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Homework Help: Moving things to the other side of the equals sign

  1. Aug 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A 20gram bullet it shot out of a 10kg gun at a speed of 300 m/s
    What is the recoil speed of the gun

    (I am putting this in maths rather than physics because I understand the physics part it is just basic maths of using an equation that I am stuck with)
    2. Relevant equations
    G= gun, B = bullet
    mGvG + mbvb = 0

    3. The attempt at a solution
    Because I know that the momentum of both has to be the same and the only thing effecting the speed is the mass
    I got the correct answer by doing this:
    10g = 600ms
    10,000g = 6000 divided by 10,000) = 0.6. Then I just add a minus because it is going in the opposite direction of the bullet. Answer: -0.6

    But I have to do it using the formula...

    so 10kg x vG + 0.02kg x 300m/s = 0

    I don't understand how I am supposed to multiply a weight by a speed, maybe they aren't next to each other and I'm not supposed to multiply them?

    I suppose what I have to do is move everything except vG over to the side of the equals sign..
    When you bring things across you have to change their sign

    vG = -10kg + 0.02kg -300m/s ( two minuses making a plus)
    Did I do that right? What do I do with it now?

    Can someone give me a clue?
    Thank you
  2. jcsd
  3. Aug 13, 2016 #2

    Ray Vickson

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    It is much, much easier to first solve the equation symbolically, then plug in numbers at the end. From ##m_g v_g + m_b v_b = 0## you can solve for ##v_g##:
    $$v_g =-\frac{m_b}{m_g} v_b$$
    Now plug in the numbers.

    Note that ##m_b/m_g## is a ratio of two masses, so is "dimensionless"---that is, it is just a number without any units. That means that ##v_g## is just some number times ##v_b## and so will have the same units (m/s or whatever).
  4. Aug 13, 2016 #3
    I'm sorry but I don't understand how you did that. How did you know to divide? Were they being multiplied on the other side? Where? I thought there were spaces between them. And how do you know which number to put on top? What do I do with the vb in the end?
  5. Aug 13, 2016 #4


    Staff: Mentor

    Starting with ##m_gv_g + m_bv_b = 0##, you want to isolated ##v_g##, right?
    Get the term that contains ##v_g## by itself by adding ##-m_bv_b## to both sides of the equation. This results in ##m_gv_g = -m_bv_b##.
    Now divide both sides of the equation by ##m_g##, resulting in ##v_g = -\frac{m_b}{m_g}v_b##.
    This is very basic algebra. If you haven't had algebra yet, you will have a difficult time in physics. If you have had algebra yet, apparently you have forgotten much of it. If that's the case, I would advise you to review the basic operations of solving an equation.
  6. Aug 13, 2016 #5
    I last did algebra about 12 years ago. So yes, I can't remember almost any of it. That's kind of what I wanted help with. I guess I am on the wrong forum because people only deal with advanced things here. Sorry about this.

    why do you add -mbvb , where does it come from? Can you add whatever you want to an equation as long as you add it to both sides, is that it? Then you divide by mg, where does that come from?

    Is there no more basic way to solve
    10kg x vG + 0.02kg x 300m/s = 0
    this seems like a lot for a introductory book that said on it that no knowledge of maths was required. :(
  7. Aug 13, 2016 #6

    Ray Vickson

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    Homework Helper

    Cancel out the "kg" from both terms, to get ##10\times v_G + 0.02 \times 300 \, m/s = 0##, or ##10 \, v_G + 6\, m/s = 0##. (Note that ##a b## means the same as ##a \times b## (unless you use "ab" as a name all on its own). Can you see what to do now?
  8. Aug 13, 2016 #7
    Thank you so much, Ray! I got it :)
  9. Aug 13, 2016 #8


    Staff: Mentor

    No, we don't deal with just advanced math here. This forum section deals with math topics that precede calculus, such as algebra and trig and some others.
    Their claim of no knowledge of math required isn't true, if that's what they claimed.

    Again, I would advise you to review the algebra that you had long ago. You should either get an algebra textbook or look at the videos on the KhanAcademy website.

    The specific operations to solvel the equation of this problem are
    1. Adding equal quantities to both sides of an equation.
    2. Multiplying both sides of an equation by the same nonzero number.
    Both of these operations are always valid, and yield a new equation with exactly the same solution.

    Regarding 1 above, to get rid of the term ##m_bv_b## from the left side, I added ##-m_bv_b## to both sides. ##-m_bv_b## is the negative, or opposite, of ##m_bv_b##. Adding these two terms together results in 0, so the left side of the equation became ##m_gv_g + 0##, or just plain ##m_gv_g##.

    Regarding 2 above, to get rid of the coefficient ##m_g##, multiply both sides by ##\frac 1{m_g}##, which is equivalent to dividing both sides by this quantity.
  10. Aug 14, 2016 #9
    You don't happen to know a website that teaches algebra through physics? I only have 1 year to do a 2 year course and if I could be practising using physics formula and helping myself to learn them while also studying algebra that would be much more efficient.
  11. Aug 14, 2016 #10


    Staff: Mentor

    I don't know of such a site, although there might be some out there.
    I'm not sure I agree. If you understand the basics of algebra, you should be able to apply them to manipulating equations that are applications of physics.
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