WEBVTT physics/ap-physics-1-2/fullerton
00:00:00.000 --> 00:00:03.000
Hi everyone and welcome back to Educator.com.
00:00:03.000 --> 00:00:08.000
This topic is collisions and conservation of momentum.
00:00:08.000 --> 00:00:19.000
Our objectives are going to be to use conservation of momentum to solve a variety of problems and also explain the difference between an elastic and an inelastic collision.
00:00:19.000 --> 00:00:26.000
Conservation of momentum -- linear momentum P is conserved in an isolated system.
00:00:26.000 --> 00:00:39.000
Total momentum of a system is constant and this is very useful for analyzing collisions and explosions -- where collision is an event in which two or more objects approach and interact strongly for a brief period of time.
00:00:39.000 --> 00:00:43.000
Or an explosion, which results when an object is broken up into several smaller fragments.
00:00:43.000 --> 00:00:59.000
And the key here -- conservation of momentum says in any of these events, the initial momentum and the final momentum have to be the same, and these are vector sums so we have to add up those momenta in a vector fashion.
00:00:59.000 --> 00:01:04.000
Now the easiest way I know to solve these -- to help keep organized -- is to create a momentum table.
00:01:04.000 --> 00:01:10.000
What we are going to do is we are going to identify all the objects in the system and list them down the left-hand side of our column.
00:01:10.000 --> 00:01:14.000
We will then determine the momenta of the objects before and after the collision.
00:01:14.000 --> 00:01:25.000
We will add them all up using variables for anything we do not know in calculating their momenta, and then we will set the total momentum before equal to the total momentum after.
00:01:25.000 --> 00:01:28.000
Sounds a lot more complicated than it is.
00:01:28.000 --> 00:01:32.000
Let us dive in and see how we would do this.
00:01:32.000 --> 00:01:36.000
As we talk about these, we have to remember that there are two types of collisions.
00:01:36.000 --> 00:01:43.000
In an elastic collision, also known as a "bouncy collision" -- kinetic energy is conserved.
00:01:43.000 --> 00:01:50.000
The total kinetic energy before is equal to the total kinetic energy after -- just like momentum is always conserved before and after.
00:01:50.000 --> 00:01:55.000
In an inelastic collision, kinetic energy is not conserved.
00:01:55.000 --> 00:02:04.000
And in a completely inelastic collision, or a "sticky collision", the objects actually stick together after they collide.
00:02:04.000 --> 00:02:13.000
Let us take a look at a 2,000 kg car traveling at 20 m/s and it collides with a 1,000 kg car at rest at a stop sign.
00:02:13.000 --> 00:02:23.000
If the 2,000 kg car has a velocity of 6.67 m/s after the collision, find the velocity of the 1,000 kg car after the collision.
00:02:23.000 --> 00:02:25.000
Here let us try out that momentum table.
00:02:25.000 --> 00:02:30.000
I am first going to list my objects over here.
00:02:30.000 --> 00:02:47.000
And the objects that I have -- let us call our first car, car A, the second car, car B and we are going to want to look at these in terms of the momentum before the collision in kg-m/s.
00:02:47.000 --> 00:02:49.000
We will put the units down here for this one.
00:02:49.000 --> 00:02:57.000
The momentum after the collision -- momentum after in kg-m/s.
00:02:57.000 --> 00:03:05.000
We will make our table -- and total.
00:03:05.000 --> 00:03:17.000
Now before the collision, car A is traveling at 20 m/s so it is 2,000 as its mass times its velocity (20) for a total momentum of 40,000 kg-m/s.
00:03:17.000 --> 00:03:28.000
Car B is at rest, so the total then -- 40,000 + 0 = 40,000 kg-m/s -- is the total momentum before the collision.
00:03:28.000 --> 00:03:47.000
Now after the collision, car A still has a mass of 2,000, but it has a velocity of 6.67 m/s, so its new momentum after the collision is 13,340 kg-m/s.
00:03:47.000 --> 00:03:54.000
Car B on the other hand, has a mass of 1,000 but we do not know its velocity, so I will put a variable in there.
00:03:54.000 --> 00:03:56.000
Let us call that the velocity of B.
00:03:56.000 --> 00:04:06.000
So when I add these up, I get 13,340 + 1,000 VB.
00:04:06.000 --> 00:04:16.000
Now here is the slick part -- now that we have made the table, conservation of momentum says that these momenta before and after must be equal because there were no external forces.
00:04:16.000 --> 00:04:20.000
Here is the equation that I have to solve in order to figure out what happened.
00:04:20.000 --> 00:04:47.000
If I subtract 13,340 from both sides, I find out that 1,000 VB is going to be equal to 4-3-9-9-9 -- 10 - oops 0, 0 -- 9 -... 10 - 4 = 6, 9 - 3 = 6, 9 - 3 = 6, 3 - 1 = 2, equals 1,000 VB.
00:04:47.000 --> 00:05:10.000
Now, divide both sides by 1,000 and VB -- the velocity of car B after the collision -- is equal to about 26.7 m/s -- Using a momentum table to help organize everything we need to know in these collisions.
00:05:10.000 --> 00:05:13.000
Let us take a look at inelastic collision.
00:05:13.000 --> 00:05:24.000
On a snow covered road, a car with a mass of 1.1 x 10³ kg collides head on with a van having a mass of 2.5 x 10³ kg traveling at 8 m/s.
00:05:24.000 --> 00:05:28.000
Note that if they colliding head on they must be going in opposite directions.
00:05:28.000 --> 00:05:32.000
As a result of the collision, the vehicles lock together and immediately come to rest.
00:05:32.000 --> 00:05:38.000
Let us calculate the speed of the car immediately before the collision, and again we can neglect friction.
00:05:38.000 --> 00:05:42.000
Let us start off by listing our objects.
00:05:42.000 --> 00:05:55.000
We have the car, we have the van, and the momentum before and the momentum after.
00:05:55.000 --> 00:06:06.000
Now before the collision, the car has a momentum of -- its mass is 1,100 times some velocity we do not know.
00:06:06.000 --> 00:06:09.000
Vcar, that is what we are trying to find.
00:06:09.000 --> 00:06:17.000
The van has a mass of 2,500 and its velocity is -8 because it is going in the opposite direction of the car.
00:06:17.000 --> 00:06:23.000
Now after the collision, they stick together and essentially they become one object.
00:06:23.000 --> 00:06:30.000
However, they are at rest so their momentum afterwards must be 0.
00:06:30.000 --> 00:06:36.000
As we make our table, let us fill out our row for total.
00:06:36.000 --> 00:06:51.000
Momentum before must be 1,100 Vcar - 2,500 × 8 and all of that must be equal to the momentum after -- 0.
00:06:51.000 --> 00:06:58.000
If I solve this equation then, 1,100 Vcar = 20,000.
00:06:58.000 --> 00:07:08.000
If I divide both sides by 1,100, then the velocity of our car must be 18.2 m/s.
00:07:08.000 --> 00:07:14.000
A nice easy way to organize our thoughts here with those momentum tables.
00:07:14.000 --> 00:07:17.000
Let us take a look at one involving recoil velocity.
00:07:17.000 --> 00:07:24.000
If you shoot a gun, initially before you do anything with it, it has a momentum of 0 -- you are holding it still.
00:07:24.000 --> 00:07:30.000
Then you shoot a bullet out one end of it -- very, very quickly -- it has a momentum.
00:07:30.000 --> 00:07:34.000
Conservation of momentum says the total momentum of the system must remain constant.
00:07:34.000 --> 00:07:40.000
So if a bullet comes out one way with some momentum, the rest of the gun must have a momentum back.
00:07:40.000 --> 00:07:42.000
That is the kick on a gun or the recoil.
00:07:42.000 --> 00:07:44.000
We call the velocity it kicks back with the recoil velocity.
00:07:44.000 --> 00:07:56.000
So a 4 kg rifle fires a 20 g shell with a velocity of 300 m/s. Find the recoil velocity of the rifle.
00:07:56.000 --> 00:08:08.000
Our objects -- here we have a rifle and we have a bullet.
00:08:08.000 --> 00:08:13.000
We have a momentum before and a momentum after.
00:08:13.000 --> 00:08:24.000
Now before this explosion that fires the bullet -- in essence, the rifle and bullet are really one object and they are at rest.
00:08:24.000 --> 00:08:27.000
So their momentum before is 0.
00:08:27.000 --> 00:08:38.000
After the incident, after they are fired, the rifle has a mass of 4 kg and it has some recoil velocity -- Vrecoil -- we do not know what it is, we are trying to find it.
00:08:38.000 --> 00:08:51.000
The bullet has a mass of 20 g -- 0.02 grams -- and it has a velocity of 300 m/s.
00:08:51.000 --> 00:08:54.000
That is going to be equal to 6.
00:08:54.000 --> 00:09:05.000
So as I make my table here -- total on the left is 0 and on the right I have 4 Vrecoil + 6.
00:09:05.000 --> 00:09:23.000
Momentum before must equal momentum after, therefore if I subract 6 from both sides, -6 = 4, Vrecoil, or Vrecoil, the recoil velocity of the rifle is -1.5 m/s.
00:09:23.000 --> 00:09:29.000
What does the negative mean? It is in the opposite direction of the bullet's velocity.
00:09:29.000 --> 00:09:40.000
So conservation of momentum in two dimensions -- in this problem we have a cue ball that Bert strikes with a mass of 0.17 kg giving it a velocity of 3 m/s to the right.
00:09:40.000 --> 00:09:51.000
When the cue ball strikes the 8-ball of mass 0.16 kg, the 8-ball gets deflected in this direction with an angle of 45 degrees and the cue ball comes to this direction with an angle of 40 degrees.
00:09:51.000 --> 00:09:56.000
We need to find the velocity of the cue ball and the 8-ball after the collision.
00:09:56.000 --> 00:10:04.000
A great problem for momentum tables but it is all about staying organized -- taking our time and doing it right.
00:10:04.000 --> 00:10:14.000
Here we go -- First off, let us say that the velocity of the cue ball is Vc -- that is after the collision, that is what we are trying to find.
00:10:14.000 --> 00:10:19.000
We will call V8 the velocity of the 8-ball after the collision.
00:10:19.000 --> 00:10:38.000
So when we make our momentum table in the x direction for momentum, our objects are -- we have a cue ball, we have an 8-ball, and of course we have our line for total.
00:10:38.000 --> 00:10:45.000
We will take a look at the momentum before the collision in the x direction and the momentum after the collision in the x direction.
00:10:45.000 --> 00:11:02.000
--Long pause--
00:11:02.000 --> 00:11:14.000
The cue ball before the collision has a mass of 0.17 and a velocity of 3, so its total momentum before is 0.51.
00:11:14.000 --> 00:11:19.000
The 8-ball is at rest so its total momentum before in the x direction is 0.
00:11:19.000 --> 00:11:26.000
Our total then is just 0.51 before the collision and that is going to equal our momentum after in the x direction.
00:11:26.000 --> 00:11:30.000
Now the cue ball has the same mass after the collision.
00:11:30.000 --> 00:11:41.000
We do not know its velocity but we are going to call that Vc and we need to take the x component of that, so that is going to be the cos 40 degrees.
00:11:41.000 --> 00:11:52.000
The 8-ball has a mass of 0.16 kg -- its velocity after the collision is V8 and we need its x component so that is going to be the cos 45 degrees.
00:11:52.000 --> 00:12:13.000
When I add these up I get 0.17 cos 40 degrees × Vc -- that is going to be 0.13 Vc + 0.16 cos 45 × V8, which is 0.113 V8.
00:12:13.000 --> 00:12:17.000
Now let us make our momentum table for the y direction.
00:12:17.000 --> 00:12:28.000
Again we have the same objects -- the momentum before in the y direction and the momentum after in the y direction.
00:12:28.000 --> 00:12:48.000
And our objects are our cue ball, our 8-ball, and a row for the total -- so objects.
00:12:48.000 --> 00:12:54.000
All right, we have a nice little momentum table here.
00:12:54.000 --> 00:13:04.000
The cue ball -- its momentum in the y direction before the collision is 0 because it has no velocity in the y direction and the 8-ball is at rest so the total must be 0.
00:13:04.000 --> 00:13:29.000
Now after the collision in the y -- P after in the y direction -- the cue ball still has a mass of 0.17, its velocity is Vc but now it is going down in that direction, so that is going to be times the sin -40 degrees.
00:13:29.000 --> 00:13:43.000
The y component of the 8-ball's momentum -- well its mass is 0.16 times its velocity, V8, times the sin 45 degrees to get its y component.
00:13:43.000 --> 00:14:02.000
So 0 is going to be equal to 0.17 sin -40 × Vc is going to be -0.109 Vc + 0.16 × sin 45 × V8 = 0.113 V8.
00:14:02.000 --> 00:14:11.000
If you look here I have two equations now and two unknowns that I can then solve to find my unknown unknown values.
00:14:11.000 --> 00:14:18.000
What I am going to do to begin with is let us solve this and see if we can find what Vc is equal to.
00:14:18.000 --> 00:14:28.000
If I add 0.109 Vc to both sides, the left-hand side becomes 0.109 Vc = 0.113 V8.
00:14:28.000 --> 00:14:38.000
If I divide both sides by 0.109, I find that Vc equals about 0.104 V8.
00:14:38.000 --> 00:14:43.000
Now what I am going to do is I am going to take that value up here to my Equation 1.
00:14:43.000 --> 00:14:47.000
So up here in Equation 1 -- let us give ourselves a little room here.
00:14:47.000 --> 00:15:04.000
We can now write that 0.51 = 0.13, and I am going to replace Vc with 0.104 V8 + 0.113 V8.
00:15:04.000 --> 00:15:14.000
A little bit of algebra -- 0.51 equals -- well that is going to give me 0.248 V8.
00:15:14.000 --> 00:15:25.000
If I divide both sides by 0.248, thenI find out that the velocity of the 8-ball is 2.06 m/s.
00:15:25.000 --> 00:15:33.000
Great start. Now that we have that, we can take that value and we can put it back in here.
00:15:33.000 --> 00:15:53.000
Velocity of our cue ball then is 1.04 × velocity of our 8-ball, 2.06 m/s, so the velocity of our cue ball then comes out to be 2.14 m/s.
00:15:53.000 --> 00:16:02.000
Same basic strategies -- now we are just using momentum tables for the x components and the y components of momentum.
00:16:02.000 --> 00:16:08.000
Let us take a look at one more -- an atomic collision -- but in this case we are dealing with an elastic collision.
00:16:08.000 --> 00:16:15.000
A proton with some mass (m) and a lithium nucleus with some mass (7m) undergo an elastic collision.
00:16:15.000 --> 00:16:24.000
Elastic collision means that kinetic energy is conserved -- the kinetic energy before the collision is equal to the kinetic energy after the collision.
00:16:24.000 --> 00:16:28.000
Find the velocity of the lithium nucleus following the collision.
00:16:28.000 --> 00:16:33.000
Well since this is coming straight on, we really only have to deal with one dimension here.
00:16:33.000 --> 00:16:35.000
So let us take a look.
00:16:35.000 --> 00:16:59.000
Our objects are our proton, our lithium nucleus, and a row for total, and the momentum before and momentum after.
00:16:59.000 --> 00:17:09.000
For our proton before the collision, it has a velocity of 1000 and a mass of m, so its momentum must be 1000 m.
00:17:09.000 --> 00:17:18.000
Lithium nucleus is just sitting there nice and happy at rest, 0, so the total before must be 1000 m.
00:17:18.000 --> 00:17:28.000
Now after the collision, the momentum after -- well the mass of the proton does not change but it has some velocity, Vp, the velocity of our proton.
00:17:28.000 --> 00:17:36.000
The lithium nucleus has some mass (7m) times the velocity of L, the velocity of our lithium nucleus.
00:17:36.000 --> 00:17:53.000
So 1000 m is going to be equal to mVp + 7 mVL and I can right away divide the mass out of all of that to say that 1000 = Vp + 7 VL.
00:17:53.000 --> 00:17:55.000
There is one equation.
00:17:55.000 --> 00:17:59.000
Now we need to bring in that this is an elastic collision.
00:17:59.000 --> 00:18:08.000
That means that the total kinetic energy before the collision equals the total kinetic energy after the collision.
00:18:08.000 --> 00:18:22.000
Kinetic energy is 1/2 mass times velocity², so kinetic energy before -- we have 1/2 times (m) times the square of its velocity.
00:18:22.000 --> 00:18:24.000
Let us call that V-initial².
00:18:24.000 --> 00:18:26.000
That must equal the kinetic energy after.
00:18:26.000 --> 00:18:40.000
Now both of these are in motion -- so that is going to be 1/2m times Vp² plus 1/2 our 7m, our lithium nucleus, times its speed squared.
00:18:40.000 --> 00:18:56.000
And once again I can make a nice simplification and divide out 1/2m out of all of those to say that V0² = Vp² + 7 VL².
00:18:56.000 --> 00:19:03.000
So we have two equations, two unknowns because really we know this V0 is 1,000, so V0² is 10^6.
00:19:03.000 --> 00:19:12.000
So, 10^6 = Vp² + 7 VL².
00:19:12.000 --> 00:19:25.000
All right to take that a little bit further then, we can then say that 10^6 -- let us come over here first and solve for Vp.
00:19:25.000 --> 00:19:49.000
If we do this, we could say Vp = 1000 - 7 VL, so over here, 10^6 now equals -- I am going to replace Vp with 1000 - 7 VL² + 7 VL².
00:19:49.000 --> 00:20:10.000
All right, so 10^6 = 1000 - 7 VL², which is going to be 10^6 - 7000 VL - 7,000 VL, so that is minus 14000 VL + 49 VL² + 7 VL².
00:20:10.000 --> 00:20:22.000
That implies then, that 56 VL² - 14,000 VL = 0.
00:20:22.000 --> 00:20:38.000
We can factor out a VL to say that VL × 56 VL - 14000 = 0, so either VL = 0 or 56 VL - 14,000 = 0.
00:20:38.000 --> 00:20:55.000
So, 56 VL - 14,000 = 0, add 14,000 to both sides, divide by 56 and I can find that VL = 250 m/s.
00:20:55.000 --> 00:21:11.000
And if VL = 250 m/s, I can take that back over here to say that Vp = 1,000 - 7 × 250...
00:21:11.000 --> 00:21:21.000
... or Vp = 1000 - 7 × 250 = -750 m/s.
00:21:21.000 --> 00:21:31.000
So what do we have here? Velocity of our lithium nucleus is 250 m/s -- velocity of our proton is -750 m/s.
00:21:31.000 --> 00:21:35.000
Why negative? It is going in the opposite direction that it first started.
00:21:35.000 --> 00:21:43.000
It comes this way to the right, bounces off, and comes back to the left with a speed of 750 m/s to the left.
00:21:43.000 --> 00:21:51.000
This brings in the fact that it is an elastic collision to give us one more equation -- this kinetic energy before equals kinetic energy after.
00:21:51.000 --> 00:21:56.000
Hopefully this gets you started with analyzing collisions and looking at conservation of momentum.
00:21:56.000 --> 00:21:59.000
I appreciate your time. Thanks for watching Educator.com