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Conservation of momentum & energy

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Good evening phsyicsforum. I am having a lot of trouble finding out what must be true for me to use conservation of momentum or energy (sometimes both) in problems involving electricity. Below I will post 2 different questions and what they conservation they use.

    1. Find the initial velocity of an alpha particle with a mass of 6.64 x 10^-27 kg and a charge of +3.2 x 10^-19 C, if it undergoes a head-on "collision" with a gold nucleus. You may assume the gold nucleus does not move at all during the interaction. The charge on the gold nucleus is +2.53 x 10^-17 C and the distance of closest approach between the two is 4.7 x 10^-15

    2. An alpha particle moving at 3.0 x 10^6 ms [east] (m2= 6.64 x 10^-27 kg and q2= +3.2 x 10^-19 C) is headed directly towards a proton moving at 5.0 x 10^6 m/s [west] (m1= 1.67 x 10^-27 kg, q1 = 1.6 x 10^-19 C). Find the distance of closest approach assuming that they start from a very far apart position.



    2. Relevant equations
    Pto=Ptf

    Eto=Etf


    3. The attempt at a solution

    Now I don't want any help answering the question itself because that is quite easy. I am only having trouble deciding when it is true to use conservation of momentum or energy or both.

    Do I only use conservation of momentum when 2 objects are moving? And is conservation of energy used when only one object is moving and 1 is at rest?
     
  2. jcsd
  3. Feb 14, 2014 #2
    Generally, that's true. The conservation laws are of course always true, it's just a matter of whether the variables they open up allow you to solve for desired quantities. For example, in the first problem, conservation of momentum is true and can be used, but it's not going to get you anywhere. You can know that the initial momentum equals the final momentum, so initial velocity equals final velocity, and you don't know either. So in that case you would look to energy conservation.

    And there are of course some exceptions to those rules. Consider this situation: two blocks travel at different velocities along a frictionless surface. The block in front travels slower, and has a spring attached to one side of it, facing the other block, which is moving faster. To solve for things here, there's both momentum conservation and energy conservation.

    Most important I think is to just do a lot of problems, and you can start to feel which laws will be most helpful based on the scenario.
     
  4. Feb 14, 2014 #3
    Ok thanks a lot for your input jackarms. I have a question to ask back to you. Let's say for a question that didn't really require conservation of momentum and you could just use conservation of energy. If I solved using both conservation of momentum and energy, would it affect my answer? or would I get the same answer, just with a lot more steps?
     
  5. Feb 14, 2014 #4
    No, as long as you don't misinterpret anything in the problem, you'll never get two different answers with two logical sequences of steps. And in almost all cases problems can't be solved using either law, and it's much more common that both laws are required. The only case I can think of where either would work would be a problem where only kinetic energy is involved -- kinetic energy tends to be the link between energy and momentum. If you throw potential energy in the mix, that can only be explained by energy, except in special cases.

    Hope this has helped at least a bit. I'd say the best thing to do for any problem is to consider conservation of anything you can think of and see where it get you.
     
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