# Ap physics 1 momentum and impulse relationship

### Impulse Momentum Cheat Sheet

The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it. +F1 = −F2, conservation of momentum. General Notes About AP Physics Scoring Guidelines. 1. 2 points. For correct use of the impulse–momentum relationship. 1 point avg s s. F. t m u. Δ. Δ. . This test covers momentum, impulse, conservation of momentum, elastic 1. An air track glider of mass M is built, consisting of two smaller connected gliders with a varies as a function of time according to the equation F = k t 2, where k is a.

Let's say a box started with an initial speed and slides from one platform up to another platform. We'll assume that frictional forces and air resistance are negligible. And for the system that's consisting of the mass and the Earth, what's happening to the total mechanical energy in this system?

So, you've got to pay special attention to what is in your system. Since my system includes the mass, which is going to be moving, my system's gonna have kinetic energy, and since my system has two objects that are interacting gravitationally, the mass and the Earth, my system's also going to have gravitational potential energy.

So, when I asked about the total mechanical energy of the system, that's really just code for the total kinetic and potential energy of the system.

So, as this mass slides up to a higher point on the ramp, the gravitational potential energy increases, but the mass is gonna slow down, so the kinetic energy's gonna decrease. However, since the Earth and the mass are in our system, and there's no dissipative forces, there's no external work done on our system.

Yes, the Earth is doing work on the box, but the Earth is part of our system so it can't do external work, and that means energy just gets transferred from one form to another within our system, and the total mechanical energy, here, is gonna remain the same for the entire trip. Now, what if we asked this same question but we consider a system that consists only of the box.

In that case, our system has a box that's moving, so it'll have kinetic energy. But, our system no longer includes two objects interacting gravitationally so our system will have no gravitational potential energy. What happens to the total mechanical energy in this case? Well, the only energy that I've got in my system, now, is kinetic energy and since that kinetic energy decreased, the total mechanical energy of the box, as a system, decreases.

How does it decrease? It decreases because now the Earth is outside of our system and the work that it is doing on the box is external work and it's taking away energy from the box.

What does work mean? In physics, work is the amount of energy transferred from one system, or object, to another. In other words, if a person lifted a box and gave it 10 Joules of gravitational potential energy, we'd say that person did positive 10 Joules of work on the box since that person gave the box 10 Joules of energy.

But since the box took 10 Joules of energy from that person, we'd say that the box did negative 10 Joules of work on the person since the box took 10 Joules of energy.

**Introduction to momentum - Impacts and linear momentum - Physics - Khan Academy**

So, you can find the work done if you can determine the amount of energy that was transferred. But, there's an alternative formula to find the work done. If something's having work done to it, there's got to be a force on that object, and that object has to be displaced.

So, if you take the force on the object times the displacement of the object, and multiply by the cosine of the angle between the force and the displacement, you'll also get the work done. In other words, one way to find the work done is by finding the amount of energy that was transferred. But, another way to find the work done is by taking the magnitude of force exerted on an object times the displacement of the object and then times cosine of the angle between the displacement and the force.

Since work is a transfer of energy, it also has units of Joules. And even though work is not a vector, it can be positive or negative. If the force on an object has a component in the direction of motion, that force will do positive work on the object and give the object energy. If the force on the object has a component in the opposite direction of the motion, the work done by that force would be negative and it would take away the object's energy.

### AP Physics 1 review of Momentum and Impulse (video) | Khan Academy

And, if the force on an object is perpendicular to the motion of the object, that force does zero work on the object. It neither gives the object energy nor takes away the object's energy. So, what's an example problem involving work look like? Let's say a box of mass M slides down a frictionless ramp of height, H, and angle two-theta, as seen in this diagram, here, and a separate box of mass two-M slides down another frictionless ramp of height, H, and angle theta, as seen in this diagram, here.

And, we want to know how work done on the object by the Earth compares for each case? The easiest way to find the work done here is by finding the change in energy. The box will gain an amount of kinetic energy equal to the amount of potential energy that it loses. So, the work done by the Earth is just gonna equal positive m-g-h. Both heights are the same.

So, the H's are equivalent, but one box has twice the mass. So, the work done by gravity on the mass, two-M, is gonna be twice as great as the work done on the mass one-M.

What's the work energy principle mean?

## Linear Momentum AP Physics Chapter 7. Linear Momentum 7.1 Momentum and Its Relation to Force.

The work energy principle states that the total work, or the net work done on an object, is gonna be equal to the change in kinetic energy of that object. So, if you add up all the work done by all forces on an object, that's got to be equal to the change in the kinetic energy of that object. In other words, one-half m-v-final, squared, minus one-half m-v-initial, squared.

So, this is a really handy way to find how the speed of an object changes if you can determine the net work on an object. In other words, if there's multiple forces on an object, and you can find the work done by each of those forces, you can determine how much kinetic energy that object gained or lost. So, what's an example of the work energy principle? Let's say a four kilogram box started with a velocity of six meters per second to the left.

Some net amount of work is done on that box, and it's now moving with a velocity of four meters per second to the right. We want to know what was the amount of new work done on the box?

Without even solving it, we can say since this object's slowed down, energy was taken from it, so the amount of net work had to be negative which means it's either B or D. To figure out which one exactly, we could use the work energy principle which says that the net work done is equal to the change in kinetic energy.

So, if we take the final kinetic energy, which is one-half times four kilograms times the final speed squared, and we subtract the initial kinetic energy, one-half times four kilograms times the initial speed, squared, six meters per second, we get negative 40 Joules of net work.

So the total initial momentum of the system would be 6Mv plus -5Mv, which is one Mv, and that's positive, which means the total momentum, initially, is to the right. That means, after the collision, the total momentum will also have to be to the right, and the only way that could be the case if these two masses joined together, is for the total combined mass to also move to the right.

What does impulse mean? The impulse is the amount of force exerted on an object or system multiplied by the time during which that force was acting. So in equation form, that means J, the impulse, is equal to the force multiplied by how long that force was acting.

And the net impulse is gonna be equal to the net force times the time during which that net force was acting. And this is also going to be equal to the change in momentum of that system or object.

In other words, if a mass had some initial momentum and ends with some final momentum, the change in momentum of that mass, p final minus p initial, is gonna equal the net impulse, and that net impulse is gonna equal the net force on that object multiplied by the time during which that force was acting.

And since impulse is a change in momentum, and momentum is a vector, that means impulse is also a vector, so it can be positive and negative.

And the units are the same as momentum, which is kilograms times meters per second. Or, since it's also force times time, you could write the units as Newtons times seconds. So what's an example problem involving impulse look like? Let's say a bouncy ball of mass M is initially moving to the right with a speed 2v. And it recoils off a wall with a speed v. We want to know, what's the magnitude of the impulse on the ball from the wall?

So the impulse, J, is going to be equal to the change in momentum. The change in momentum is p final minus p initial, so the final momentum is gonna be the mass times the final velocity, but this velocity's heading leftwards, so you can't forget the negative sign, minus the initial momentum, which would be M times 2v, which gives a net impulse of -3Mv.

The net impulse has to point in the same direction as the net force. This wall exerted a force to the left, that means the impulse also points left, and has a magnitude of 3Mv. If you get a force versus time graph, the first thing you should think about is that the area under that graph is going to equal the impulse on the object.

So if you graph the force on some object as a function of time, the area under that curve is equal to the impulse. Just be careful, since area above this time axis is going to count as positive impulse, and area underneath the time axis would count as negative impulse, since those forces would be negative. Why do we care that the area's equal to the impulse? Well, if we can find the area, that would equal the impulse, and if that's the net impulse on an object, that would also equal the change in momentum of that object.

Which means we could figure out the change in velocity of an object. So what's an example problem involving impulse as the area under a graph look like? Let's say a toy rocket of mass two kilograms was initially heading to the right with a speed of 10 meters per second, and a force in the horizontal direction is exerted on the rocket, as shown in this graph, and we want to know, what's the velocity of the rocket at the time t equals 10 seconds?

To figure that out, we'll figure out the area under the curve. This triangle would count as positive area.

This triangle would count as negative area. And since this triangle is just as positive as this triangle is negative, these areas cancel completely. And the only area we'd have to worry about is the area between eight seconds and 10 seconds.

This is going to end up being a negative area, since the height of the rectangle isand the width of the rectangle is going to be two seconds.

This gives an impulse of Newton seconds. So if the impulse on this object is Newton seconds, that's going to equal the change in momentum of that object. How much momentum did this object start with? The initial momentum of this object is going to be two kilograms times the initial velocity, which was 10 meters per second to the right, which is positive 20 kilogram meters per second.

So if the initial momentum of the rocket is positive 20 and there was a change in momentum ofthe final momentum just has to be Or in other words, since the change in momentum would have to be the final momentum minus the initial momentum, which was positive 20, we could find the final momentum by adding 20 to both sides, which would give us plus 20, which is What's the difference between an elastic and an inelastic collision?

What we mean by an elastic collision is that the total kinetic energy of that system is conserved during the collision. In other words, if a sphere and a cube collide, for that collision to be elastic, the total kinetic energy of the sphere plus the kinetic energy of the cube before the collision would have to equal the kinetic energy of the sphere plus the kinetic energy of the cube after the collision.

If the total kinetic energy before the collision is equal to total kinetic energy after the collision, then that collision is elastic. It's not enough for the system to just bounce of each other.

If two objects bounce, the total kinetic energy might not be conserved. Only when the total kinetic energy is conserved can you say the collision is elastic. For an inelastic collision, the kinetic energy is not conserved during the collision.

In other words, the total initial kinetic energy of the sphere plus cube would not equal the total final kinetic energy of the sphere plus cube. Where does this kinetic energy go? Typically, in an inelastic collision, some of that kinetic energy is transformed into thermal energy during the collision.

## Linear momentum and collisions

While masses could bounce during an inelastic collision, if they stick together, the collision is typically called a perfectly inelastic collision, since in this collision you'll transform the most kinetic energy into thermal energy.

And when two objects stick together, it's a surefire sign that that collision is definitely inelastic. So what's an example problem that involves elastic and inelastic collisions look like? Let's say two blocks of mass 2M and M head toward each other with speeds 4v and 6v, respectively. After they collide, the 2M mass is at rest, and the mass M has a velocity of 2v to the right.

And we want to know, was this collision elastic or inelastic?

Now you might want to say that, since these objects bounced off of each other, the collision has to be elastic, but that's not true. If the collision is elastic, then the objects must bounce, but just because the objects bounce does not mean the collision is elastic. In other words, bouncing is a necessary condition for the collision to be elastic, but it isn't sufficient. If you really want to know whether a collision was elastic, you have to determine whether the total kinetic energy was conserved or not.

And we can figure that out for this collision without even calculating anything. Since the speed of the 2M mass decreased, the kinetic energy of the 2M mass went down. And since the speed of the M mass also decreased after the collision, the kinetic energy of the mass M went down, as well. So if the kinetic energy of both masses go down, then the final kinetic energy after the collision has to be less than the initial kinetic energy.

Which means kinetic energy was not conserved, and this collision had to be inelastic. One final note, even though kinetic energy wasn't conserved during this process, the momentum was conserved. The momentum will be conserved for both elastic and inelastic collisions.

It's just kinetic energy that's not conserved for an inelastic collision.