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Conservation of Momentum

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
  • Question 1 0:15
  • Question 2 2:08
  • Question 3 4:03
  • Question 4 4:10
  • Question 5 6:08
  • Question 6 6:55
  • Question 7 8:26

Transcription: Conservation of Momentum

Hi everyone and welcome back to 0000

In this mini-lesson, we are going to talk about conservation of momentum by doing the first page of the APlusPhysics worksheet on conservation of momentum. 0003

You can find the link to it down below the video. 0011

Taking a look at Number 1, we have a 1.2 kg block and a 1.8 kg block initially at rest on a frictionless horizontal surface. 0015

When a compressed spring between the blocks is released, the 1.8 kg block moves to the right at 2 m/s as shown. 0023

What is the speed of the 1.2 kg block after the spring is released?0030

The way I like to solve these is with momentum tables, so let us call this block (A) and that one block (B) and as we list our objects we have (A), (B), and our total momentum. 0036

Up here we will have our initial momentum, momentum before the collision or event, and momentum after. 0048

Initially, they are both at rest, so the initial momentum, the momentum before for (A) and (B) is 0, so the total is 0. 0060

Now after the event, (A) -- well what do we know? 0066

We are looking for the speed of the 1.2 kg block after the spring is released, so its momentum after mass × velocity will be 1.2 × v, and we are looking for that (v).0070

The momentum of (B) is going to be 2 m/s × 1.8 kg, which is going to be 3.6 kg-m/s, so the total momentum after is going to be 1.2 v + 3.6. 0080

The conservation of momentum says that these must be equal, so if 0 = 1.2 v + 3.6, that means -3.6 = 1.2 v, or V = -3.6/1.2, therefore V = -3 m/s... 0097

...where that just means the negative is it is in the opposite direction of what we called positive previously, so the correct answer -- What is its speed? -- 3 m/s.0116

Taking a look at Number 2 -- We have an 8 kg ball fired horizontally from a 1,000 kg cannon initially at rest. 0128

After it has been fired, the momentum of the ball is 2400 kg-m/s east. 0137

Find the magnitude of the cannon's velocity after the ball is fired. 0142

I am going to do this again by making a momentum table. 0146

Our objects go down on the left -- we have a ball, we have a cannon, and we will make a line here or a row for total.0149

We will write down the momentum before the event and after the event. 0158

All right, so initially, the cannon and the ball are both at rest, so their initial momentum is going to be 0 and the total therefore is going to be 0. 0166

Now after it is fired, the momentum of the ball is 2400 kg-m/s east. We will call east the positive direction. 0178

Find the magnitude of the cannon's velocity. 0186

After the event, the cannon has a mass of 1,000 kg and it has some unknown velocity (v), so our total here for momentum after will be 2400 + 1,000 v. 0189

The law of conservation of momentum says that these two must be equal -- the total momentum before and the total momentum after.0202

So all I have to do is solve that equation, 0 = 2400 + 1,000 v or -2400 = 1,000 v, which implies that V = -2400/1,000 or -2.4 m/s, so the magnitude of the cannon's velocity will be 2.4 m/s. 0211

In the next question it is going to ask us to identify the direction of the cannon's velocity. 0234

Well, if the ball went east, the cannon of course is going to go west. 0239

Moving on to Number 4 -- ball (A) of mass 5 kg moving at 20 m/s collides with ball (B) of unknown mass moving at 10 m/s in the same direction. 0249

After the collision, ball (A) moves at 10 m/s and ball (B) at 15 m/s, both still in the same direction. 0259

Find the mass of ball (B). Let us see if we cannot do this with the momentum table again. 0266

We have (A), (B), and total and we have our momentum before and our momentum after. 0272

Now, initially ball (A) has a mass of 5 and it is moving at 20 m/s, so its momentum is going to be 20 × 5 or 100. 0286

(B) has some unknown mass moving at 10 m/s in the same direction, so this will be 10 m for its initial momentum, so our total initial momentum will be 100 + 10 m. 0296

Now after the collision, ball (A) moves at 10 m/s and its mass is still 5, so its momentum after must be 50 and ball (B) is moving at 15 m/s, so that will be 15 × (m) for its momentum after the collision. 0310

The total here then will be 50 + 15 m and again the law of conservation of momentum says that these two must be equal, so all we have to do now is solve that. 0327

If I subtract 50 from both sides, I end up with 50 + 10 m = 15 m. 0340

If I subtract 10 m from both sides and I have that 50 = 5 m or therefore the mass must be equal to 10 kg -- Answer Number 3. 0351

Number 5 -- In the diagram below scale vectors represent the momentum of each of two masses (A) and (B) sliding toward each other on a frictionless, horizontal surface. 0367

Which scale vector best represents the momentum of the system after the masses collide? 0378

Well, by conservation of momentum, you must have the same total momentum at any point unless you have an outside force. 0384

If you have a long momentum vector to the right and the little one to the left, when we add them together we are going to get something like that. 0392

That has to be the same after the collision, so I would say our best answer is whatever that length is minus that little length, which is going to be answer Number 2. 0402

Number 6 -- At the circus, a 100 kg clown is fired 15 m/s from a 500 kg cannon. 0415

What is the recoil speed of the cannon? 0423

Here we have a clown, a cannon and we will make our row for total. 0427

We have momentum before and momentum after. 0436

Now, initially, the clown is at rest, the cannon is at rest, so our total is at rest. 0444

Now, after the event, the clown has a momentum of 100 × 15 or 1500 kg-m/s. 0451

The cannon has a momentum of 500 times whatever its velocity is, so our total momentum after the collision is 1500 + 500 v. 0461

Now again using conservation of momentum, we can state that the total momentum before must be equal to the total momentum after. 0472

Now to solve that, if I subtract 1500 from both sides, -1500 = 500 v, and divide both sides by 500 to find that V = -3 m/s, where that just means the cannon is going in the opposite direction of the clown. 0480

So what is the recoil speed of the cannon? 3 m/s. 0497

Just one more here -- A woman with a horizontal velocity (v1) jumps off a dock into a stationary boat. 0505

After landing in the boat, the woman in the boat moves with velocity (v2). 0513

Compared to velocity (v1), velocity 2 must -- well let us think about it. 0517

We have a woman who has a horizontal velocity (v1) jumping into a stationary boat. 0522

After she lands in the boat, the woman and the boat, both move with some velocity (v2). 0528

Well because the woman initially had some momentum, when she jumps in the boat, the mass goes up, so the velocity must go down by conservation of momentum. 0539

We will have the same direction, but the velocity must be smaller, so I have a smaller magnitude in the same direction. 0549

Hopefully, those went pretty well for you. 0560

If they did not, go back and check out the conservation of momentum lecture and if they did -- Terrific -- you are in good shape, so let us keep moving on. 0563

Thanks so much for your time everybody. Make it a great day! 0570