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Math Help - Trouble Simplifying/Factorising

  1. #1
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    Trouble Simplifying/Factorising

    I have a rather straightforward Physics problem to solve, only I can't for the life of me simplify the answer to what it needs to be. What am I doing wrong?

    <br />
v_1=\frac{m}{m+M}v_o<br />

    \frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_  1^2+\frac{1}{2}kd^2

    The answer is in the form d= and it's a simplification of the information I provided here in the post.

    I get as far as d=\sqrt{\frac{1}{k}(mv_o^2-(m+M)v_1^2)}

    The book plugs in v_1

    and simplifies to d=\sqrt{\frac{mM}{k(m+M)}}\times{v_o}

    How?
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  2. #2
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    Quote Originally Posted by dkaksl View Post
    <br />
v_1=\frac{m}{m+M}v_o<br />

    \frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_  1^2+\frac{1}{2}kd^2
    Changing your v1 to v and v0 to w, then your 2 equations are:

    v = mw / (m + M) [1]

    mw^2 = mv^2 + Mv^2 + kd^2 [2]

    Simplifying:
    m + M = mw / v [1]
    mw^2 - kd^2 = v^2(m + M) [2]

    Substitute [1] in [2] ; OK?
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  3. #3
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    Very useful, easy-to-understand solution. Thanks a lot.
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  4. #4
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    Hello again.

    Just got to checking the textbook and unfortunately the question was to define d in the terms m, M, k and v_o.

    Cancelling out (m + M) got me an answer with v_1.

    Edit:

    kd^2=mv_o^2-(m+M)v_1^2

    if v_1=\frac{mv_o}{m+M} then what is v_1^2?

    \frac{mv_o}{m+M}\times\frac{mv_o}{m+M} which is

    \frac{m^2v_o^2}{(m+M)^2}

    kd^2=mv_o^2-(m+M)\times\frac{m^2v_o^2}{(m+M)^2}

    kd^2=mv_o^2-\frac{m^2v_o^2}{(m+M)}

    kd^2=\frac{mv_o^2(m+M)}{(m+M)}-\frac{m^2v_o^2}{(m+M)}

    kd^2=\frac{mv_o^2(m+M)-m^2v_o^2}{(m+M)}

    kd^2=\frac{m^2v_o^2+mMv_o^2-m^2v_o^2}{(m+M)}=\frac{mM}{(m+M)}\times{v_o^2}

    d=\sqrt{\frac{1}{k}\times{\frac{mM}{(m+M)}}}\times  {v_o}

    Last edit: Thanks a lot. I see where I made a mistake in expanding. Learned a lot today. Merry Christmas.
    Last edited by dkaksl; December 25th 2009 at 05:42 AM.
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  5. #5
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    Quote Originally Posted by dkaksl View Post
    Just got to checking the textbook and unfortunately the question was to define d in the terms m, M, k and v_o.
    Ahhh; well then, we need to get rid of your v1 (my v); the 2 equations:

    v = mw / (m + M) [1]

    mw^2 = mv^2 + Mv^2 + kd^2 [2]

    square [1]: v^2 = m^2w^2 / (m + M)^2
    rearrange [2]: v^2 = (mw^2 - kd^2) / (m + M)

    m^2w^2 / (m + M)^2 = (mw^2 - kd^2) / (m + M)

    m^2w^2 / (m + M) = mw^2 - kd^2

    m^2w^2 = mw^2(m + M) - kd^2(m + M)

    kd^2(m + M) = mw^2(m + M) - m^2w^2

    kd^2(m + M) = mw^2(m + M - m)

    kd^2(m + M) = mMw^2

    d^2 = mMw^2 / (k(m + M))

    d = wSQRT[mM / (k(m + M))] ........Merry Xmas!
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