Newtons 2nd law states that force equals mass times acceleration, but the actual law states that a force is the rate of change of momentum over time. Momentum is the natural tendency for objects to continue moving despite having no net force applied to them. Because of moment, a pencil continues to roll after you push it along a table and impulse makes you catch the pencil after it falls causing your hand to move down slightly.
Momentum is a vector quantity describing how hard it is to stop a moving object. Momentum is equal to the product of an object's mass and its velocity.
The change in linear momentum is the product of the mass and change in the velocity of the center of mass.
Change in momentum is a vector known as impulse. The impulse vector is in the direction of the net force and occurs over a time interval.
Velocity of the center of mass cannot be changed by an interaction within the system.
Forces that systems exert on each other are due to interactions between objects in the system. If the interacting objects are part of the same system, there will be no change in the velocity of the center of mass of the system.
Impulse & Momentum
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Having said that then, if we wanted to find out the momentum of Big Red, that is just going to be -- since it has twice the mass of Little Blue, it is going to be two times Little Blue's mass times whatever the velocity is.0137
What that allows me to do now is, if I look at this -- this piece here -- δv over δt, the rate of change of velocity with respect to time -- that is the definition of acceleration (a).0623
So then I can write this as impulse is equal to mass times acceleration multiplied by δt. 0645
But there is another transformation we can make here.0655
Now we have this ma -- Newton's Second Law, net force equals mass times acceleration, so I can now rewrite this again as J (impulse) equals mass times acceleration of force times your time interval. 0660
So putting it altogether, impulse is a change in momentum, which is equal to a force applied over some time interval.0680
How do you apply an impulse? You apply a force for some amount of time.0693
How do you change an object's momentum? You apply a force for an amount of time.0697
It has some momentum, so if you want to change it, you have to apply a force.0702
The longer you apply the force, the greater the change in momentum.0706
The larger the force you apply, the larger the change in momentum.0710
That is what is known as the impulse-momentum theorem.0713
What it is saying again, when an unbalanced force acts on an object for a period of time, a change in momentum is produced and that is known as an impulse.0722
Here is our example problem -- A tow truck applies a force of 2,000N on a 2,000 kg car for a period of 3 s.0735
First off, what is the magnitude of the change in the car's momentum?0744
In an automobile collision, a 44 kg passenger moving at 15 m/s is brought to rest -- V is 0 -- by an airbag during a 0.1 s time interval.1175
That is kind of the point of airbags -- not only do they spread out where the force is applied to decrease the pressure at any given point, but they increase the time interval in which that force is applied.1187
If you have an impulse, you are coming to rest at some point -- no matter what, that impulse is going to be applied, so you would rather have it applied over a longer period of time so that you have a smaller force.1199
Before the days of airbags, you would hit the windshield -- smack -- really short time, really big force. Game over.1210
I am trying to avoid those game overs, those really big forces over short time intervals.1218
So increase the time interval, lower the force -- greater survival rates.1222
So we are going to find the average force exerted on the passenger during that time.1227
Impulse is change in momentum, which is force times time.1231
Therefore, force equals change in momentum divided by time which is the final momentum minus the initial momentum divided by time.1239
Final momentum (0) came to rest minus the initial momentum (44) mass times velocity (15 m/s) in 0.1 s means that the force exerted on the passenger during that time is -6600 N.1253
If there was not an airbag, imagine that was 0.01 s.1274
Now we are talking 66,000N exerted on the passenger. Yay, airbags!1278
All right. Let us take a look at a couple of center of mass problems.1286
Find the center of mass of an object modeled as two separate masses on the x axis.1291
The first mass is 2 kg in an x coordinate of 2 and the second mass is 6 kg in an x coordinate of 8.1295
Just by looking at this, we should be able to guesstimate where this is going to be.1303
First off, it is going to be somewhere between those two objects.1307
Since the object over here on the right, the 6 kg mass, is bigger, I would anticipate that we are going to be somewhere closer to the 6 kg mass than they are the 2 kg mass when we find the center of mass.1311
So let us go back to our formula for center of mass.1324
In the x direction, x center of mass is m1x1 plus m2x2 plus for however many masses we have divided by all of the masses (m1 + m2 + ....).1328
In this case, mass 1 is 2 kg, so that is 2 and its x position is 2 -- m2 is 6 kg and its x position is 8.1344
We are going to divide that by the sum of the masses, 2 kg + 6 kg.1354
So I get 4 + 48/8 -- that is 52/8 which is going to be -- 6.5 should be the position on the scale of our center of mass.1361
1, 2, 3, 4, 5, 6 -- So we could draw it in here, right about there.1383
We could replace these two masses and treat it as if it is one 8 kg mass centered at a point that has an x value of 6.5.1391
Let us finish off by doing one that is a two-dimensional center of mass.1406
Find the coordinates of the center of mass for this system where we have three masses.1411
So we are going to have an x and a y coordinate now.1415
And right away, let us look at it again and go, "You know, chances are, in terms of the x position, we're going to be between this mass and this mass and in the y position, we're going to be between that one and that one."1419
So we are probably looking for a center of mass somewhere in that area.1429
That will help us see if we get this right or not.1433
The formula for the x center of mass again is m1x1 plus m2x2 and so on divided by the sum of all the individual masses.1436
In the y direction, the y center of mass works the same way -- m1y1 plus m2y2 plus however many more divided by all of the masses.1449
So let us figure out exactly where the center of mass should lie.1465
Start with the x coordinate. That is going to be...1469
Well, we have got 3 kg in x coordinate of 1 + 4 kg in x coordinate of 5 + 1 kg in x coordinate of 7 divided by the total mass, 3 + 4 + 1 = 8.1472
So that is going to be 3 + 20 + 7 -- that is going to be 30/8 or 3.75 for the x coordinate.1490
For the y coordinate -- now we are going to take our 3 kg × the y value (2) + 4 kg × the y coordinate (3) + 1 kg × the y coordinate (1) divided by the the total mass is 8.1500
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This book is written by our very own Professor Fullerton and features more than 600 worked-out problems with full solutions and deeper understanding questions. AP Physics 1 Essentials covers all major topics included in the AP Physics 1 course, including: kinematics, dynamics, momentum, impulse, gravity, uniform circular motion, rotation, work, energy, power, mechanical waves, sound, electrostatics, and circuits.