Physics is the science of give and take. Except that with physics, you always take precisely the amount you give. For example, did you know that when a semi-truck and a sedan collide, they both feel the same amount of force? Newton's third law, or the Law of Impulse, is the principle that two objects exert equal and opposite forces on each other. It seems hard to believe, but even a tiny pebble hitting the Earth feels the same force as the Earth hitting the pebble.
Man, if only physics were similar to relationships, then you would always get what you give! (Maybe you should share this with that special someone to see if they'll start conforming to the laws of nature. Then, if they ever complain again, tell them that Newton said you cannot take more than you give!)
In this article, we explore the notion of impulse, which is the change of momentum of a system (recall that a system is a defined set of objects; for example, a basketball going through a hoop would have a system including the ball, the hoop, and the Earth exerting the force of gravity on the ball). We will also go over the formula for impulse, talk about the rate of change of momentum and even practice some examples. So let's dive right in!
Change of Momentum Formula
To understand what a change of momentum is, we must first define momentum. Remember that momentum is a quantity given to an object due to its velocity \(\vec{v}\) and mass \(m\), and a lowercase \(\vec p\) represents it:
$$\vec p = m \vec v\mathrm{.}$$
The greater the momentum, the harder it is for an object to change its state of motion from moving to stationary. A moving object with significant momentum struggles to stop and on the flip side, a moving object with little momentum is easy to stop.
The change of momentum, or impulse (represented by the capital letter \(\vec J)\), is the difference between an object's initial and final momentum.
Therefore, assuming the mass of an object doesn't change, the impulse is equal to the mass times the change in velocity. Defining our final momentum,
$$\vec p_\text{f}=m\vec v_\text{f}\mathrm{,}$$
and our initial momentum,
$$\vec p_\text{i}=m\vec v_\text{i}\mathrm{,}$$
allows us to write an equation for the total change in momentum of a system, written as:
$$\vec{J}=\Delta \vec p = \vec p_\text{f}- \vec p_\text{i}=m(\vec v_\text{f}- \vec v_\text{i})=m\Delta \vec v,$$
where \(\Delta \vec p\) is our change in momentum, \(m\) is our mass, \(\vec v\) is our velocity, \(\text{i}\) stands for initial, \(\text{f}\) stands for final, and \(\Delta \vec v\) is our change in velocity.
Rate of Change of Momentum
Now, let's prove how the rate of change of momentum is equivalent to the net force acting on the object or system.
We've all heard that Newton's second law is \(F = ma\); however, when Newton was first writing the law, he had in mind the idea of linear momentum. Therefore, let's see if we can write Newton's second law a little differently. Starting with
$$\vec F_\text{net}= m \vec a$$
allows us to see a correlation between Newton's second law and linear momentum. Recall that acceleration is the derivative of velocity. Therefore, we can write our new force formula as
$$\vec F_\text{net}= m \frac{\mathrm{d}\vec v}{\mathrm{d}t}\\\mathrm{.}$$
It is essential to note the change that was made. Acceleration is just the rate of change in velocity, so to replace it with \(\frac{\mathrm{d} \vec v}{\mathrm{d} t}\) is valid. As the mass \(m\) stays constant, we see that the net force is equal to the rate of change of momentum:
$$\vec F_\text{net} = \frac{\,\mathrm{d}(m\vec v)}{\mathrm{d}t} = \frac{\mathrm{d} \vec p}{\mathrm{d} t} .$$
We can rearrange this to get
\[\mathrm{d}\vec{p}=\vec{F}_\text{net}\,\mathrm{d}t.\]
With this new outlook on Newton's second law, we see that the change of momentum, or impulse, can be written as follows:
\[\vec{J}=\Delta\vec{p}=\int\,\mathrm{d}\vec{p}=\int\vec{F}_\text{net}\,\mathrm{d}t.\]
- The change of momentum, or impulse (represented by the capital letter \(\vec J)\), is the difference between a system's initial and final momentum. Therefore, it is equal to the mass times the change in velocity.
- Newton's second law is a direct result of the impulse-momentum theorem when mass is constant! The impulse-momentum theorem relates the change of momentum to the net force exerted:
$$\vec F_\text{net} = \frac{\mathrm{d} \vec p}{\mathrm{d} t} = m\frac{\mathrm{d}\vec v}{\mathrm{d} t} = m\vec a.$$
As a result, the impulse is given by\[\vec{J}=\int\vec{F}_\text{net}\,\mathrm{d}t.\]
In physics, we often deal with collisions: this doesn't necessarily have to be something as big as a car crash – it can be something as simple as a leaf brushing past your shoulder.
A collision is when two objects with momentum exert an equal but opposite force on each other through short physical contact.
The momentum of a collision system is always conserved. Mechanical energy, however, does not necessarily have to be conserved. There are two types of collisions: elastic and inelastic.
Elastic Collisions and Momentum
First, we'll talk about elastic collisions. "Elastic" in physics means that the system's energy and momentum are conserved.
Elastic collisions occur when two objects collide and bounce off each other perfectly.
This entails that the total energy and momentum will be the same before and after the collision.
Fig. 3 - The interactions of billiard balls are great examples of collisions that are very close to being perfectly elastic.
Two billiard balls exemplify a near-perfect collision. When they collide, they bounce so that energy and momentum are almost completely conserved. If this world were ideal and friction were not a thing, their collision would be perfectly elastic, but alas, billiard balls are only a near-perfect example.
Fig. 4 is a great example of an elastic collision in action. Notice how the motion transfers completely from the left object to the right one. This is a fantastic sign of an elastic collision.
Inelastic Collisions and Momentum
Now to the far-from-perfect evil twin.
Inelastic collisions are collisions where objects stick rather than bounce. This means that kinetic energy is not conserved.
An example is throwing a piece of gum into a trash can floating in space (we specify that it is in space because we do not want to deal with the rotation of the Earth in our calculations). Once the gum takes flight, it has a mass and a velocity; therefore, we are safe to say that it also has momentum. Eventually, it will hit the surface of the can and will stick. Thus, energy is not conserved because some of the kinetic energy of the gum will dissipate to friction when the gum sticks to the can. However, the system's total momentum is conserved because no other outside forces had the chance to act on our gum-trash can system. This means that the trashcan will gain a bit of speed when the gum collides with it.
The Variable Change of Momentum of a System
All of the examples of collisions above involve constant impulse. In all collisions, the system's total momentum is conserved. A system's momentum is not conserved, however, when that system interacts with outside forces: this is a critical concept to understand. Interactions within a system conserve momentum, but when a system interacts with its environment, the system's total momentum is not necessarily conserved. This is because in this case, there can be a non-zero net force acting on the system, giving the whole system a non-zero impulse over time (through that integral equation we wrote down earlier).
Examples of Change in Momentum
Now that we know what the change of momentum and collisions are, we can start applying them to real-world scenarios. This wouldn't be a collision lesson without car crashes, right? Let's talk about how the change of momentum plays a role in collisions – first, an example.
Jimmy just got his license. All excited, he takes out his dad's brand new \(925\,\mathrm{kg}\) convertible for a test drive (but with Jimmy inside, the convertible is \(1.00\times 10^3\,\mathrm{kg}\)). Traveling at \(18\,\mathrm{\frac{m}{s}\\}\), he hits a stationary (obviously) mailbox that has a mass of \(1.00\times 10^2\,\mathrm{kg}\). This doesn't stop him much, however, and he and the mailbox continue together at a speed of \(13.0\,\mathrm{\frac{m}{s}\\}\). What is the magnitude of the car-Jimmy-mailbox system's impulse over the collision?
Remember that impulse is the same as change of momentum.
Recall that impulse is the difference between initial momentum and final momentum. Therefore, we write down that
$$p_\text{i} = 1.00\times 10^3\,\mathrm{kg} \times 18\,\mathrm{\frac{m}{s}\\}+1.00\times 10^2\,\mathrm{kg}\times 0\,\mathrm{\frac{m}{s}} = 18\,000\,\mathrm{\frac{kg\,m}{s}\\}$$
is equal to the magnitude of our initial momentum, whereas
$$p_\text{f} = (1.00\times 10^3\,\mathrm{kg}+1.00\times 10^2\,\mathrm{kg})\times 13.0\,\mathrm{\frac{m}{s}\\} = 14\,300\,\mathrm{\frac{kg\,m}{s}\\}$$
is equal to the magnitude of our final momentum. Finding the difference between them yields
$$\Delta p = p_\text{f}-p_\text{i} = 14300\,\mathrm{\frac{kg\,m}{s}\\} - 18000\,\mathrm{\frac{kg\,m}{s}\\} =-3700\,\mathrm{\frac{kg\,m}{s}\\}\mathrm{.}$$
Therefore, the impulse of the car-Jimmy-mailbox system has a magnitude of
$$J = 3700\,\mathrm{\frac{kg\,m}{s}\\}\mathrm{.}$$
The system's total impulse tells us what happened between Jimmy speeding down the street at \(18\,\mathrm{\frac{m}{s}\\}\) and flying along with a mailbox at \(13.0\,\mathrm{\frac{m}{s}\\}\). We know that the total momentum of the car-Jimmy-mailbox system changed by
$$3700\,\mathrm{\frac{kg\,m}{s}\\}\mathrm{.}$$
We have the whole story now!
Right now, you are probably wondering how this example works out. Above, we described inelastic collisions as conserving momentum, but this example seems to show that a system's total momentum can change after an inelastic collision.
However, it turns out momentum is still conserved in the scenario above. The excess momentum was simply transferred to the Earth. Since the mailbox was attached to the surface of the Earth, hitting it caused Jimmy to exert a force on the Earth. Think of sticking a pencil into a soccer ball and then flicking it. Even if the pencil came off the ball, the ball would still feel a force in the direction of the flick.
When Jimmy hit the mailbox, it was equivalent to flicking a very small "pencil," if you will, off of the gigantic "soccer ball" of the Earth. Remember that exerting a force over a time interval is equivalent to saying there was a momentum change. Therefore, by exerting a force on the Earth over a short time, some of the system's momentum was transferred to the Earth. Thus, the momentum of the entire system (including the Earth) was conserved, but the individual momenta of Jimmy, the car, and the mailbox changed, as did their joint momentum.
Change of Momentum - Key takeaways
- The change of momentum is the same thing as impulse. It is equal to the mass times the change of velocity and is the difference between the final and initial momentum.
- Impulse is a vector quantity in the same direction as the net force exerted on the system.
- Here is our equation for the total change in momentum of a system:
$$\Delta \vec p = \vec p_\text{f}- \vec p_\text{i}=m(\vec v_\text{f}- \vec v_\text{i})=m\Delta \vec v.$$
A net force is equivalent to the rate of change of momentum:
$$\vec F_\text{net} = m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = \frac{\mathrm{d} \vec p}{\mathrm{d} t} .$$
Newton's second law is a direct result of the impulse-momentum theorem when mass is constant! The impulse-momentum theorem relates the change of momentum to the net force exerted:
$$\vec F_\text{net} = \frac{\mathrm{d} \vec p}{\mathrm{d} t} = m\frac{\mathrm{d}\vec v}{\mathrm{d} t} = m\vec a.$$
- Impulse is the area under a force over time curve, thus, it is equal to the force exerted times the time interval that the force was exerted over.
- Therefore, the impulse is the time integral of the force and is written as:
$$\vec J=\int_{t_\text{i}}^{t_\text{f}} \vec F(t)\,\mathrm{d}t\mathrm{.}$$
- Elastic collisions "perfectly bounce" and have conservation of kinetic energy and momentum.
- Inelastic collisions "stick" and only have conservation of momentum.
- The impulse, or the change of momentum, tells us "the middle of the story" when we talk about collisions.
References
- Fig. 1 - Force vs. Time Graph, StudySmarter
- Fig. 2 - Stick Figure Playing Soccer, StudySmarter Originals
- Fig. 3 - Billiard Balls (https://www.peakpx.com/632581/snooker-colored-billiards-game-balls-sport-pool-ball) by Peakpx (https://www.peakpx.com/) is licensed by Public Domain
- Fig. 4 - Elastic Collision, StudySmarter Originals.
- Fig. 5 - Inelastic Collision, StudySmarter Originals.
Explanations 
Exams 
Magazine 
