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Elastic and Inelastic Collisions

  1. Jan 2, 2014 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/c5pOIe8.png & http://i.imgur.com/zdxfXLE.png

    Sitting in a parking lot you witness a minor, but still noteworthy collision between two cars. You wondered, "Is momentum really conserved in collisions?" It didn't seem like it was in the real world so you made the hypothesis, "Momentum is not conserved in all collisions. Collisions with less elasticity will lose more momentum than collisions with higher elasticity."

    To test your hypothesis you used a simulation to gather data for two colliding objects. You looked at both linear (straight line) and nonlinear collisions for three different values of elasticity; 100% elastic, 50% elastic, and 0% elastic. You are to use the data as your basis to show the degree of momentum conservation and whether your hypothesis is validated or not.

    2 & 3
    Any help at all would be appreciated! I have been stuck on this for over a day now and it's starting to get me a little flustered. I mean, I know how to figure out the final velocity of entangled masses or just one object after a collision, but how do I do it with this data? I'm really lost. What does Vx, Vy mean and how do we use distance to determine if momentum is conserved? What does 100% or 50% elasticity mean? I thought there were only Perfectly Inelastic, Elastic or in between, but I never found a way to measure the elasticity?
    I thought that I could make use of these equations;
    p = mv ; FΔt = mvf - mvi ; Δx = 1/2(vi + vf)Δt ; m1V1i + m2v2i = (m1 + m2)vf ; 1/2(m1V1i)^2 + 1/2(m2V2i)^2 = 1/2(m1V1f)^2 + 1/2(m2V2f)^2
    But as I said, I'm not sure how to interpret the data.. I'm not really sure what to write down either. Do I just do equations for each one?
    Any help would be so very appreciated. Thanks in advance.
     
  2. jcsd
  3. Jan 2, 2014 #2
    You do not have to worry about "elasticity". All you have to do is test how well theoretical conservation of momentum agrees with the experiment. Compute total momentum before and after from the data. Is it conserved? How big is the discrepancy?
     
  4. Jan 2, 2014 #3

    haruspex

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    They're just the velocity components in the two directions. X is left to right, Y is from bottom to top (in the diagrams).
    You are not given any distances.
    You have enough information to calculate the momentum in the x direction and the momentum in the y direction, before and after, in each case.
     
  5. Jan 2, 2014 #4
    But for the 1st two, there is "position (m)" and it has 1.00, 2.00, and 1.55, and 2.16, respectively.
    What am I supposed to do with that?

    And how do I use that data set? Does y mean North / South, and how would I interpret it? How would I compute the momentum with two velocities though? It has Vx AND Vy. There are two values for the velocity, when there should only be one. This is where I'm stuck at, I don't know what to input for initial speed, final velocity, etc. I'm fairly confused right now. I really do appreciate your help.
     
  6. Jan 2, 2014 #5
    I get that, but how am I supposed to compute total momentum with two velocities? It has Vx AND Vy. There are two values for the velocity, when there should only be one I thought. This is where I'm stuck at, I don't know what to input for initial speed, final velocity, etc. I'm fairly confused right now. I really do appreciate your help.
     
  7. Jan 2, 2014 #6
    Recall that the velocity is vector. So is momentum. They have components.
     
  8. Jan 2, 2014 #7
    Yes, I know that, but which formula would I use? Would I use the formula to find ΔKE?
    and wouldn't I have to use 1/2(m1V1i)^2 + 1/2(m2V2i)^2 & 1/2(m1V1f)^2 + 1/2(m2V2f)^2?
    But what would I use for v1i, v2i, v2f, v1f? Unless I'm just thinking incorrectly.
     
  9. Jan 2, 2014 #8
    Why would KE be relevant here? The question is about momentum?
     
  10. Jan 2, 2014 #9
    Because my instructor gave me sample problems that have me check work to make sure no energy was lost. I'm sorry I'm very confused.
     
  11. Jan 2, 2014 #10

    haruspex

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    Ok, I don't know why they give you those position values. I would just ignore them. In that first set up, it's all in a straight line, so the Vy values are zero.
    Yes, you can think of the y direction as NS and the x direction as EW.
    They are the components of the velocity in the two given directions. If you don't understand the principle of expressing a vector in its components then you need to go back and revise some earlier work.
    Since momentum is also a vector, that too can be resolved into its x and y components. The conservation law then applies to each component set independently, giving you one equation for momentum conservation in the x direction and another for its conservation in the y direction.
     
  12. Jan 3, 2014 #11
    Alright, thanks.
    That second comment though, it almost sounded demeaning. I'm really sorry I'm stuck on this. I don't know why I'm in such a roadblock. I've been doing so well in this class, now I'm stuck and I'm not sure why. I think I'm just over analyzing things.
    So would we just use those two value for the formula 1/2(m1V1i)^2 + 1/2(m2V2i)^2 = 1/2(m1V1f)^2 + 1/2(m2V2f)^2?
    So for the first one; Before m1 = 0.5 kg, m2 = 1.5 kg, v1i = 1 m/s to the right, v2i = 0, v1f = -0.5 m/s to the left, v2i = 0.5 m/s right?
    Is that what I'm supposed to be doing? :-)
    That's really all I was trying to get clarified. I'm not sure which formula to use or how to interpret the variables into the functions.

    P.S. For the one's after, am I supposed to use the given values for velocity in the formula: V = (Vx^2 + Vy^2)? And then use the result to input into the initial velocity / final velocity?
     
  13. Jan 3, 2014 #12

    haruspex

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    I certainly don't mean to put you down, but from your questions it does appear that you are not familiar with the concept of a velocity having components in different co-ordinate directions:
    Do you understand how to add vectors? Do you understand that a vector can be treated as equivalent to a set of vectors which add up to it? In particular, given a co-ordinate system, you can express a vector as a sum of vectors, one along each co-ordinate axis? If you have not absorbed these notions then answering this problem is going to be a struggle.
    There is no such formula that I'm aware of. If you, mean 1/2 m1(V1i)^2 + 1/2 m2(V2i)^2 = 1/2 m1(V1f)^2 + 1/2 m2(V2f)^2, that's an energy equation. As has already been mentioned, you should forget about energy here and concentrate on momentum.
    Given the components of the velocity and the mass, do you know how to write down the components of the momentum?
     
  14. Jan 4, 2014 #13
    Yes I believe I do. And yes. I meant that equation..
    It's just that my instructor hasn't necessarily been the best help. In all honesty, I find at times that he gives us assignments that weren't covered yet or that are misleading. That's the formula he gave us to solve these. Would you have to use P = mv? Or m1(V1i) + m2(V2i) = m1(V1f) + m2(V2f)?

    You figure out velocity with this equation right: V = (Vx^2 + Vy^2)?
     
  15. Jan 4, 2014 #14
    This is better, if this really means ##m_1 \vec V_{1i} + m_2 \vec V_{2i} = m_1 \vec V_{1f} + m_2 \vec V_{2f} ##. Note the arrows on V's, those are vectors.

    No. It would be better written as ## V = \sqrt {V_x^2 + V_y^2 } ##, but that would give you speed. The velocity is already given by the components ##V_x## and ##V_y##, you do not need any other computations to obtain it.
     
  16. Jan 4, 2014 #15

    haruspex

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    ... or, if it means m1(V1i) + m2(V2i) = m1(V1f) + m2(V2f) as a scalar equation applied separately in the x and y directions:
    ##m_1V_{1ix} + m_2V_{2ix} = m_1V_{1fx} + m_2V_{2fx}##
    ##m_1V_{1iy} + m_2V_{2iy} = m_1V_{1fy} + m_2V_{2fy}##
     
  17. Jan 4, 2014 #16
    Yes! That's exactly what I meant!
    So would I just have to input the values and compare both results.
    That easy?
    I knew it wouldn't be difficult, I was just unsure of which formula or equation to use to figure this out.
    Thanks for the help you two!
     
  18. Jan 4, 2014 #17

    haruspex

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    In principle, yes. There is one small catch. The numbers are unlikely to come out perfectly, since the given data only have a few significant figures. You need a way to decide whether the data are nevertheless consistent with conservation of momentum.
     
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