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How to use the shortcut for solving elastic collisions

In this video, David solves an example elastic collision problem to find the final velocities using the easier/shortcut approach. Created by David SantoPietro.

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  • aqualine ultimate style avatar for user George Balderas
    so V1i + v1f = V2i + V2f only used for Elastic Collision?
    (10 votes)
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  • mr pink red style avatar for user Akash Suryavanshi
    What if there is elastic collision in two dimension?
    Will we take components and then apply the formula?
    (5 votes)
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    • leaf green style avatar for user Mark Zwald
      Yes, apply conservation of momentum for each dimension separately. That means breaking up the momentum vectors into components parallel to each dimension.
      If the collision is elastic, kinetic energy will also remain unchanged before and after the collision, but for inelastic collisions, this is not true. Momentum however will always be conserved for both elastic and inelastic collisions.
      (6 votes)
  • blobby green style avatar for user S26Patel
    Dear Tutors:
    Please explain how the golf ball gained speed after the collision? Initial velocity of the golf ball was 50 m/s. Final velocity of golf ball was calculated to be 51.36 m/s. Is this an error?
    (4 votes)
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    • starky ultimate style avatar for user Erfan Shoaie
      @S26Patel, as explained here by Andrew M here above, it means in other words that the energy of the objects is transferred to one another. The law of conservation of energy says that energy is never really 'lost', it only is transferred to another object or another form of energy. Either within the system of the problem or to outside of the problem, but that is another story.

      Basicly this means that a huge mass(M1) with high velocity (and thus high energy) when colliding with a way smaller mass(M2) with a certain velocity, M1 will transfer some of it's energy to M2. And because E_kin = 1/2 * mass * velocity^2; this means that M2 will gain more velocity because it has less mass. If the exactly same amount of velocity lost by M1 would be gained by M2, there must have been some energy transferred into heat or something else.
      (2 votes)
  • purple pi purple style avatar for user Francis Hopkins
    can you solve of either velocity originally or should you pick a certain one to make it easier? Say if one object was at rest already would it be easier to solve for it's velocity since it's initial KE is 0?
    (2 votes)
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  • orange juice squid orange style avatar for user ME_Matt
    So am I correct in assuming that V1ix + V1fx = V2ix + V2fx and the same for any other dimension if I needed to decompose velocity vectors in a higher dimensional problem?
    (2 votes)
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    • female robot ada style avatar for user Rodrigo Campos
      The best way to start a collision problem in 2 or 3 dimensions (3 dimensional collisions problems are very rare) is to select an appropriate set of coordinates. If you have two dots or spheres colliding, which is almost always the case, then the best coordinates will have an axis connecting their centres at the moment of collision, and an axis perpendicular to that. This system will give you the easiest equations. In one of the axis, the one connecting the centres, the solution is exactly like a one dimensional collision, and the other 2 axis show no change in each of the objects' velocity (if there's no friction).
      (3 votes)
  • hopper cool style avatar for user JPhilip
    Is there a way of roughly estimating how much final velocity is lost, for example, by heat and sound? How would I deduct this from my final answer?
    (2 votes)
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  • blobby green style avatar for user Rohit G
    is there any video explaining coeffeicient of restitution
    (1 vote)
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  • aqualine ultimate style avatar for user carefarrar
    what happened to the kilograms when he multiplied it
    (1 vote)
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  • blobby green style avatar for user Luis Rodriguez
    Hello! I had an exam today and the first question was asking to find the INITIAL and FINAL velocities of the second ball, when the first ball's initial and final velocity values were given. How would one go about solving this problem?

    Ball 1: m=3.0kg
    vi= 6.0m/s
    vf= 0.5m/s

    ELASTIC COLLISION

    Ball 2: m= 2.0kg
    Vi= ?
    Vf= ?
    (1 vote)
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    • male robot hal style avatar for user Anirudh
      >Let Ball 1(A) have a mass(m) of 3kg velocity of vi m/s initially and vf m/s finally in the +x direction.
      >Let Ball 2(B) have a mass(M) of 2kg velocity of Vi m/s initially and Vf m/s finally in the +x direction.

      So, applying the conservation of linear momentum:
      => m(vi)+M(Vi)=m(vf)+M(Vf)
      OR:
      => 3(6)+2Vi=3(1/2)+2Vf
      => 18+2Vi=(3+4Vf)/2
      => 36+4Vi=3+4Vf
      => 4Vf-4Vi=33 ...(i)

      Now, using the formula for the coefficient of restitution:
      e=(v2-v1)/(u1-u2)
      OR:
      (Vf-0.5)/(6-Vi)=1 [Elastic Collision]
      => Vf-(1/2)=6-Vi
      => 2Vf-1=12-2Vi
      => 2Vf+2Vi=13
      => 4Vf+4Vi=26 ...(ii)

      Adding equations (i) and (ii):
      =>8Vf=59 =>Vf=7.375m/s
      Putting Vf in (ii):
      =>4*7.375+4Vi=26
      =>Vi=-3.5m/s
      I hope it helped!
      (2 votes)
  • duskpin sapling style avatar for user Alina Chen
    How would you solve a problem that involves external forces (like gravity and such during points of contact) in cases where it is not negligible?
    (1 vote)
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Video transcript

- [Narrator] Alright, here's pretty much the fastest way you can solve one of these elastic collision problems, when you don't know two of the velocities. In this case we don't know the final velocities. We know the initial velocity of the tennis ball and its mass. We know the initial velocity of the golf ball and its mass, but we don't know the final velocities of either ball, and the trick to make these calculations go faster for an elastic collision is to use this equation, which says the initial velocity of one of the objects before the collision, plus the final velocity of that same object after the collision should equal, if it's an elastic collision, it'll equal the initial velocity of the second object before the collision plus the final velocity of that second object after the collision. If you want to see where this comes from, we derived it in the previous video. But now that we know it, we can just use it for any elastic collision. So this expression right here only holds for elastic collisions. But you should love this formula right here, because what this does is it allows you to avoid having to use conservation of energy, and it avoids having to square terms and create horrible messes of algebra. This formula's gonna be much cleaner, much nicer. Let's see how to use it. Let's just say object one is the tennis ball and object two here is the golf ball. So in that case, the initial velocity of the tennis ball, that'd be 40, and it'd be positive 40. I'm gonna write positive just so I know that's to the right, and it should be positive. Plus the final velocity of the tennis ball, I'll write that as Vt. So instead of writing one, I'm gonna get confused which one was number one, so I'll write a Vt. That way I get to know which object's velocity I'm talking about. So Vt final, that's what Vt final's gonna mean, final velocity of the tennis ball. And that should equal the initial velocity of the second object, our second object is the golf ball. The initial velocity of the second object, of our golf ball, is not 50. It's negative 50, you've gotta be careful. These are velocities in this formula. So if you've got a velocity, that's directed in the negative direction, you better make it negative. And if you solve in here, you might get a negative, so these are vector values up here. You gotta plug them in with the proper sign. So this initial velocity of the golf ball would be negative 50 meters per second, 'cause we're gonna assume leftward is negative, and rightward is positive. Plus the final velocity of the second object, second object is our golf ball. I'll call this Vg instead of V2. Vg will be V of the golf ball and then f for final, i.e. after the collision. So we can solve this for Vt final. I can subtract 40 meters per second from both sides and I get that the final velocity of the tennis ball is gonna equal, I'll have this Vg final just sittin' over here, final velocity of the golf ball after the collision. And then negative 50 minus a 40 is gonna be negative 90 meters per second. So this formula alone was not enough, 'cause I've still got two unknowns. I can't solve for either. I've gotta use another equation, and the other equation we're gonna use is conservation of momentum. 'Cause during this collision, the momentum should be conserved, assuming the collision happens so fast, any net external impulse is negligible. So we can say that the total initial momentum is equal to the total final momentum. We basically do this for every single collision because we make that assumption, that the net external impulse during this collision is gonna be small. That means the momentum should be conserved. So the formula for momentum is mass times velocity. So the momentum of this tennis ball initially is the mass of the tennis ball, .058, times the initial velocity, would be positive 40. Positive because it's directed to the right, and I'm gonna consider rightwards positive. Plus the initial momentum of the golf ball would be .045 times the initial velocity of the golf ball and that'd be negative 50. Again, you have to be careful with the negative signs. Momentum is also a vector, so if these velocities are ever negative, you've got to plug them in with their negative sign. And that initial momentum should equal the final momentum. So the final momentum of the tennis ball is gonna be 0.058 times our final velocity of the tennis ball. And I'm gonna use the same nomenclature. I'm gonna use the same symbol over here that I used over here. This Vt final, final velocity of the tennis ball is the same as this Vt final, final velocity of the tennis ball, plus .045 times same thing, final velocity of the golf ball. I'll use the same symbol which is gonna be Vg final. I could multiply out this entire left hand side, and what I get is 0.07 kilogram meters per second. And that equals on the right hand side, this entire expression right here. I'll just copy that. So we've got two unknowns in this equation as well, so we can't solve this directly or either of the final velocities. But we do have two equations and two unknowns now. And whenever you have that situation, you can solve one of the equations for one of the variables and plug that expression into the other equation. In other words, I know that the final velocity of the tennis ball is equal to the final velocity of the golf ball minus 90, so I can take this entire term right here, since it's equal to Vt final, and just plug that in for Vt final. And what that would do for me is give me one expression, all in terms of the final velocity of the golf ball. So let's do that. We've still got 0.07 kilogram meters per second on the left. And that's gonna equal 0.058, that's still here, kilograms, times Vt final. I'm plugging in this entire expression for Vt final. So that gets multiplied by Vg final, the final velocity of the golf ball, minus 90 meters per second. That was the term I plugged in for Vt final, and it got multiplied by this mass here, so I can't forget about that mass. And then I still have to add this final momentum of the golf ball, .045 kilograms times the final velocity of the golf ball. So at this point you might be feeling ripped off. You might be like, easy way to do this? This isn't easy, this is hard. I've gotta plug one equation into another, and then solve? Well, this easy approach does not avoid having to plug one equation into the other, that's true. But the reason that it's easy is because the equations that we're plugging into each other are a whole lot simpler than the kinetic energy formula that you would have to use if you didn't know this expression here. Because we have this one, we do not have to plug conservation of momentum into conservation of energy. That would square the term we put in, that would get nasty, the algebra would be a lot worse. These formulas that we're dealing with in this process only have velocity. None of these velocities are squared, so the algebra doesn't get nearly as bad. And we're actually almost done over here. Let me show you how close we are to finishing this thing. I just need to multiply through this term right here. So what am I going to get, this left hand side stays the same. And then when I multiply through .058, I'm gonna get 0.058 times the final velocity of the golf ball, and then negative 90 times .058 is negative 5.22, and that would be units of kilogram meters per second. And I still just have this term right here, so I'll copy that one. And now here's a key step. The whole reason we plugged one formula into the other is so that we had the same unknown in that formula. Only one unknown, there's only one unknown variable in here, which is the final velocity of the golf ball. It's located in two spots, but at least it's the same variable. And what that allows us to do is to combine these terms. If on this right hand side I have this much of the final velocity of the golf ball, and that much of the final velocity of the golf ball, when I add them up, I can just add these two factors out front. In other words I'll have 0.07 equals, I can rewrite this as 0.058 kilograms plus 0.045 kilograms times the final velocity of the golf ball. And then I still have this minus 5.22. So if that looked like mathematical witchcraft, all I did was I combined the terms that had Vg final. Cause if you had A times Vg final plug B times Vg final, that's the same as A plus B, times Vg final. When you multiply this through, you'll just get both of these terms back again. So we keep going, don't divide by this first, sometimes people try to divide by this whole term right now. You don't want to do that, you gotta go in the right order. I need to add this 5.22 to both sides to get rid of it first. So when I do that, when I add 5.22 to both sides it'll cancel this term, and on the left hand side I'll get 5.29 kilogram meters per second. And that's gonna equal, if I add these two terms together, if I just add .058 and .045, I get 0.103 kilograms times the final velocity of the golf ball. And let me move these initial velocities down so we can see. At this point, if I divide both sides by .103, I'll get 51.36 meters per second on the left hand side. And that equals the final velocity of the golf ball. So we did it, we found one of the final velocities of these objects. We found the final velocity of the golf ball. But what about the final velocity of the tennis ball? How do we figure out what that is? 'Cause that's also an unknown in this problem. Well that one's really easy. Now that we know the final velocity of the golf ball, I can just take that value, plug it right back into here, this expression that we plugged into conservation of momentum. And I can figure out what the velocity of the tennis ball is. So in other words, the velocity of the tennis ball I know, finally, should equal the final velocity of the golf ball, which was 51.36 meters per second, and then minus 90 meters per second, and I'll get the final velocity of the tennis ball was negative 38.64 meters per second. And it came out to be negative, that means that this tennis ball got deflected backwards. It was heading leftward, 38.64 meters per second after the collision. So recapping, we used this nice formula to get one equation that involved the velocities that we didn't know for an elastic collision, which you can only use for an elastic collision. If you want to see where this equation comes from, we derived it in the last video. We solved for one of the velocities and plugged it into conservation of momentum. We combined the like unknowns, and solved algebraically to get one of the final unknown velocities. Then we plugged that back into the first equation to get the other unknown velocity. And this is easier than the alternative, because the alternative involves kinetic energies, which means when you take one of these expressions, you'd have to square them, and the algebra would be significantly more difficult.