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3 problems involving linear momentum

  1. Nov 2, 2005 #1
    1) Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of 4.0 x 10^4 m/s (@ the moment of take off) Answer: 6.0 x 10^7 N

    Okay I can not figure how to get an equation to use... Though I did noticed that the rate x the speed in this problem does in fact give you 6.0 x 10^7. Does that play a role? :confused:

    2) A 12,600 kg railroad car travels alone on a level frictionless track with a constant speed of 18 m/s. A 5350 kg load, intially at rest, is dropped on to the car. What will be the car's new speed? Answer: 12.6 m/s

    So... here was my approach...

    (Before) KE1 + KE2 ( = 0) = (after) KE1 + KE2
    (1/2)(12600 kg)((18 m/s)^2) + 0 = (12600 +5350)(v^2)
    sq root of 114 = 10.6 m/s... I am off by 2 numbers... did I do something wrong?


    3) An atomic nucleus intially moving @ 420 m/s emits an alpha particle in the direction of it's velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u what speed does the alpha particle have when it is emitted. Answer: 4.2 x 10^3 m/s

    I did the same approach as above with Kinetic energies... Cannot figure out what I am doing wrong. Please help me.
  2. jcsd
  3. Nov 2, 2005 #2


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    For the first one, it helps to remember that a newton is a kgm/s^2 (think m*a). In other words, you just multiply them together.
  4. Nov 2, 2005 #3


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    For part two, momentum before = momentum after

    mv + mv = (m+m)v

    You should be able to figure out number 3 now.
  5. Nov 3, 2005 #4
    But vertigo, is 1500 kg/s considered a mass? I was confused because masses aren't usually per anything.

    2) mv + mv = (m+m)v
    (12600 kg)(18m/s) + (5350 kg)(0 m/s) = (17950 kg)(v)
    v= 12.6

    So for number 3 do I apply the same logic, with momentums?
  6. Nov 3, 2005 #5


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    1500 kg/s is a rate of change of mass, like power is W = J/s, a rate of change of energy. (kg/s)(m/s) = kgms^-2 = N

    Do you understand that formula for number 2? Numbers 2 and 3 are very similar, read them again.
    Last edited: Nov 3, 2005
  7. Nov 3, 2005 #6
    Yes, I know the formula in # 2 with momentum.

    So for number three I get this...

    mv + mv = mv +mv
    (222 u)(420 m/s) + (4.0 u)v = (222 u)(350 m/s)

    But that doesn't work...
  8. Nov 3, 2005 #7
    I also tried the problem like this:

    (222 u)(420 m/s) + (4.0 u)v = (222 -4 u)(350 m/s)
    And I get -4.2 x 10^3 m/s... it is the correct answer just negative... Must I consider the v to be negative, that way I get a positive answer?
  9. Nov 3, 2005 #8


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    What is the total momentum before? What is the total momentum after?
  10. Nov 3, 2005 #9
    Isn't that what I just stated? I alpha particle has none after.

    (222 u)(420 m/s) + (4.0 u)v = (222 -4 u)(350 m/s)
  11. Nov 3, 2005 #10


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    The alpha particle does have a mometum afterwards.

    You have a mass, 222, which splits into two separate masses, 218 and 4, with separate velocites.

    The single mass is the before case. The two masses is the after case.
  12. Nov 4, 2005 #11
    (222 u)(420 m/s) = (218 u)(350 m/s) + (4.0 u)(v)
    I understand it now. Thank you!

    v = 4.2 x 10^3 m/s!
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