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Intro to Conservation of Momentum | 20 mins | 0 completed | Learn |

Push-Away Problems | 29 mins | 0 completed | Learn |

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Newton's Second Law and Momentum | 11 mins | 0 completed | Learn |

Momentum & Impulse in 2D | 25 mins | 0 completed | Learn |

Push-Away Problems With Energy | 12 mins | 0 completed | Learn |

Elastic Collisions | 8 mins | 0 completed | Learn |

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Example #1: Push-Away With Energy

**Transcript**

Hey guys so a few push away problems may also ask you to solve for energy at some point now this is pretty straightforward but we haven't done one yet so let me show you how it happens how it works. So here for example I have a bomb that exploded in the air into two fragments, OK? So this is a classic type of push away problem where you have an object that has two parts and it breaks apart much like a bomb explosion which is the case here one fragment's going to go to one side, the other fragments is going to go the opposite way so this is a push away where 2 parts of the bomb are pushing away from each other as a result of the explosion so I'm going to draw the two parts of the bomb and let's call this part 1 and let's call this part 2, one fragments 3 kilograms in mass and moves with 100 meters per second now it says that its positive so I'm going to assume that positive is to the right so let's say it's this side here, 100....I'm sorry that's 3 kilograms and it's going to move this way with 100meters per second after the explosion, the other fragment is 4 kilograms and it would presumably move to the left it has to move to the left if one moves to the right the other moves to the left but we don't know that velocity so we're going to call it V1 final and that's what we want to know, if the second fragment is 4 kilograms in mass what velocity will it attain, OK? We're going to use conservation of momentum to do this, the big equation here and M1V1+M2V2=M1V1 final+M2V2 final, we use conservation of momentum because we have two parts of an object that are pushing away from each other that's a push of a problem so we use this, let's put the masses I have 4 and 3, notice that there is no references to the initial velocity of this thing it is in the air but it doesn't say if it's moving left or right so you assume that it's not moving otherwise you have to make up a velocity and that's not cool you can't just make up numbers, so we'll assume that it's not moving therefore the initial velocity is 0, right? If it's falling that's fine but it can't be moving left or right it doesn't tell you that so you have to assume it's 0, after the explosion the 3 kilogram moves to the right with a 100 and the 4 is what we want to know so V1final is what we're looking for so let's solve for that, this is 0 and 0 for V1final + 300 and I'm going to move everything out of the way, we get -300/4 is our Vfinal so that's -75 meters per second, I got a negative which means it's going to the left and it makes sense if one fragment explodes the right the other one goes to the left so there's nothing new here, the part that's new is this idea of asking how much energy was stored in the bomb, OK? Now here we have assumed that energy is conserved because it doesn't tell us otherwise so we assume that the energy is conserved so the idea is that the initial energy this is before the explosion equals the final energy which is after the explosion, now again initially it's not moving so it only has potential energy, and at the end it is moving so you have kinetic energy the idea is that the energy stored in is your potential energy, right? you can think of it as your potential energy before the explosion and that's how bombs work it's got some stored energy in there that turns into an explosion and then all of the potential energy becomes kinetic energy that's how all these problems work when it's asking for the store energy in a bomb so all you have to do is solve for K final, 1/2 M1V1 final squared + 1/2 M2V2 final squared, that's it we're going to plug in numbers and we're done, so this is 1/2 (4)(-75 squared) + 1/2(3)(100 squared), those are all the numbers and if you plug everything you get 26250 Joules that is the store energy so the idea is that by knowing what happens at the end of the explosion how fast each piece is moving you can sort of work backwards and deduce and calculate what the initial energy of the explosion was or the stored energy, OK? Stored energy potential energy whatever, cool? Another question another way that this could have been asked it could have been asked as how much work is done during the explosion, it would have been the same thing so the work done during the during the explosion is 26250, in this case the idea is that we would be thinking this is like a separate way we would be thinking of the explosion as work instead of potential energy, OK? But the basic idea is that whatever initial energy will be equal whether you call it stored energy whether you call work it work it will be just kinetic final which is kinetic final of the two parts, however this thing gets called. Hope that makes if you have any questions let me know.

Practice: Two blocks (3 kg and 4 kg) on a smooth floor are pressed against a light spring (force constant 800 N/m) between them. When the blocks are released, the 3 kg is launched with 10 m/s.

(a) What speed is the 4 kg launched with?

(b) How much was the spring compressed by before the blocks were released?

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Example #1: Push-Away With Energy

Practice #1: Push-Away With Energy

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