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## Meaning of Collisions and Momentum Conservation

We see collisions occur all around us; a couple of examples are car collisions and the collision of a ball and a bat. In physics, when two objects experience a **collision**, each object exerts a very strong contact force on the other in a short amount of time. Generally, the forces that the objects exert on each other are very strong in comparison to any other forces acting on the system. Because of this, we can ignore the external forces and only consider the forces that the objects exert on each other. A good example of this is how in a car collision, the forces that cars exert on each other are much greater than the friction force from the road on the tires.

A **collision** is an interaction between objects in which they exert very strong contact forces on each other over a short time period.

The mechanical energy of a system, or the sum of the kinetic and potential energy, is conserved when only conservative forces are acting on objects in the system. If a nonconservative force, such as friction, acts on an object, the total mechanical energy is different than what it was initially. We categorize collisions as either elastic or inelastic based on whether or not the mechanical energy is conserved during the collision.

An **elastic collision** is a collision in which the mechanical energy is conserved.

An **in****elastic collision** is a collision in which the mechanical energy is not conserved.

Atoms and molecules experience elastic collisions frequently. The macroscopic collisions that we see every day are inelastic as kinetic energy converts into other forms of energy, like thermal energy, when macroscopic objects collide. We can, however, approximate that some macroscopic collisions are elastic, like collisions between billiard balls when playing pool. We can justify treating a collision as elastic only when the energy loss is negligible and there is virtually no deformation during the collision.

Usually, during a collision, the objects in the collision get damaged from the large forces they experience. The forces experienced by the objects also vary in most collisions. If we use momentum conservation and energy conservation instead of Newton’s laws of motion to describe the collision, the problem becomes much simpler. We do not need to consider the varying force or how much damage is done to each object if we use the conservation of momentum.

## The Law of Conservation of Momentum

When there are no external forces acting on a system, it is called an **isolated system**. When a collision occurs in an isolated system, momentum is conserved. The law of conservation of momentum tells us that momentum is constant before and after the collision. So, while mechanical energy is not always conserved during a collision, momentum in an isolated system is always conserved. As mentioned above, the forces exerted by the objects on each other are much greater than the external forces. So, for the examples we will consider, we will ignore any external forces and consider the system as isolated.

The principle of momentum conservation states that in an isolated system, momentum is constant during a collision.

## Formulas for Collisions with Momentum Conservation

Using the laws of conservation of momentum and energy, we can write down equations to describe the motion of objects in a collision. Here, we will focus on linear momentum, although this concept can also be applied in the form of angular momentum. Let’s consider separately the two types of collisions we mentioned above: elastic collisions and inelastic collisions.

### Formula for Elastic Collisions

For elastic collisions, both mechanical energy and momentum are conserved. Thus we have an equation that we can write for each of these. The conservation of mechanical energy tells us that the energy before the collision,${E}_{i}$, must equal the energy after the collision,${E}_{f}$:

${E}_{i}={E}_{f}$

The mechanical energy is the sum of the kinetic and potential energy of the system, but for now let’s consider a system that has just kinetic energy, like a billiard ball about to collide with another billiard ball. We will assume that both billiard balls are moving in a straight line and the collision is one-dimensional. The total energy before the collision is given by

${E}_{i}={K}_{1i}+{K}_{2i}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{m}_{1}{{v}_{1i}}^{2}+\frac{1}{2}{m}_{2}{{v}_{2i}}^{2},$

where${m}_{1}$and ${m}_{2}$are the masses of the billiard balls, and${v}_{1i}$and${v}_{2i}$are their initial velocities. Billiard balls typically have the same mass, but for generality we will assume they are slightly different. The total energy after the collision is given by

${E}_{f}={K}_{1f}+{K}_{2f}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{m}_{1}{{v}_{1f}}^{2}+\frac{1}{2}{m}_{2}{{v}_{2f}}^{2},$

where${v}_{1f}$and${v}_{2f}$are the final velocities of the billiard balls. Equating these gives us:

$\begin{array}{rcl}\frac{1}{2}{m}_{1}{{v}_{1i}}^{2}+\frac{1}{2}{m}_{2}{{v}_{2i}}^{2}& =& \frac{1}{2}{m}_{1}{{v}_{1f}}^{2}+\frac{1}{2}{m}_{2}{{v}_{2f}}^{2}\\ & & \\ \overline{)\frac{1}{2}}{m}_{1}{{v}_{1i}}^{2}+\overline{)\frac{1}{2}}{m}_{2}{{v}_{2i}}^{2}& =& \overline{)\frac{1}{2}}{m}_{1}{{v}_{1f}}^{2}+\overline{)\frac{1}{2}}{m}_{2}{{v}_{2f}}^{2}\\ & & \\ {m}_{1}{{v}_{1i}}^{2}+{m}_{2}{{v}_{2i}}^{2}& =& {m}_{1}{{v}_{1f}}^{2}+{m}_{2}{{v}_{2f}}^{2}.\end{array}$

This is the equation we have arrived at using the conservation of mechanical energy. Now, let’s look at the equation we can get from the conservation of momentum. The initial linear momentum,${P}_{i}$, must equal the final linear momentum, ${P}_{f}$:

${P}_{i}={P}_{f}$

The linear momentum before the collision is

${P}_{i}={P}_{1i}+{P}_{2i}\phantom{\rule{0ex}{0ex}}={m}_{1}{v}_{1i}+{m}_{2}{v}_{2i}$and the linear momentum after the collision is

${P}_{f}={P}_{1f}+{P}_{2f}\phantom{\rule{0ex}{0ex}}={m}_{1}{v}_{1f}+{m}_{2}{v}_{2f}.$

We get our equation from the conservation of linear momentum by equating these:

${m}_{1}{v}_{1i}+{m}_{2}{v}_{2i}={m}_{1}{v}_{1f}+{m}_{2}{v}_{2f}$

These equations look like they have a lot of unknown variables, but most problems give us the mass and initial velocities of the objects in the system. Recall that whenever we have two equations, we can solve for at most two unknown variables. Thus, we can then use both the conservation of energy and the conservation of momentum equations to solve for the final velocity of each of the two colliding objects.

The equations for momentum and energy conservation used above only apply to objects that are traveling in one dimension and have an elastic collision. If the collision happens at an angle, we must consider the vertical and horizontal components of the velocities to find the components of the momentum.

### Formula for Inelastic Collisions

For inelastic collisions, the mechanical energy is not conserved so we cannot equate the initial and final mechanical energies. We can, however, still use the conservation of momentum. Let’s consider a one-dimensional situation similar to the example above, except this time when the balls collide, some kinetic energy is lost to other forms of energy like thermal and sound energy. The equation to describe the conservation of momentum is the same as before:

$\begin{array}{rcl}{P}_{i}& =& {P}_{f}\\ & & \\ {P}_{1i}+{P}_{2i}& =& {P}_{1f}+{P}_{2f}\\ & & \\ {m}_{1}{v}_{1i}+{m}_{2}{v}_{2i}& =& {m}_{1}{v}_{1f}+{m}_{2}{v}_{2f}\end{array}$

Since we only have one equation, we can only solve for one variable. We can do this if the problem gives us the values of the other variables, or we can solve for the ratio of the final velocities when given the values of the masses and initial velocities.

A special case for inelastic collisions is when the objects stick together after the collision. This is called a **perfectly**** inelastic collision**. In this case, they share the same final velocity. Imagine that one of the balls from the example above is sticky and sticks to the other ball after the collision. Equating the initial momentum and the final momentum would then give us:

$\begin{array}{rcl}{P}_{i}& =& {P}_{f}\\ & & \\ {P}_{1i}+{P}_{2i}& =& {P}_{1,2f}\\ & & \\ {m}_{1}{v}_{1i}+{m}_{2}{v}_{2i}& =(& {m}_{1}+{m}_{2}){v}_{f}\end{array}$

For a perfectly inelastic collision, there is just one final velocity to find, which we can do if we are given the other variables.

Other names for a perfectly inelastic collision are a **completely**** **inelastic collision, **totally** inelastic collision, **maximally** inelastic collision, etc. Different authors use different adjectives, but all these expressions mean the same thing.

## Collisions and Momentum Conservation Examples

Since collisions happen frequently in our daily lives, we need to be able to set up energy and momentum conservation equations based on each situation. Below are a couple of examples to help you practice.

A large block of mass${m}_{1}$collides with a smaller block of mass${m}_{2}$that is initially at rest. The blocks move together after the collision. What is the ratio of the initial and final velocities?

The collision between the blocks is a completely inelastic collision because kinetic energy is lost during the collision and the blocks move together after the collision. We will use the conservation of momentum to find the ratio of the velocities. The initial momentum of the larger block is

${P}_{1i}={m}_{1}{v}_{i}$.

The smaller block is at rest before the collision and thus has zero initial momentum:

${P}_{2i}=0$So, the total initial momentum is

$\begin{array}{rcl}{P}_{i}& =& {P}_{1i}+{P}_{2i}\\ & =& {m}_{1}{v}_{i}.\end{array}$We find the final momentum of the system by taking the combined mass multiplied by the final velocity so that

${P}_{f}=({m}_{1}+{m}_{2}){v}_{f}$Equating these gives us:

$\begin{array}{rcl}{P}_{i}& =& {P}_{f}\\ & & \\ {m}_{1}{v}_{i}& =& ({m}_{1}+{m}_{2}){v}_{f}\end{array}$

Dividing the combined mass over to the left side and then the initial velocity to the right side gives us our ratio:

$\begin{array}{rcl}{v}_{f}& =& \frac{{m}_{1}}{{m}_{1}+{m}_{2}}{v}_{i}\\ & & \\ \frac{{v}_{f}}{{v}_{i}}& =& \frac{{m}_{1}}{{m}_{1}+{m}_{2}}\end{array}$

A billiard ball collides with another billiard ball that is initially moving with velocity${v}_{2i}=1\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$. The two billiard balls go off at an angle of$\theta =30\xb0$after the collision. The initial speed of the first billiard ball is${v}_{1i}=4\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$, and they have equal masses. Find the final speeds of both billiard balls.

Two-dimensional collision between billiard balls, StudySmarter Originals

Let’s use the conservation of momentum to find the final velocities of the billiard balls. Since the billiard balls go off at an angle after the collision, we will need to split the initial and final momentum up into the vertical and horizontal components. We will start with the horizontal component. The horizontal component of the initial momentum is given by

${P}_{ix}=m{v}_{1i}+m{v}_{2i}$

There is no initial velocity in the vertical direction, so the initial momentum is zero:

${P}_{iy}=0$

To find the horizontal component of the final momentum, we need to split the velocity into vertical and horizontal components.

The vertical components of each velocity are:

${v}_{1fy}={v}_{1f}\mathrm{sin}\theta $

${v}_{2fy}={v}_{2f}\mathrm{sin}(-\theta )\phantom{\rule{0ex}{0ex}}=-{v}_{2f}\mathrm{sin}\left(\theta \right)$

. The horizontal components are:${v}_{1fx}={v}_{1f}\mathrm{cos}\theta $

${v}_{2fx}={v}_{2f}\mathrm{cos}(-\theta )\phantom{\rule{0ex}{0ex}}={v}_{2f}\mathrm{cos}\left(\theta \right)$

Notice that we have a negative angle for the green billiard ball. The horizontal component for the final momentum is then:

${P}_{fx}=m{v}_{1fx}+m{v}_{2fx}\phantom{\rule{0ex}{0ex}}=m({v}_{1f}\mathrm{cos}\theta +{v}_{2f}\mathrm{cos}\theta ),$

and the vertical final momentum component is

${P}_{fy}=m{v}_{1fy}+m{v}_{2fy}\phantom{\rule{0ex}{0ex}}=m({v}_{1f}\mathrm{sin}\theta -{v}_{2f}\mathrm{sin}\theta ).$

Now we can equate them to the components of the initial momentum:$\begin{array}{rcl}{P}_{ix}& =& {P}_{fx}\\ m{v}_{1i}+m{v}_{2i}& =& m({v}_{1f}\mathrm{cos}\theta +{v}_{2f}\mathrm{cos}\theta )\\ & & \\ {P}_{iy}& =& {P}_{fy}\\ 0& =& m({v}_{1f}\mathrm{sin}\theta -{v}_{2f}\mathrm{sin}\theta )\end{array}$

From the vertical component of the momentum, we can cancel the mass and the sine function:

$\begin{array}{rcl}0& =& \begin{array}{l}m({v}_{1f}\mathrm{sin}\theta -{v}_{2f}\mathrm{sin}\theta )\end{array}\\ 0& =& m\mathrm{sin}\theta ({v}_{1f}-{v}_{2f})\end{array}$

Since we know that the scattering angle and the mass are not zero, we can divide them over to the other side to get:

$\begin{array}{rcl}\frac{0}{m\mathrm{sin}\theta}& =& ({v}_{1f}-{v}_{2f})\\ 0& =& ({v}_{1f}-{v}_{2f})\\ {v}_{1f}& =& {v}_{2f}\end{array}$

If the scattering angle is zero, we would not be able to divide the sine term to the other side.

We found that the magnitude of the velocity for the billiard balls is the same. So now let’s call it${v}_{f}={v}_{1f}={v}_{2f}$. We can use this in the equation for the horizontal momentum component to solve for the magnitude of the final velocity of both balls:

$\begin{array}{rcl}m{v}_{1i}+m{v}_{2i}& =& m({v}_{1f}\mathrm{cos}\theta +{v}_{2f}\mathrm{cos}\theta )\\ \overline{)m(}{v}_{1i}+{v}_{2i})& =& \overline{)m}({v}_{f}\mathrm{cos}\theta +{v}_{f}\mathrm{cos}\theta )\\ {v}_{1i}+{v}_{2i}& =& 2{v}_{f}\mathrm{cos}\theta \\ {v}_{f}& =& \frac{{v}_{1i}+{v}_{2i}}{2\mathrm{cos}\theta}\\ & =& \frac{\left(4\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.+1\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\right)}{2\mathrm{cos}(30\xb0)}\\ & =& 2.89\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\end{array}$

The billiard balls both go off at an angle of$30\xb0$with a velocity of$2.89\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

## Collisions and Momentum Conservation - Key takeaways

- In physics, a collision is an interaction between objects in which they exert very contact strong forces on each other in a short time period.
- Momentum is always conserved when a collision occurs in an isolated system.
- Mechanical energy is
**not**always conserved during a collision. - An elastic collision is one in which the mechanical energy and momentum are conserved.
- An inelastic collision is one in which the mechanical energy is not conserved, and there is a decrease in kinetic energy.

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##### Frequently Asked Questions about Collisions and Momentum Conservation

What is the meaning of collisions and momentum conservation?

When there is a collision between objects in an isolated system in which they exert very strong forces on each other in a short time period, momentum is conserved.

How do collisions with momentum conservation occur?

A collision occurs when two objects exert strong forces on each other in a short time period. Momentum is conserved during the collision if the system is an isolated system.

What is a good example of a collision with momentum conservation?

An example of a collision in which momentum is conserved is when two billiard balls collide during a game of pool.

What is the law of momentum conservation during collisions?

The law of momentum conservation states that momentum is conserved during a collision in an isolated system in which there are no external forces acting on the system.

How do you prove conservation of momentum in a collision?

We can prove the conservation of momentum in a collision by finding the momentum of the system before and after the collision. The momentum in the system is constant and will be the same before and after the collision in an isolated system.

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