity vector is

And the figure looks like this:

### Impulse

The last thing to talk about with momentum and collisions is called*Impulse.*It’s typically noted with an upper-case and it is defined as force through a given time period. Early in this chapter, we said that Newton’s Second Law can be written as: the sum of forces (net force) is the change in momentum over time. So we have: Re-arranging, we getor, using our new definition, Recall that means “a change in,” or “final minus initial,” so our final version is

I cover this here because you may see it in other books. I’ve never found it to be of much use.

### Center of Mass

The center of mass is a concept that is used to make calculations easier. If you have several masses that move together, or a mass that is “spread out,” it

is possible to reduce the system to a single point, called the center of mass.To calculate the center of mass, we multiply the mass of each item times its distance from the origin, then add up all these products, and divide by the total mass. Let’s look at an example. The following diagram show 4 particles arranged around the origin of our coordinate system. Since we have masses distributed in two dimensions, we have to calculate and directions separately.

Let be the total mass (per axis). On the axis, we have the green and red particles, and their center of mass is

and for the axis, we have the yellow and blue particles,

So the center of mass is located at point (-1, 1), shown in the figure as a brown “X.”

Now, say we had a metal disc with a circular hole punched out. The disc has radius and the hole has radius

We can calculate the center of mass in pieces. First, we’ll say that the disc has uniform thickness, and density Remember that mass is volume density. The first piece is the whole disc, with no piece punched out. That is a disc of radius and thickness so the mass of the disc is volume density, or area thickness density. We can write the mass of the disc, asThen, the mass of the “hole,” — the part of the disc that was removed, is

Because of symmetry, the center of mass of each of these pieces is at the center of the disc (or hole). So we have the center of mass for the disc, and the center of mass for the “hole” is at We can calculate the total mass by calculating the mass of the entire disc minus the piece that was removed. We can also calculate the center of mass in a similar way. Think of these two pieces, the complete disc and the hole, as two “particles” at their center of mass. But we use “negative mass” for the hole (it gets subtracted out). So the center of mass for this disc-with-a-hole is

Substituting

Simplifying

So the center of mass for the “punched out” disc is on the axis, a distance of to the right of the origin.

### Newton’s Law for a Collection of Particles

We can use Newton’s laws for a group of particles. Remember that Newton’s Second Law states that the net force acting on a mass equals mass acceleration, or Say we had a group of bees, trained to fly in formation. If a wind is blowing, what is the net force on the bees? We can calculate the acceleration of the center of mass, similar to calculating the position of the center of mass.Re-writing

The mass acceleration for each of the individual bees () sums up to total mass, times the acceleration of the center of mass, Or another way to think of it, the force on each individual bee () sums up to the net force, So this gives us the final definition

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Hello Doc. This is a timely entry, thanks!

We are teaching momentum right now, and ran into something interesting. In the case of a ball hitting a stationary wall (one-dimensional, perfectly elastic collision), the conservation of momentum equation ALONE does not work. One needs to add conservation of energy, in order for things to make sense. But the kid is not there yet: we have not talked about kinetic energy formula — we proceeded directly to momentum because we had covered m and v, touched upon vectors, so we were ready to mesh them together and introduce momentum, p.

So we are here:

m1v1 (initial) + m2v2 (initial) = m1v1 (final) + m2v2 (final)

But the wall (2) is always stationary, so its v2 = 0, both in the initial and final states.

It follows that

m1v1 (initial) = m1v1 (final)

Which only works in the scalar sense, as the velocity vector has changed direction.

Yes, yes: Conservation of energy makes sense of it. But the child is not there yet, and we are looking for a way to make sense of momentum WITHOUT introducing energy into the picture. Is there a way to explain what happens in this perfectly elastic, one-dimensional, collision of a ball vs. stationary wall, so that it makes intuitive sense to the kid while demonstrating the conservation of momentum?

Thanks,

Syd

Syd — thanks for the comment. Glad you are finding my material useful.

Sorry for the long delay in replying — I had to think about this question for a few days.

The problem is that a ball hitting a wall is “strange math.” It works, but you have to “interpret” the results.

Here’s the problem: It’s true that the wall has velocity v=0, but it’s also true that (compared to the ball) it has infinite mass. So, when you calculate its momentum, you have (infinity x 0). That’s troublesome for a direct calculation. The “full” answer requires calculus and limits, but we can approximate it.

Let the wall’s mass be very large — say 1000 kg. Then let its velocity be something very small — say 0.001 m/s. Now, do the elastic collision calculation with these numbers, but interpret anything that comes out very small as being 0, and anything that comes out very large as being infinity. I did it, and I get just the answer you’re looking for. I let the ball have a mass of 3 kg and velocity 5 m/s. I got a final velocity for the ball of -4.97 and the wall 0.031.

Let me know if you want further clarification.

Hope this helps! Thanks for the questions —

Doc