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Momentum and Elastic Collisions

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.06kg tennis ball, moving with a speed of 2.50 m/s, collides head-on with a 0.09kg ball initially moving away from it at a speed of 1.15m/s.Assuming a perfectly elastic collision,what are the speed and direction of each ball after the collision?


    2. Relevant equations
    Ok well I know P=mv
    Also my teacher gave me the equation: V1-V2=-(V1prime-V2prime)
    and kinetic energy along with momentum is conserved.


    3. The attempt at a solution
    What I tried was setting the second ball's velocity to 0 m/s meaning the first ball would have a velocity of 1.35 m/s instead of the original 2.50m/s and I used the equation V1prime=(m1-m2)/(m1+m2) x (v1), however the answer didn't come out. I have no idea how to start this problem and Im also wondering why setting one ball's velocity to zero wouldn't work? Please help!! Thank you and any help is greatly appreciated =D
     
  2. jcsd
  3. Dec 8, 2008 #2

    turin

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    extracted relevant info:
    0.06kg tennis ball, speed of 2.50 m/s
    collides head-on with
    0.09kg ball, moving away from it at 1.15m/s.
    perfectly elastic collision,

    question:
    what are the speed and direction of each ball after the collision?



    I agree that this is a relevant equation, but where did you use it?



    Did he/she really?! That is not good. I am sorry. Are you sure there isn't a typo(s) here?



    I agree that this is relevant, but where did you use it?



    Why?



    Where did you get this equation? It wasn't in your list. Did you derive or did your teacher give it to you also?



    You can't set it to whatever you want. The physics will determine both velocities.
     
  4. Dec 8, 2008 #3



    I found the momentum of the first ball and the momentum of the second ball, but there are two variables to be solved for.



    Yes he gave us this equation in the case that it is elastic and both objects are in motion.




    Im considering that I would need to use the conservation of momentum and conservations of kinetic energy as a system of equations. For instance (m1)(v1i)+(m2)(v2i)=(m1)(v1f)+(m2)(m2f) and (.5)(m1)(v1i^2)+(.5)(m2)(v2i^2)=(.5)(m1)(v1f^2)(.5)(m2)(v2f^2)




    So that I can use the equation V1prime=(m1-m2)/(m1+m2) x (v1)



    Yes this was also given by my teacher. Along with two other equations he said that we can use this to solve for the velocity after the collision ONLY if it is elastic and one object is stationary. That is why I decided to set one ball's velocity to 0 so I can use this equation. By setting it to 0 I subtract 1.15m/s from both ball's velocity meaning the second ball's velocity is 0 then I can use the previous equation my teacher gave me.

    I'm sorry I dont really understand the concept of conservation of energy and momentum. Do you know of a simpler way to solve this problem? Also your reply/help is greatly appreciated!
     
  5. Dec 9, 2008 #4

    turin

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    Great. Yes. Use these equations. How's your algebra. These equations can seem a bit daunting if you are not comfortable with systems of equations. BTW, I'm assuming that the missing "+" on the RHS of the KE equation is just a typo.

    You need to answer (for yourself) three questions (now that you have decided what equations to use):
    - what variables in the equations do you know?
    - what variables in the equations are you trying to find?
    - what variables seem irrelevant?

    For all of the variables that seem irrelevant, you need to answer the followup question:
    - can I eliminate the irrelevant variable(s), and, if so, how?



    Here you have the opportunity to learn perhaps the most valuable lesson from phys 101. Even if you don't want to be a physicist, and even if you don't ever want to think about another physics problem for the rest of your life after finals, I think that it is still good that you are taking physics, and even better that you are reading this right now. Do you know why they make students take physics, even if that is not their major, and even if that is the last subject in the world that they care about. Because, they aren't really expected to learn physics, they are expected to learn problem solving skills.

    Your logical reasoning here is a big no no in problem solving. The equations are your tools, and the given information is the thing that you are working on. Imagine that your job was to open up a box that contains a delicate, valuable item, and this box is held together by screws. However, the first tool you pull out of your toolbox is a hammer, so you decide to ignore the screws and hammer the box open, because you want to use the hammer because it is the first tool you pulled out of your toolbox. You destroy the object inside, and so defeat the purpose of opening the box. This is like what you have done in this physics problem. You need to get to know your tools and critically assess which ones are best for the job.

    One hint is to compare the variables in the formulae that you know to the variables and given quantities in the problem. Then, try to use the most fundamental equations that are still practical for solving the problem. CONSERVATION OF MOMENTUM is definitely an important one, and very fundamental. CONSERVATION OF ENERGY is another biggy. Equations like the two that your instructor gave you are like specialty tools. They apply in a very limited category of problems, and, in my opinion, you shouldn't even try to use them until you have rigorously derived them yourself. That is my second hint. Deriving such specialty equations for yourself will help you immensely in deciding when they should be used. Furthermore, when you derive them, you are forced to use variables rather than numbers, and so you will more clearly appreciate the relationships among the physical quantities and how those relationships are stated precisely in the language of math.

    For test purposes, you must find your own personal balance between using specialty equations and deriving from fundamental laws. The advantage of using fundamental laws is that you don't need to remember many of them, and you are more likely to maintain a logically consistent solution. However, this is unrealistic for some problems, because the derivation could be pages and pages long. The advantage of specialty equations is that, if you do happen to know the one that you need, the solution comes in a single step, and the actual time to solve the problem is much shorter. By test time, you should already have a good idea what is your balance.
     
    Last edited: Dec 9, 2008
  6. Dec 9, 2008 #5
    Yes. Sorry that was a typo. However I am still a little unsure of how to solve systems of equations. Lemme try answering your questions:
    -I know the masses and initial velocities of both objects.
    -I am trying to find the final velocities of both objects.
    -I think the masses seem irrelevant because they cancel each other out?
    I am thinking that I can eliminate all the masses in that equation then I would end up with:
    (v1i)+(v2i)=(v1f)+(m2f) and (v1i^2)+(v2i^2)=(v1f^2)+(v2f^2)
    thus:
    (2.50m/s)+(1.15m/s)=(v1f)+(v2f) and (2.50m/s^2)+(1.15m/s^2)=(v1f^2)+(v2f^2)
    Then would I solve for one variable such as:
    v1f=(2.50m/s)+(1.15m/s)-(v2f)
    and plug that back into the second equation making it:
    (2.50m/s^2)+(1.15m/s^2)=(((2.50m/s)+(1.15m/s)-(v2f))^2)+(v2f^2)
    From here I simplified to:
    (7.5725m/s)=(3.65m/s-v2f)^2+(v2f^2)
    Then I got confused, would I expand the (3.65m/s-v2f)^2 or would i subtract it making it
    (7.5725m/s)-(3.65m/s-v2f)^2=(v2f^2)?? I guess im not too good at algebra. Am I on the correct track or is this completely off??

    Oh now I understand why these equations don't always work. hmm I understand what you are saying. Are you a physics professor because you explain it quite well. And again I thank you for all your help!!
     
  7. Dec 9, 2008 #6

    turin

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    I just want to make sure you're clear on two things, so that you don't get confused in other problems. The following two issues (paragraphs) don't apply to this problem.

    When I say that conservation of momentum and conservation of energy are fundamental, you should realize that the equations for these laws can, in general, carry a term that accounts for "violation". For momentum, you already know what this term is: an external force. For energy, you already know one of these kinds of terms: work. You will encounter another "energy violating" term later in thermodynamics (if your class gets there): heat. These aren't really violations of the conservation laws; they just seem like violations when you first encounter them. They are "flow" terms that quantify the amounts of the conserved quantities that "flow" into or out of the physical system. Or, they indicate the fact that the system you usually consider is not the entire universe, so they quantify how momentum and energy are shared with the rest of the universe. We still believe that momentum and energy are conserved in the entire universe as a whole.

    The second issue for you to realize is that the equation that you have for conservation of momentum is actually a specialty equation! However, it is very close in form to the most general version (without external force). In order to generalize, you simply put vector arrows on top of the four velocity variables, so that you really have three equations for momentum conservation: one for each of the three spatial directions. Then, you can think of your momentum equation as the x-component (or the y-component, or the z-component, or whatever is your favorite direction) of the general version, and it is OK. However, be careful when doing a 2-dimensional problem (like billiard balls, hocky pucks, and stuff). All you need for momentum is a carbon copy of that same equation for the y-components. But for kinetic energy, you must add the components in the same equation. The kinetic energy equation is not a vector equation, but, at the same time, it must account for all components of velocity.



    You are having a pesky little algebra problem that you can't see because you have been staring at it too long. You cancel masses in these equations by dividing them out. Do one at a time and see what happens. Make sure to divide everywhere that you need to. Hint: each of the two equations has four terms. Hint: there are two UNEQUAL masses. After you have convinced yourself that THE MASSES DON'T CANCEL, then convince yourself not to worry about it anyway, because you know the masses.

    Next, realize that YOU KNOW ALL OF THE QUANTITIES except the two final velocities. And, you have two equations: one for momentum and one for energy. Do you know how to handle two equations of two variables?



    No. I am a grad student who has been one semester away from my PhD for, let's just say, more than one semester. What that means is that I have taught my fair share of freshman physics. Grad students make much better teachers than profs because the pain of learning is very fresh in their memories. And believe me, almost all grad students, and even profs, of physics had almost all of the same problems understanding this stuff as you. Grad students just remember how they got past it better than profs.
     
  8. Dec 9, 2008 #7
    Ahh I see what you mean. I can solve for one of the variables and plug it back into the other equation then I would only have one variable instead of two different variables. Thank you!! I understand now!!


    Ahh thank you!! saying that even profs and grad students have the same problems made me feel a lot better! I'm extremely grateful for your guidance and help because today I took my test and I felt very confident in my work. Hopefully I did as well as I think I did! Again thank you so much!! =D
     
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