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Just an algebra question

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    And so here is the derivation, and i am quickly checking here to see if the algebra is fine.

    which is the same as
    divide by v from both sides
    then take away M from both sides
    [tex]M^2=\frac{k}{v} -1[/tex]
    take the square root
    [tex]M=\sqrt{\frac{k}{v} -1}[/tex]
    which is the same as
    so by rearranging
    [tex]M \sqrt{v}=\sqrt{k-v}[/tex]

    See any mistakes i should be aware of?
    Last edited: Jan 10, 2009
  2. jcsd
  3. Jan 10, 2009 #2
    by the way, equation 3 seems to be coming out wrong. Even though i have changed it, the right hand side should read, k/v
  4. Jan 10, 2009 #3
    The error is when distributing M with (v + 12). 12 is 1 so this gives M(v + 1) = Mv + M. So, solving for M does not, in this case, require distributing but rather dividing both sides by (v + 1).
  5. Jan 10, 2009 #4

    I thought this, thanks... but otherwise, the rest would have been fine?
  6. Jan 10, 2009 #5
    In other words, i could state k=M(M + v), and that would solve it yes?
  7. Jan 10, 2009 #6
    I'll just assume then it would be correct. There was more.

    Near enough where i last left this;
    Removing the radical sign
    And then multiply by v would give
    So now divide by 2 gives
    Rearranging gives, using again the notion that [tex]MMv^2=(M\sqrt{v})^2[/tex]

    But it is these last steps i was unsure about. Can you notice any visible mistakes for me to correct?
    Last edited: Jan 10, 2009
  8. Jan 10, 2009 #7
  9. Jan 10, 2009 #8
    I'm not sure what the objective of this question is, but your algebra is incorrect in Step 3 and step 4. What should the final answer look like? you want to find the kinetic energy of the system?

    As chrisk suggested, the easy way is to divide both sides by (v+1) in first step.

    Ultimately, this
    is wrong. Check your steps carefully.
  10. Jan 10, 2009 #9
    Could you explain how it is wrong, carefully, because i thought it was right...
  11. Jan 10, 2009 #10
    The reason why i never devided, by the way, both sides by (v+1), because i wanted to isolate it first.
  12. Jan 10, 2009 #11
    Isolate what?

    Dividing both sides by v does not give you
    and taking M away from both sides won't give you
    M^2=\frac{k}{v} -1
  13. Jan 10, 2009 #12
    Dividing by both sides, i thought would have removed the v in the Mv part. This is wrong i take it?

    Also, taking M from both sides, i was going to write on the right hand side k/v-M, but since M was gained from the 1 part in (v+1), i thought i could bring the value back... but i can't?
  14. Jan 10, 2009 #13
    Starting from
    Dividing by v from both sides:
    which simplifies to

    And starting from
    Taking M away from both sides:
    Which simplifies to
    M on the right hand side cannot be simplified to 1. The 1 part in (v+1) is lost when you re-wrote the first equation and expanded it.

    Make sure you fill in the middle step when you are doing the algebra (like I did above). This makes mistakes easier to spot.
  15. Jan 10, 2009 #14
    Oh i see now. I apologize, i wasn't thinking straight. Yes, that makes much more sense. Is the rest right?
  16. Jan 10, 2009 #15
    I have a question as well. In equation 3, the v vanishes from the left hand side. How did this happen?
  17. Jan 10, 2009 #16
    I'm assuming you're referring to the original post:

    That was your mistake. If you did the correct math, the v does not vanish from left hand side as I showed in my last post:
  18. Jan 10, 2009 #17


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    Homework Helper

    The error that Crisk found means that there is no M^2 and therefore no square root in the problem! You have to go back to the beginning and avoid that mistake which totally changes the expression you have to work with for the rest of the problem.

    It might be an idea to check the original question - that 1 squared is rather an odd thing to put in a question. Why not just plain "1" ?
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