WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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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.
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You can find the link to it down below the video.
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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.
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When a compressed spring between the blocks is released, the 1.8 kg block moves to the right at 2 m/s as shown.
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What is the speed of the 1.2 kg block after the spring is released?
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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.
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Up here we will have our initial momentum, momentum before the collision or event, and momentum after.
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Initially, they are both at rest, so the initial momentum, the momentum before for (A) and (B) is 0, so the total is 0.
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Now after the event, (A) -- well what do we know?
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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).
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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.
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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...
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...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.
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Taking a look at Number 2 -- We have an 8 kg ball fired horizontally from a 1,000 kg cannon initially at rest.
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After it has been fired, the momentum of the ball is 2400 kg-m/s east.
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Find the magnitude of the cannon's velocity after the ball is fired.
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I am going to do this again by making a momentum table.
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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.
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We will write down the momentum before the event and after the event.
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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.
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Now after it is fired, the momentum of the ball is 2400 kg-m/s east. We will call east the positive direction.
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Find the magnitude of the cannon's velocity.
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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.
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The law of conservation of momentum says that these two must be equal -- the total momentum before and the total momentum after.
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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.
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In the next question it is going to ask us to identify the direction of the cannon's velocity.
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Well, if the ball went east, the cannon of course is going to go west.
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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.
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After the collision, ball (A) moves at 10 m/s and ball (B) at 15 m/s, both still in the same direction.
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Find the mass of ball (B). Let us see if we cannot do this with the momentum table again.
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We have (A), (B), and total and we have our momentum before and our momentum after.
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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.
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(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.
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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.
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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.
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If I subtract 50 from both sides, I end up with 50 + 10 m = 15 m.
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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.
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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.
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Which scale vector best represents the momentum of the system after the masses collide?
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Well, by conservation of momentum, you must have the same total momentum at any point unless you have an outside force.
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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.
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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.
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Number 6 -- At the circus, a 100 kg clown is fired 15 m/s from a 500 kg cannon.
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What is the recoil speed of the cannon?
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Here we have a clown, a cannon and we will make our row for total.
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We have momentum before and momentum after.
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Now, initially, the clown is at rest, the cannon is at rest, so our total is at rest.
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Now, after the event, the clown has a momentum of 100 × 15 or 1500 kg-m/s.
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The cannon has a momentum of 500 times whatever its velocity is, so our total momentum after the collision is 1500 + 500 v.
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Now again using conservation of momentum, we can state that the total momentum before must be equal to the total momentum after.
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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.
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So what is the recoil speed of the cannon? 3 m/s.
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Just one more here -- A woman with a horizontal velocity (v1) jumps off a dock into a stationary boat.
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After landing in the boat, the woman in the boat moves with velocity (v2).
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Compared to velocity (v1), velocity 2 must -- well let us think about it.
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We have a woman who has a horizontal velocity (v1) jumping into a stationary boat.
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After she lands in the boat, the woman and the boat, both move with some velocity (v2).
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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.
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We will have the same direction, but the velocity must be smaller, so I have a smaller magnitude in the same direction.
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Hopefully, those went pretty well for you.
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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.
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Thanks so much for your time everybody. Make it a great day!