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Momentum: explosion of a block

  1. Oct 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A block of mass 8.45 kg in outerspace is moving at 2.47 m/s with no external forces acting on it. After an explosion, the block is split into two parts both having mass equal to half the mass of the original block. The explosion supplies the two masses with an additional 19.3 J of kinetic energy. Neither mass leaves the line of original motion. Calculate the magnitude of the velocity of the mass that is moving at a greater velocity.

    2. Relevant equations

    3. The attempt at a solution
    (1/2)(8.45)(2.47)^2 =ke
    9.28 joules = ke

    What should i do?
  2. jcsd
  3. Oct 24, 2010 #2
    How about:
    conservation of energy, i.e. the initial energy = the final energy + 19.3 J; how can that help you?
  4. Oct 24, 2010 #3
    9.28 =(1/4)mv^2 +(1/4)mv^2 +19.3
  5. Oct 24, 2010 #4
    Good, but remember that the velocities are different, i.e. they should be [tex]v_1[/tex] and [tex]v_2[/tex].

    Now, you also know conservation of momentum: initial momentum = final momentum. This gives you another equation.
    At this point you should have two equations with two unknowns (the two velocities), so you can solve them.
  6. Oct 25, 2010 #5
    can you please show me how to do this. i would be very grateful.
  7. Oct 25, 2010 #6
    Set up the two equations:
    You have conservation of energy, [tex]E_{initial} + 19.3 = E_{final}[/tex] and conservation of momentum, [tex]p_{initial} = p_{final}[/tex].

    When you write them out, you'll see you have only two unknowns: the velocities. If you have trouble solving them, post them here and I'll give you some pointers.
  8. Oct 25, 2010 #7
    (1/2)(8.45)(2.47)^2+19.3= (1/4)mv1^2+(1/4)mv2^2

  9. Oct 25, 2010 #8
    Awesome. In the momentum equation, solve the equation for either of the velocities---then plug that result into the energy equation. This will eliminate one of the variables, and you can find the other (remember to plug in the block mass for m). You can then plug back into either of the original equations to find the other velocity.
  10. Oct 25, 2010 #9
    from momentum

    21.35 = v1^2 +v2^2
    4.62 = v1+v2

    i did something wrong
  11. Oct 25, 2010 #10
    [tex] (a+b)^2 \neq a^2 + b^2[/tex]
    [tex] \sqrt{a^2 + b^2} \neq a+b [/tex]
  12. Oct 25, 2010 #11
    err.... 4.62=v1^2+(4.9-v1)
    0 = v1^2-v1+0.32

    there is a negative under the square root when i use quad formula
  13. Oct 25, 2010 #12
    Thank you su much for you help, i got the answer!!! you are good at helping!
  14. Oct 26, 2010 #13
    :D thanks, happy to help
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