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Momentum of a projectile and ship

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A battle ship (4.09*10^7 kg) fires a salvo of 3 rounds from its foward turret in the direction of the bow at an angle of 10 degrees above the horizontal. Each projectile weighs 1220 kg, each barrel is 20.9 m long and the muzzle velocity of the projectile is 770 m/sec. Acceleration of each projectile down the barrel is 1.418 x 10^4 m/s2 at 0.0543 seconds.

    If the momentum must be conserved in the x direction, 1) what's the velocity of the ship in the x direction and 2) in what direction is it moving along the x axis after the projectiles have left the ship's guns?

    2. Relevant equations
    p(linear momentum of an object)=mv
    M1V1 + M2V2=0

    3. The attempt at a solution
    1) I'm not sure whether I should get the velocity of the projectile in the x direction, but I just used it's velocity in the x direction in solving the problem. I did the trigonometric work on paper.

    M1V1 + M2V2=0
    V2= -M1/M2(V1)
    V2= -1220/4.09*10^7(758.3)
    V2= -.0226 m/sec

    2) The ship recoils to the left of the x axis.
     
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 25, 2007 #2
    "1.418 x 104 m/s2 at 0.0543 seconds" what do you mean by that? Is it an increase of 1.418 x 104 m/s every 0.0543 seconds?
     
  4. Mar 25, 2007 #3
    sorry, its suppsoe to be 1.418*10^4; 1.418*10^4 is the acceleration of each projectile in the barrel and .0543 sec is the time it takes for the projectile to travel down the barrel.
     
    Last edited: Mar 25, 2007
  5. Mar 25, 2007 #4
    Ok but how is this useful if we are given the muzzle velocity?
     
  6. Mar 25, 2007 #5
    i was told that it was supposidly the final velocity/velocity at which the projectile leaves the barrel.
     
  7. Mar 25, 2007 #6
    okay; so does that mean the muzzle velocity is the the velocity I would use in the momentum equation for V1, but instead the velocity vectors in the x direction which is 758.3 m/sec?
     
  8. Mar 25, 2007 #7
    Ok, it actually does affect the problem. What we are interested in is the moment when the projectile leaves the ship. This moment would be right when the projectile begins to travel down the barrel. So with the information given, we must calculate the initial velocity. We have

    [tex]{v_{i}}^{2} = {v_{f}}^{2} - 2ad [/tex]

    We know vf, a and d, so we can calculate vi. Once we get vi, we calculate its horizontal component and use it in the conservation of momentum formula.
     
  9. Mar 25, 2007 #8
    [tex]{v_{i}}^{2}[/tex]= (770)^2 - 2(14180)(20.9m)
    Vi=13.27 m/sec!!!!

    thank you, sir.

    I do have one more question, if you don't mind...
    What is the total momentum of the 3 projectiles in the x direction parrell to the water?
    Would I just multiply the momentum of one of the projectiles by three or is it zero since total momentum is always zero??
     
  10. Mar 25, 2007 #9
    If the three projectiles are in the same direction, their momentum adds ups since the sign of momentum depends on the sign of the velocity. If you add the momentum of the ship into the equation, the sum would be 0. In any explosion or implosion (in this case it's an explosion - things get separated) the momentum gained in a direction by a fragment is equal to the momentum gained by the opposite fragment in the other direction. This is basically the law of conservation of momentum.
     
  11. Mar 25, 2007 #10
    so is the following correct? (I now used the new initial velocity in the x-direction:
    M1V1(3)=?
    (1220)(13.07)(3)= 47830.34 J

    ?
     
  12. Mar 25, 2007 #11
    Yes. The ship would have the opposite momentum.
     
  13. Mar 25, 2007 #12
    If I had the power to bestow upon you a salmagundi of riches beyond belief, you would be the first person to share in this wealth.
    THANK YOU!
     
  14. Mar 25, 2007 #13
    Not to rain on ayones parade, from the earlier thread you posted, I wrote:

    "This is right, but perhaps a more straightforward approach assumes vf=770, and Vi=0,

    Given that Vf^2-Vi^2=2ax, then a=770^2/(2*20.9)
    a=14184m/s^2 and t would just be the distance divided by ave velocity:
    20.9/385=0.0543s Different ways to skin a cat. Best to have many knives.:"

    In other words you just have introduced some roundoff error and vinit=0, this was the assumption after all, and acceleration was calculated on this basis.

    Again, this operates under several assumptions that acceleration is linear from the detonation of the charge, to the time the shell leaves the barrel. It is entirely possible that this is not the case due to friction, leakage of gas, the fact that the volume occupied by the gas is becoming larger as the shell progresses down the barrel, etc.

    The question about which velocity to consider is an interesting one. If we ignored the ship altogether, I think most would agree that say in the case of the rifle, its the muzzle velocity (Vf) that determines the momentum, even though the physical recoil may be experienced sooner.

    So I would disagree, and suggest by extension to the case with the rifle, that the ships negative velocity*mass=3*cos(10)*770*mass(shell).
     
  15. Mar 25, 2007 #14
    you do indeed have a very strong point when noting this new initial velocity has turned the momentum problem into a bit of an error. However, i am a bit perplexed as to whether the equation you presented above clearly depicts the shell's (or projectile's) velocity if one multiplies along with it 3 and cos(10). is such thing possible to correctly calculate the momentum?
     
  16. Mar 25, 2007 #15
    actually, I would like to answer my own question...
    Multiplying cos(10) and (770) gives one the velocity in the x direction, which is what I want. Multiplying three simply takes into account the 3 projectiles.

    You are as well a life saver, denverdoc and you too are deemed greatful to partake of these chimerically hypothetical riches.
     
    Last edited: Mar 25, 2007
  17. Mar 26, 2007 #16
    any of u guys know how to do electric potential?
     
  18. Mar 26, 2007 #17
    negatory, sir.
     
  19. Mar 26, 2007 #18
    never mind--answered already by op.

    well, i think your point is well taken, we are ignoring the vertical component of the momentum, as this would simply try to push the ship deeper in the water, which would be opposed by buoyancy, etc. It is essential tho that we only consider the x component of the shells velocity, as one fired vertically would have no tendency to cause the boat to recoil along the x axis, thus the cos 10 degree factor. As to algebraically adding the volley of shells, I can see some concern on your part that this isn't fair and proper. One reason might be that the ship has already come to rest between shots. But this is physics homework where all kinds of outlandish assumptions are made routinely in the service of introducing principles. In other words, for the purpose of this problem since there is no consideration of friction, we can superimpose the three shots whether minutes apart or a single shell of 3600Kg was shot with the same muzzle velocity, or whether three guns all aimed in parallel lines were fired simultaneously. (coure then you might actually get some torque and cause the ship to rotate--couple chapters down the line, not to worry). Hope this didn't confuse matter furthers, if you have some specific concerns re the legitimacy, holler.
     
    Last edited: Mar 26, 2007
  20. Mar 26, 2007 #19
    so would you say that the ship, for my homework's sake, is moving left of the x direction after all of this firing of shells has been made?
     
  21. Mar 26, 2007 #20
    denverdoc, I don't think they gave information on acceleration and distance if we were not meant to consider the initial velocity. This said, I think the whole idea of a projectile accelerating down a barrel is flawed. The projectile gets sent by absorbing some momentum of an explosion that happens at the lower end of the barrel. I believe the projectile stops accelerating there and then, as in a instantaneous collision.
     
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