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Math Help - A ballistic...

  1. #1
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    A ballistic...

    A ballistic pendulum is a device used to measure the velocity of a projectile-for example, the muzzle velocity of a rifle bullet. The projectile is shot horizontally into, and becomes embedded in, the bob of a pendulum. The pendulum swings upward to some height h, which is measured. The masses of the block and the bullet are known. Using the laws of momentum and energy, show that the initial velocity of the projectile is given by
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    Last edited by CaptainBlack; July 22nd 2006 at 10:21 AM.
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    Super Member Rebesques's Avatar
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    I cannot see what the required relation says and hope it's not just me. Is it \sqrt{\frac{2gh(M+m)}{m}} or something else?
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    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Rebesques
    I cannot see what the required relation says and hope it's not just me. Is it \sqrt{\frac{2gh(M+m)}{m}} or something else?
    It says:

    v_{\circ}=\left[\frac{(m+M)}{m}\right]\sqrt{2gh}
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    Super Member Rebesques's Avatar
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    Thanks Q. My Mac hates word documents (and I hate macs )



    So let's see... Reminds me of my days in high school. Lets use preservation of momentum, to find (the measure of) the speed with which the total mass M+m body will begin moving:

    \sum p_{b}=\sum p_{a}

    where \sum p_{b}(total momentum before impact)= mv_0, and
    \sum p_{a}(total momentum after impact)= (M+m)v,

    v here is unknown. So we equate and get

    mv_0=(M+m)v, or

    <br />
v=\left[\frac{mv_0}{M+m}\right]<br />
.

    Let's try conservation of energy, at the instants where the total mass begins moving (phase A) and at when it just stops moving at height h (phase B):

    (kinetic energy at A)+(dynamic energy at A)=(kinetic energy at B)+(dynamic energy at B),

    where

    (kinetic energy at A)= \frac{1}{2}(m+M)v^2=\frac{1}{2}(m+M)\left[ \frac{mv_0}{M+m}\right]^2,

    (dynamic energy at A)=0 (no height, no energy!)

    (kinetic energy at B)=0 (the total mass has just stopped moving)

    (dynamic energy at B)= (m+M)gh.

    So by equating these,

    \frac{1}{2}(m+M)\left[\frac{mv_0}{M+m}\right]^2=(m+M)gh

    and solve for v_0 to get the result.
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