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Lecture Comments (50)

2 answers

Last reply by: Professor Dan Fullerton
Sat Jul 9, 2016 6:11 AM

Post by Elman Ahmed on July 7 at 07:27:14 PM

Professor, i was talking Physics 1 in the summer. I listened to ALL of your lectures from Physics 1. The example no. 2 in this lecture was very helpful! I got something very Similar. As a matter of fact, i got a lot of similar problems from your sample problems. Thank you so much. I did great in that class.

1 answer

Last reply by: Professor Dan Fullerton
Mon Jun 27, 2016 7:58 AM

Post by Peter Ke on June 25 at 07:23:51 PM

For example 4, at 15:00 you used the substitution method but I used the elimination method by getting rid of the .113V_8 by subtracting. The answer I got was 2.13 m/s for V_cue. You got 2.14 m/s. Will the AP exam grader accept my 2.13 m/s as my answer or not? Just curious.

.51 = .13V_cue + .113V_8
-(0 = -.109V_cue + .113V_8)
----------------------------
.51 = .239V_cue
 V_cue = 2.13 m/s


0 = -.109(2.13) + .113V_8
V_8 = 2.06 m/s

0 answers

Post by Jayden Luis on May 7 at 11:45:35 AM

Great!

1 answer

Last reply by: Professor Dan Fullerton
Mon Jan 4, 2016 6:14 AM

Post by Mustafa Lambay on January 3 at 01:26:36 PM

Heyo, for example 5: Atomic Collisions, why did you use 1000m as the momentum before the elastic collision for the proton instead of just 1000 m/s?

Thanks amigo.

2 answers

Last reply by: Anh Dang
Fri Jul 24, 2015 2:36 PM

Post by Anh Dang on July 23, 2015

Velocity needs speed and direction, right?  So, for example 4, what would be the directions for the cue ball and the eight ball?

2 answers

Last reply by: Professor Dan Fullerton
Tue Jun 16, 2015 7:54 PM

Post by Denise Aguilar on June 16, 2015

hello professor,
I am missing where the 14000 VL came from when you expanded (1000-7Vl)^2 by writing it as 10^6-14000Vl+49VL^2. Please help : )
Thank you

3 answers

Last reply by: Professor Dan Fullerton
Mon Apr 6, 2015 6:55 AM

Post by Anna Ha on April 1, 2015

Hi,
How would you do this question?
1. A frictionless trolley, m1, of mass 2.00kg, moving to the right at 6.00m/s, collide with and sticks to an initially stationary frictionless trolley, m2, of mass 4.00kg.
a) calculate the magnitude and direction of the impulse exerted ON m1 BY m2 during the collision?

Thanks!

3 answers

Last reply by: Professor Dan Fullerton
Wed Nov 5, 2014 8:27 AM

Post by Caleb Martin on November 5, 2014

Hi,
what equation best fits these questions? Problem: "A railroad car of mass 2.53  104 kg is moving with a speed of 4.06 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.03 m/s."
Question:(a) What is the speed of the four cars after the collision?
Question:(b) How much mechanical energy is lost in the collision?
Thanks

1 answer

Last reply by: Professor Dan Fullerton
Sat Oct 11, 2014 8:10 PM

Post by Foaad Zaid on October 11, 2014

So the only purpose of the kinetic energy equation was to give us a second equation to allow us to solve for 2 unknowns? I'm having difficulty understanding why the kinetic energy equations were implemented. Thank you again.

1 answer

Last reply by: Professor Dan Fullerton
Sat Oct 11, 2014 6:47 AM

Post by Foaad Zaid on October 11, 2014

I don't seem to understand why including kinetic energy was necessary? Thank you

1 answer

Last reply by: Professor Dan Fullerton
Sat Oct 11, 2014 6:47 AM

Post by Foaad Zaid on October 11, 2014

For the atomic collision, how come it just wouldn't be possible to use set momentum final = momentum initial?

1 answer

Last reply by: Professor Dan Fullerton
Sat Aug 2, 2014 1:27 PM

Post by Jungle Jones on August 2, 2014

In Ex. 5, how do you know VL can't be zero?

3 answers

Last reply by: Professor Dan Fullerton
Mon Jun 16, 2014 6:17 AM

Post by Pedro Gallegos on June 15, 2014

19:57 19:58 I dont understand how -7VL = 14,000

1 answer

Last reply by: Professor Dan Fullerton
Sun Feb 23, 2014 9:59 AM

Post by rumeh mandible on February 22, 2014

A 1-kg chunk of putty moving at 1 m/s collides with and stick to a 5-lkg bowling bail initially at rest.The bowling ball and putty then move with a momentum of ......

1 answer

Last reply by: Professor Dan Fullerton
Sun Feb 23, 2014 9:58 AM

Post by rumeh mandible on February 22, 2014

when bullets are fired from an airplane in the forward direction, the momentum of the the airplane will be

1 answer

Last reply by: Professor Dan Fullerton
Wed Jan 29, 2014 6:28 AM

Post by Hyun Cho on January 28, 2014

Hey, in example 5, when you are doing the conservation of kinetic energy part, you said 1/2mv^2=1/2mvp^2+1/2mvl^2.  but should it be 1/2mvp^2-1/2mvl^2 since the velocity of lithium is opposite the velocity of proton?

2 answers

Last reply by: Hyun Cho
Wed Jan 29, 2014 5:07 PM

Post by Hyun Cho on January 28, 2014

Hey, in example 5, when you are doing the conservation of kinetic energy part, you said 1/2mv^2=1/2mvp^2+1/2mvl^2. but should it be 1/2mvp^2-1/2mvl^2 since the velocity of lithium is opposite the velocity of proton?

2 answers

Last reply by: Gaurav Kumar
Thu Aug 29, 2013 9:51 PM

Post by Gaurav Kumar on August 29, 2013

I know this is related to the previous comment, but I still don't understand one thing. How come for PAx you use cos(40), but for PAy you use sin(-40). Shouldn't they both be -40 because the angle is under the horizontal?

1 answer

Last reply by: Professor Dan Fullerton
Fri May 3, 2013 7:04 AM

Post by Ardeshir Badr on May 1, 2013

At 13:50, why do you use sin(-40) as opposed to sin(40)for Pay column? and you did not use sin(-45), you used sin(45). where is the logic behind knowing whether it is negative or positive sign for the degrees. sin(-40) vs. sin(40). I followed the previous segments on this but still cannot figure it out! thanks!

1 answer

Last reply by: Professor Dan Fullerton
Sat Apr 20, 2013 3:43 PM

Post by Muna Lakhani on April 20, 2013

For example 5, are you using the KE equation where KE= 1/2mv(Squared)?

Collisions

  • Linear momentum is conserved in a closed system.
  • Kinetic Energy is conserved in an elastic collision.
  • Momentum tables can be used to solve for unknown quantities in collisions.

Collisions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:09
  • Conservation of Momentum 0:18
    • Linear Momentum is Conserved in an Isolated System
    • Useful for Analyzing Collisions and Explosions
  • Momentum Tables 0:58
    • Identify Objects in the System
    • Determine the Momenta of the Objects Before and After the Event
    • Add All the Momenta From Before the Event and Set Them Equal to Momenta After the Event
    • Solve Your Resulting Equation for Unknowns
  • Types of Collisions 1:31
    • Elastic Collision
    • Inelastic Collision
  • Example 1: Conservation of Momentum (1D) 2:02
  • Example 2: Inelastic Collision 5:12
  • Example 3: Recoil Velocity 7:16
  • Example 4: Conservation of Momentum (2D) 9:29
  • Example 5: Atomic Collision 16:02

Transcription: Collisions

Hi everyone and welcome back to Educator.com.0000

This topic is collisions and conservation of momentum.0003

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.0008

Conservation of momentum -- linear momentum P is conserved in an isolated system.0019

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. 0026

Or an explosion, which results when an object is broken up into several smaller fragments.0039

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.0043

Now the easiest way I know to solve these -- to help keep organized -- is to create a momentum table.0059

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.0064

We will then determine the momenta of the objects before and after the collision. 0070

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.0074

Sounds a lot more complicated than it is.0085

Let us dive in and see how we would do this.0088

As we talk about these, we have to remember that there are two types of collisions.0092

In an elastic collision, also known as a "bouncy collision" -- kinetic energy is conserved.0096

The total kinetic energy before is equal to the total kinetic energy after -- just like momentum is always conserved before and after.0103

In an inelastic collision, kinetic energy is not conserved.0110

And in a completely inelastic collision, or a "sticky collision", the objects actually stick together after they collide.0115

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.0124

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.0133

Here let us try out that momentum table.0143

I am first going to list my objects over here.0145

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.0150

We will put the units down here for this one.0167

The momentum after the collision -- momentum after in kg-m/s.0169

We will make our table -- and total.0177

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.0185

Car B is at rest, so the total then -- 40,000 + 0 = 40,000 kg-m/s -- is the total momentum before the collision.0197

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.0208

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.0227

Let us call that the velocity of B.0234

So when I add these up, I get 13,340 + 1,000 VB.0236

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.0246

Here is the equation that I have to solve in order to figure out what happened.0256

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.0260

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.0287

Let us take a look at inelastic collision.0310

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.0313

Note that if they colliding head on they must be going in opposite directions.0324

As a result of the collision, the vehicles lock together and immediately come to rest.0328

Let us calculate the speed of the car immediately before the collision, and again we can neglect friction.0332

Let us start off by listing our objects.0338

We have the car, we have the van, and the momentum before and the momentum after.0342

Now before the collision, the car has a momentum of -- its mass is 1,100 times some velocity we do not know.0355

Vcar, that is what we are trying to find.0366

The van has a mass of 2,500 and its velocity is -8 because it is going in the opposite direction of the car.0369

Now after the collision, they stick together and essentially they become one object.0377

However, they are at rest so their momentum afterwards must be 0.0383

As we make our table, let us fill out our row for total.0390

Momentum before must be 1,100 Vcar - 2,500 × 8 and all of that must be equal to the momentum after -- 0.0396

If I solve this equation then, 1,100 Vcar = 20,000.0411

If I divide both sides by 1,100, then the velocity of our car must be 18.2 m/s.0418

A nice easy way to organize our thoughts here with those momentum tables.0428

Let us take a look at one involving recoil velocity.0434

If you shoot a gun, initially before you do anything with it, it has a momentum of 0 -- you are holding it still.0437

Then you shoot a bullet out one end of it -- very, very quickly -- it has a momentum.0444

Conservation of momentum says the total momentum of the system must remain constant.0450

So if a bullet comes out one way with some momentum, the rest of the gun must have a momentum back.0454

That is the kick on a gun or the recoil.0460

We call the velocity it kicks back with the recoil velocity.0462

So a 4 kg rifle fires a 20 g shell with a velocity of 300 m/s. Find the recoil velocity of the rifle.0464

Our objects -- here we have a rifle and we have a bullet.0476

We have a momentum before and a momentum after.0488

Now before this explosion that fires the bullet -- in essence, the rifle and bullet are really one object and they are at rest.0493

So their momentum before is 0.0504

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.0507

The bullet has a mass of 20 g -- 0.02 grams -- and it has a velocity of 300 m/s.0518

That is going to be equal to 6.0531

So as I make my table here -- total on the left is 0 and on the right I have 4 Vrecoil + 6.0534

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.0545

What does the negative mean? It is in the opposite direction of the bullet's velocity.0563

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.0569

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.0580

We need to find the velocity of the cue ball and the 8-ball after the collision.0591

A great problem for momentum tables but it is all about staying organized -- taking our time and doing it right.0596

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.0604

We will call V8 the velocity of the 8-ball after the collision.0614

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.0619

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.0638

--Long pause--0645

The cue ball before the collision has a mass of 0.17 and a velocity of 3, so its total momentum before is 0.51.0662

The 8-ball is at rest so its total momentum before in the x direction is 0.0674

Our total then is just 0.51 before the collision and that is going to equal our momentum after in the x direction.0679

Now the cue ball has the same mass after the collision.0686

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.0690

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.0701

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 V8.0712

Now let us make our momentum table for the y direction.0733

Again we have the same objects -- the momentum before in the y direction and the momentum after in the y direction.0737

And our objects are our cue ball, our 8-ball, and a row for the total -- so objects. 0748

All right, we have a nice little momentum table here.0768

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 0.0774

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.0784

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.0809

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 V8.0823

If you look here I have two equations now and two unknowns that I can then solve to find my unknown unknown values.0842

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.0851

If I add 0.109 Vc to both sides, the left-hand side becomes 0.109 Vc = 0.113 V8.0858

If I divide both sides by 0.109, I find that Vc equals about 0.104 V8.0868

Now what I am going to do is I am going to take that value up here to my Equation 1.0878

So up here in Equation 1 -- let us give ourselves a little room here.0883

We can now write that 0.51 = 0.13, and I am going to replace Vc with 0.104 V8 + 0.113 V8.0887

A little bit of algebra -- 0.51 equals -- well that is going to give me 0.248 V8.0904

If I divide both sides by 0.248, thenI find out that the velocity of the 8-ball is 2.06 m/s.0914

Great start. Now that we have that, we can take that value and we can put it back in here.0925

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.0933

Same basic strategies -- now we are just using momentum tables for the x components and the y components of momentum.0953

Let us take a look at one more -- an atomic collision -- but in this case we are dealing with an elastic collision.0962

A proton with some mass (m) and a lithium nucleus with some mass (7m) undergo an elastic collision.0968

Elastic collision means that kinetic energy is conserved -- the kinetic energy before the collision is equal to the kinetic energy after the collision.0975

Find the velocity of the lithium nucleus following the collision.0984

Well since this is coming straight on, we really only have to deal with one dimension here.0988

So let us take a look.0993

Our objects are our proton, our lithium nucleus, and a row for total, and the momentum before and momentum after.0995

For our proton before the collision, it has a velocity of 1000 and a mass of m, so its momentum must be 1000 m.1019

Lithium nucleus is just sitting there nice and happy at rest, 0, so the total before must be 1000 m.1029

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.1038

The lithium nucleus has some mass (7m) times the velocity of L, the velocity of our lithium nucleus.1048

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. 1056

There is one equation.1073

Now we need to bring in that this is an elastic collision.1075

That means that the total kinetic energy before the collision equals the total kinetic energy after the collision.1079

Kinetic energy is 1/2 mass times velocity2, so kinetic energy before -- we have 1/2 times (m) times the square of its velocity.1088

Let us call that V-initial2.1102

That must equal the kinetic energy after.1104

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.1106

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 VL2.1120

So we have two equations, two unknowns because really we know this V0 is 1,000, so V02 is 106.1136

So, 106 = Vp2 + 7 VL2.1143

All right to take that a little bit further then, we can then say that 106 -- let us come over here first and solve for Vp.1152

If we do this, we could say Vp = 1000 - 7 VL, so over here, 106 now equals -- I am going to replace Vp with 1000 - 7 VL2 + 7 VL2.1165

All right, so 106 = 1000 - 7 VL2, which is going to be 106 - 7000 VL - 7,000 VL, so that is minus 14000 VL + 49 VL2 + 7 VL2.1189

That implies then, that 56 VL2 - 14,000 VL = 0.1210

We can factor out a VL to say that VL × 56 VL - 14000 = 0, so either VL = 0 or 56 VL - 14,000 = 0.1222

So, 56 VL - 14,000 = 0, add 14,000 to both sides, divide by 56 and I can find that VL = 250 m/s.1238

And if VL = 250 m/s, I can take that back over here to say that Vp = 1,000 - 7 × 250... 1255

... or Vp = 1000 - 7 × 250 = -750 m/s.1271

So what do we have here? Velocity of our lithium nucleus is 250 m/s -- velocity of our proton is -750 m/s.1281

Why negative? It is going in the opposite direction that it first started.1291

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.1295

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.1303

Hopefully this gets you started with analyzing collisions and looking at conservation of momentum.1311

I appreciate your time. Thanks for watching Educator.com1316