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A space vehicle splitting into two parts (velocity question)

  1. Nov 6, 2014 #1
    Hi, so i have finished this question but im not sure if I did it right. Could you guys check it and tell me if its right or show me where i went wrong.
    This is the question:
    A space vehicle travelling At a velocity of 1400m/s separates by a controlled explosion into two sections of mass 859kg (m1) and 250 kg (m2). The two parts then carry on in the same direction with the heavier rear section moving 120m/s slower than the lighter front section. Determine the final velocity of each section.



    2. Relevant equations
    Mass1=859 kg
    Mass2:250kg
    M1+M2=1109 kg
    1109kg x 1400m/s = 1552600 N

    3. The attempt at a solution
    So the vehicle is creating 1552600N of force, the faster part shouldbe moving slower by 120m so
    1552600-120=1552480
    1552480/1109 (m1+m2)
    =1399.89 m/s
    speed of M1 is 1399.89m/s
    So the speed of M2 will be 1399.89+120
    M2=1519.89m/s
     
  2. jcsd
  3. Nov 6, 2014 #2

    jtbell

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    Staff: Mentor

    This is not force (N = kg·m/s2). It's momentum (kg·m/s), which unfortunately doesn't have a special unit name for it.

    Even after correcting N to kg·m/s, you can't subtract velocity (m/s) from momentum (kg·m/s). The units don't match.

    Have you studied conservation of momentum yet?
     
  4. Nov 6, 2014 #3
    Damn it, could you explain how to work it out? My deadline is tomorrow and this is the only question im stuck with now.
     
  5. Nov 6, 2014 #4

    jtbell

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    Staff: Mentor

    Conservation of momentum means "total momentum after explosion = total momentum before explosion." Can you set up an equation for that?
     
  6. Nov 7, 2014 #5
    So I have worked it out like this:
    Momentum before explosion
    mass1 (m1)= 859kg
    mass2 (m2)=250kg
    859kg+250kg=1109kg
    1109kg x 1400ms= 1552600 N
    they are moving with two diferrent velocities with the velocity of the smaller mass(v2) moving faster by 120m/s so
    v2-v1=120m/s
    v2=v1+120 m/s
    I can calculate how much more force the second mass will travell with.
    250kg x 120ms=30000 N
    1552600N-30000N=1522600N
    1522600N/1109kg=1372.948602 ms
    because i have subtracted the 120ms(30000N) from the total force that means im calculating v1 so
    v1=1372.948602 m/s
    v2=1372.948602 +120=1492.948602 m/s
    momentum before separation (m1+m2) v = m1 x v1 + m2 x v2 (momentum after separation)
    CHECK:
    859(m1) x 1372.948602(v1) + 250(m2) x 1492.948602(v2)= 1552600N
    I think this is right , took me a long time to figure it out.
     
  7. Nov 7, 2014 #6

    jtbell

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    Staff: Mentor

    You got it! :D

    The only other thing I would do is round off the final results to a more realistic number of significant figures, corresponding to the numbers that you were originally given. (But it's good to keep the extra digits in the intermediate steps to avoid roundoff error.)
     
  8. Nov 7, 2014 #7
    I used these extra digits because when i was checking the anwser it was wrong by about 9-10 m/s, then i realized that those numbers after the dot make a difference. Anyways thanks for help I really appreciate it :)
     
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