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Conservation of Momentum problem

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A gun is fired vertically into a 1.40kg block of wood at rest directly above it. If the bullet has a mass of 21.0g and a speed of 310m/s, how high will the block rise into the air after the bullet becomes imbedded into it?

    2. Relevant equations
    m1v1+m2v2= m1v11+ m2v12


    3. The attempt at a solution
    So I first converted 21.0 grams into .021kg because I know you use kg for this equation.
    After substituing in a few things this is as far as I have gotten:

    m1v1+m2v2= m1v11+ m2v12
    (.021)(0)+(1.40)(0)=(.021)(310)+(1.40)(V)
    V= -4.65m/s as the final velocity of the block

    so from here I have a few questions first did I do this part correctly and from here how would I proceed in order to determine how high the block will go?
     
  2. jcsd
  3. Jan 13, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Not sure where your - sign is from, but for positive y up, 4.6 m/s looks about right.
    (Your equations look like there are typos.)

    Now you know your mv for the combined block/bullet and that Kinetic energy = ½mv² , and that energy will eventually be consumed by gravity with mgh which is the height you are looking for.
     
  4. Jan 13, 2009 #3
    Thank you ever so much! That makes so much sense!
     
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