Understanding car crashes: it’s basic physics – This YouTube video explains the physics of car crashes and safety features in cars. If you’ve ever wondered why seatbelts and airbags work, this video will answer your questions!
Hello there! Welcome to lecture 6: momentum!
Momentum is inertia in motion. It is a property that quantifies the motion of an object. Between momentum and energy, which we’ll discuss in lecture 7, we can learn a lot about how the universe works by using these concepts. But even just considering momentum by itself, we can start to better understand the motion of objects, and what happens when we have collisions between two or more things.
Each of the following concepts will be discussed in this video: momentum, impulse, conservation of momentum, and collisions.
As previously mentioned, momentum is a property that quantifies the motion of an object. It is a vector, indicating that it has both a magnitude and a direction. The symbol for momentum is the lowercase letter p. The units for momentum are kilogram-meters per second, or Newton-seconds. Both of those units are interchangeable, meaning that either one can be used.
Momentum can help us to understand what happens to an object when it collides with something. In just a few moments, I’m going to show you an experiment where I change two properties of a cart, so that we can start to derive an equation for momentum. The two properties that are varied are mass and speed. Considering that we’ve defined momentum to be inertia in motion, it makes sense that momentum will be defined by mass (which is where inertia comes from) and velocity (which defines the motion of the object).
But before we quantify this, I’d like you to just think about it first. Consider that you have a friend standing a few meters away from you who gently throws a softball to you. Think about the amount of impact, or force, the softball will have when you catch it. Then consider what would happen if the softball were thrown faster.
Next, think about the same friend throwing a bowling ball to you. The impact of the bowling ball, even thrown at the same speed as the softball, would definitely be higher. So we can start to get a feel that both mass and velocity are directly proportional to momentum. Let’s take a look at that experiment now.
Each of the following collisions will look at what happens when a moving cart collides with a force probe. We can get a feel for the momentum the cart must have based on how much force the probe records when the cart hits it.
First, we’ll see what effect the mass of the cart has on its momentum. We’ll keep the speed that the cart has at the time of the impact constant by releasing the cart at a constant height. The cart initially has a mass of three hundred forty grams. When it hits the block, the force probe records a force of twenty-nine point four Newtons.
Then, we can add mass to the cart. Now it has a mass of four hundred forty grams. The force probe now records a force of thirty-two point one Newtons, an increase from the previous case.
Let’s add a little more mass to the cart. The cart has a mass of five hundred forty grams. The force probe records a force of thirty-four point four Newtons.
From these trials, we can see that momentum is directly proportional to mass. Remember, the force probe isn’t telling us the momentum of the object, but is simply a related property that we can measure to get an idea about how the momentum changes when the mass changes.
Next, we’ll see what effect the speed of the cart has on the momentum. We’ll keep the mass constant at five hundred forty grams. We’ll change the speed that the cart has when it impacts the force probe by changing the height of the platform that it’s released from.
In the slowest speed case, we saw that the force probe records a force of thirty-four point four Newtons. When we increase speed, the force probe records a force of forty-five point four Newtons.
This shows that momentum is directly proportional to speed.
One more thing. Momentum is a vector, and velocity is a vector. By looking at speed, we ignored the direction part of velocity. Here, a blue cart sits at rest on a low-friction track. When it gets impacted by the red cart, the momentum transfers from the red cart to the blue cart, causing the blue cart to move. The direction of that motion is in the same direction as the red cart.
Now that we see where momentum comes from, let’s look at the equation for momentum. Momentum equals mass times velocity. Momentum and velocity are both vectors, and they will both point in the same direction.
Let’s calculate the momentum of the cart just as it impacts the force probe in the case where the mass was three hundred forty grams and the impact occurred at the slower speed. We need to use kilograms to calculate momentum. The mass is zero point three four kilograms. I used a motion detector to measure the velocity of the cart, which was zero point four two meters per second at the exact moment it impacted with the force probe. Momentum is equal to zero point three four kilograms times zero point four two meters per second, giving a momentum of zero point one four three kilogram-meters per second.
What about two objects with the same mass, traveling in different directions? Do they have equal momentums? Let’s think about it. Momentum is a vector quantity, so we know that direction is involved. So if direction is not the same in the two cases, then regardless of the mass and speed, the momentums are not equal.
Impulse is the measure of the change in momentum of an object or of a system of objects. Because we’re looking at the change in a quantity, we know that we’re going to see an uppercase Greek letter delta in the symbol. Indeed, the symbol for impulse is Delta p. The equation for impulse is Delta p equals the final momentum minus the initial momentum.
Because momentum is a vector, the change in momentum (impulse) is also a vector. Make sure you pay close attention to the sign of your momentum variables when you do calculations.
I’m going to do an experiment to see how we can change the momentum of an object.
The cart on the track is at rest. Its velocity is zero so its momentum is zero. How do I make that momentum change? I have to push it or pull it somehow. In other words, I need to apply a force to the cart.
Let’s look at a couple different scenarios. First, I’m going to apply different forces to the cart, but make sure that the force is applied over roughly equal intervals of time. The force is applied by placing a hanging mass on the cart using a string and a pulley. The gravitational force of the hanging mass creates a constant force that acts on the cart. The interval of time is kept mostly constant by placing a barrier at a fixed distance away from the starting point.
First, a small force is applied to the cart for a short period of time. Its momentum, measured by the motion detector placed at the end of the ramp, changes slightly. When I apply a slightly larger force to the cart (by increasing the hanging mass), its momentum changes to a larger extent. Applying an even larger force to the cart (by increasing the hanging mass again) makes its momentum change even more.
Impulse is therefore directly proportional to force. The more force that’s applied, the greater the change in momentum.
Next, I’m going to keep the applied force equal by keeping the hanging mass value constant, but I’m going to apply the force over different intervals of time. Applying a force very quickly, over a short period of time, gives a small change in momentum to the cart. When I increase the amount of time over which that force is applied, the momentum of the cart changes even more. An even longer duration of force increases the momentum even more so.
Impulse is therefore directly proportional to the amount of time that the force is in contact with the object. The more time over which the force is applied, the greater the change in momentum.
Therefore, a second equation that we can use for impulse is that impulse equals force times time. Delta p equals F delta t. Remember that force is a vector, and impulse is also a vector. Those two vectors will point in the same direction.
If I want to make the greatest change in the momentum of an object, I would therefore need to exert a large force on that object over a long period of time.
Because impulse means change in momentum, and momentum is a vector, it doesn’t only apply to when an object speeds up. It also applies when an object slows down or changes its direction. To speed an object up, the force applied to the object needs to be applied in the same direction as the velocity. To slow an object down, the force needs to be applied in the opposite direction as the velocity.
We can see this by looking at the impulse generated by two nearly identical balls falling down a track and hitting a block of wood. One ball bounces, and the other does not. This helps us to understand how impulse changes with and without a bounce involved in the collision.
When the non-bouncing ball is released from the track and hits the block of wood, the block is deflected slightly due to the force the ball applies to the wood during the interaction. Newton’s third law tells us that the wood will apply an equal and opposite force to the ball, slowing it down to a stop.
When the bouncing ball is released from the track, the block of wood completely falls over. This is because the block had to not only stop the ball, but give it the impulse to start moving in the opposite direction. That force involved in the impulse was large enough to make the block of wood fall over.
We can look at this one more way. Both balls were released from roughly the same height onto a plate that measures force. The force plate records a smaller force when the non-bouncing ball hits, and a larger force when the bouncing ball hits. Note that there is a slight delay between the time that both balls hit the force plate and the time the read-out shows up on the laptop screen.
Let’s use some estimated values to quantify this. Both balls have a mass of ten grams, or zero point zero one kilograms. They both hit the force plate at a velocity of just about negative two point five meters per second. (We’re using a negative sign because the vector points down.)
The non-bouncing ball comes to a complete stop after it hits the force plate, so the final velocity is zero. The initial momentum is zero point zero one kilograms times negative two point five meters per second, or negative zero point zero two five kilogram-meters per second. The final velocity is zero kilogram-meters per second. Impulse is equal to final momentum minus initial momentum, which is positive zero point zero two five kilogram meters per second.
The bouncing ball rebounds with a speed of approximately one point five meters per second. (The value has a positive sign because it’s now moving up.) The initial momentum is the same as that of the non-bouncing ball: negative zero point zero two five kilogram-meters per second. The final momentum is zero point zero one kilograms times one point five meters per second, or zero point zero one five kilogram-meters per second. The impulse is final momentum minus initial momentum, which is positive zero point zero four kilogram-meters per second.
That provides us with quantitative evidence that the impulse is indeed greater when an object bounces compared to when it does not bounce.
Conservation of momentum
In the absence of external forces, the momentum of a system will be conserved. A system can be defined however we want.
Let’s look at a very simple example of this. In both of these situations, the system will be defined as just the red cart.
Once the cart is pushed along a low friction track, it travels at just about the same speed until it stops. There’s a little bit of friction, which is an external force, which slows it down very slightly. But if we ignore that, we can reasonably conclude that momentum of the cart is conserved in the absence of external forces.
When a high-friction pad is placed on the cart, the external forces are increased a lot. The cart slows down greatly and it’s very clear that the momentum of the cart is not conserved because of that external force being present.
Next, let’s look at a slightly more complicated scenario. There are now two carts: a blue cart and a red cart. The blue cart is pushed into the red cart where they connect with Velcro and move together afterward.
If we define the system as consisting of both the blue AND red carts, we can look at the momentum of each object before they impact and after they impact and compare. Both carts have a mass of zero point three four kilograms.
Before the collision, the system consists of the momentum of the blue cart plus the momentum of the red cart. Both can be calculated using the momentum equation. The blue cart momentum is zero point zero eight three kilogram-meters per second. The red cart has zero momentum as it is not moving. Therefore the initial system momentum is zero point zero eight three kilogram-meters per second.
After the collision, the system consists of the momentum of both carts traveling together. The combined mass of both carts is zero point six eight kilograms, and the final velocity was recorded as zero point one one seven meters per second. The final momentum of the system is therefore zero point zero eight zero kilogram-meters per second.
These two numbers are pretty nearly the same. Are they exactly the same? No. What kind of external forces might exist in this scenario that could cause that difference? Think about it. Either way, I think this example makes a very compelling case that the momentum of a system is conserved in the absence of external forces.
Now let’s change the definition of our system. Let’s say our system consists only of the blue cart. We won’t consider the red cart in our calculations any more. The momentum of the blue cart started at zero point zero eight three kilogram-meters per second. After the collision the cart has a momentum of zero point zero four kilogram-meters per second. Momentum definitely wasn’t conserved in this system! What gives? Well, the red cart, because it’s not part of our system anymore, now acts as an external object that gives the blue cart an external force. That external force slows the blue cart down and changes its momentum.
Similarly, we can consider the scenario where our system only consists of the red cart. The momentum of the red cart starts at zero kilogram-meters per second, and ends up as zero point zero four kilogram-meters per second. It’s definitely not conserved, and that’s because the blue cart acts as an external force, speeding up the red cart and changing its momentum.
These examples have provided us with evidence that momentum is conserved in one dimension, as all of the scenarios we considered took place along a straight line. What happens in two or even three dimensions? Is momentum still conserved in the absence of external forces?
The answer is – yes it is! When a collision or interaction occurs in two dimensions, we can independently consider both the horizontal and the vertical directions, or x and y. All of the momentum in the x direction must be conserved, and all of the momentum in the y direction must be conserved.
This video shows collisions between two orange pucks sliding freely along a low-friction air cushion table. The pucks don’t actually touch when they collide, but rebound off of each other due to the fact that they are made of repelling magnets.
The momentum of each puck can be represented as a vector. Summing up both of the momentum vectors before and after the collision leads to identical results. Momentum is conserved in two dimensions.
In three dimensions, we can consider a very similar outcome, but now with momentum being conserved in width, depth, and height. All three dimensions will have a conservation of momentum, in the absence of external forces.
The energy that an object has due to its motion is called kinetic energy. The symbol is KE. The units of kinetic energy are Joules. The equation for kinetic energy is one half m v-squared, where m is the mass of the object, and v is the velocity of the object. We’ll discuss kinetic energy more in lecture 7, but it’s important to define it here and now so that we can use it to better understand collisions.
We can use the conservation of momentum and kinetic energy to understand what happens when two objects collide with each other. Collisions can be categorized as being inelastic or elastic. In an inelastic collision, some of the kinetic energy of the system will be converted into other forms of energy: heat, sound, or possibly something else. In an elastic collision, the kinetic energy of the system will be conserved. In both types of collisions, momentum is conserved.
In a perfectly inelastic collision, the maximum amount of kinetic energy is lost. A perfectly inelastic collision occurs when two objects collide and then travel together after the collision.
Let’s look at an example of a perfectly inelastic collision. In this demonstration, a blue cart collides with a red cart. Both carts stick together with Velcro when they impact, so they will travel together afterward. The system consists of both carts. The mass of the blue cart is zero point five four kilograms, and the mass of the red cart is zero point three four kilograms. The initial momentum of the system is equal to the initial momentum of the red cart plus the initial momentum of the blue cart. Mass of the red times velocity of the red plus mass of the blue times velocity of the blue equals zero point one nine seven kilogram-meters per second. After the collision, the combined system mass is zero point eight eight kilogram-meters per second, and both carts travel together at a velocity of zero point two one eight meters per second. The final momentum is therefore equal to zero point one nine two kilogram-meters per second. The momentum of the system is pretty much conserved.
We can also look at the kinetic energy of the system before and after the collision. The initial kinetic energy is equal to the initial KE of the red cart plus the initial KE of the blue cart. ½ mass red times velocity red squared plus ½ mass blue times velocity blue squared equals zero point zero three six Joules. Afterward, the KE of the system is ½ mass times velocity squared, which is zero point zero two one Joules. The KE is definitely not conserved in this inelastic collision. The energy of motion was very likely converted to heat when the two objects came together, acoustic energy that created the clacking sound when they collided, and possibly some other forms as well.
As mentioned, an elastic collision occurs when both momentum and energy are conserved. We can anticipate that a collision will be inelastic if the two objects colliding won’t create any heat from rubbing together, if they won’t create any sound, and if the molecules inside the objects won’t physically deform when they impact together.
This video shows an elastic collision. The carts are rotated such that repelling magnets cause the carts to bounce off of each other and travel independently after the collision. This means that energy shouldn’t be lost due to heat, sound, or other things as the two carts will not actually touch when they collide.
In this case, the red cart has a mass of zero point five four kilograms, and the blue cart has a mass of zero point three four kilograms. The initial momentum of the system is mass times initial velocity of the red cart plus mass times initial velocity of the blue cart, which is zero point zero nine nine kilogram-meters per second. After the collision, the momentum of the system is mass times final velocity of the red cart plus mass times final velocity of the blue cart, which is zero point one zero one kilogram-meters per second. Momentum is conserved.
Similarly, we can calculate KE before and after the collision for the system. The initial KE of the system is ½ mass times initial velocity squared for the red cart plus ½ mass times initial velocity squared for the blue cart. This equals zero point zero one four Joules. After the collision, the system KE is ½ mass times final velocity squared for the red cart plus ½ mass times final velocity squared for the blue cart. This is equal to zero point zero one three Joules. KE is conserved, giving us proof that this is an elastic collision.
The concepts of conservation of momentum and conservation of energy can be used to learn about the universe. I interviewed Doctor Bennett to find out more.
Using conservation of momentum and energy to learn about the universe
Based on what we’ve learned in this lecture, we can find a way to engineer a safer collision. We’ve seen that the momentum of an object is going to create an impact force on an object when it collides. We’ve also learned that the change in momentum of an object is equal to force times time.
We want to have safe collisions in cases where we’re driving a car, or riding a bike, or otherwise could be injured when we collide with an object. In these scenarios, we want the force to be as minimal as possible in the collision. Because when we get into a collision, we’re unable to change the mass or velocity that we were traveling at, and we’re going to come to a complete stop after the collision, the initial momentum, final momentum, and therefore impulse are going to be fixed. We’d ideally like the number to be small, but we can’t control it after the fact. Because we know that impulse is equal to force times time, and we want the force to be as small as possible, we can see mathematically that extending the duration of a collision, which has the result of making delta t a large number, will cause the force to be small, given that Delta p is fixed.
How can we use this information to throw an egg without breaking it? In this video, I’m going to throw and egg and make sure that the force acting on the egg is as small as possible. By throwing it at a sheet, the egg doesn’t break. The time duration of the collision is extended for a long period of time and therefore the force is small enough for the egg not to break.
It’s very easy to break the egg. Just throw it at a wall, and the time duration of the collision will be so small that the force will be very large, and the egg shell will crack. What a mess!
Let’s look at one more example of this, an experiment where we can quantitatively assess the impulse, force, and time of the collisions.
A cart rolls down a ramp, giving it a fixed momentum at the instant it impacts with a force probe at the bottom of the ramp. The collision with the force probe happens very quickly. By looking at the readout from the force probe, the collision occurs over about zero point zero two seconds, which is a very short period of time. The maximum amount of force that was recorded is forty-eight point four Newtons.
I want to point out here that calculating impulse in this scenario is tricky, because the force isn’t constant over time. We would have to use calculus to calculate the impulse here. Real-world examples like this can be complicated. We can use force times time only when the force is constant throughout the entire duration of the collision. Thankfully, Logger Pro has a feature I used to determine the impulse, which is zero point four Newton-seconds.
Second, the cart is rolled down the ramp again, giving it a fixed momentum that is equal to that of the previous scenario. This time there’s a piece of foam in between the cart and the force probe. Looking at the data from Logger Pro, we can see that the collision occurs over a longer period of time, closer to zero point one seconds. The maximum amount of force recorded is twenty point six Newtons. The impulse calculated by Logger Pro is zero point four Newton-seconds.
Which of these collisions was safer? The scenario with the piece of foam certainly was, because the extended duration of the collision caused a decrease in force.
This is why cars are engineered to have crumple zones and air bags. Anything we can do to increase the time of a collision in the case of a car crash, for example, will reduce the impact on the occupants of the car and keep them as safe as possible.
Thanks for taking the time to learn about momentum! Until next time, stay well.