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Kinetic Energy and Momentum

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A Vulcan spaceship has a mass of 65,000 kg and a Romulan spaceship is twice as massive. Both have engines that generate the same total force of 9.5 x 10^6 N. Neglecting any changes in mass due to whatever is expelled by the engines, calculate the kinetic energy and momentum of each spaceship if each spaceship fires its engines for 100s, starting from rest.


    2. Relevant equations
    K=1/2mv^2
    p (momentum)=mv


    3. The attempt at a solution
    First I dealt with the Vulcan:
    F=ma--> F/m=a 9.5 x 10^6 N/ 6.5 x 10^4 kg= 146 m/s^2
    Vf-Vi=a(delta-t) Vf=146m/s^2(100s)=14600 m/s^2
    K=1/2mv^2=1/2(6.5 x 10^4kg)(14600 m/s)^2= 6.93 x 10^12 J

    p=mv=(6.5 x 10^4kg)(14600m/s)=9.49x10^8 kg m/s

    I just need to know... Am I on the right track? Are there factors I'm not taking into account?
     
  2. jcsd
  3. Jul 12, 2008 #2

    Hootenanny

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    I'm not checking your arithmetic, but your method seems good to me :approve:.
     
  4. Jul 12, 2008 #3
    Thanks!
     
  5. Jul 12, 2008 #4
    touche yellowgators!
    I only have 1 reproche; the method that you used makes the percent of error higher than it schoul really be. In addition your method is time consuming. Try using this method next time:
    1)P=F*t=(m*v1)-(m*v2) ****v1 is initial velocity and v2 is final velocity*******
    The impulse P = 9.5 x 10^6 N * 100s
    P=9.5 x 10^8 N*s

    2) v2 is found by:
    P=(m*v2)-(m*v1) *******v1 equals to 0 *******
    P=m*v2
    v2=9.5 x 10^8 N*s/6.5 x 10^4 kg = 14615.38m/s

    3)Potential energy(PE) = F*d
    PE=9.5 x 10^6 N* (14615.38m/s/2) *100s ****d=average velocity*time********
    PE=6.94*10^12J

    A difference of 0.01*10^12 might not seem much but look at it again :10000000000. Wow, that is a big number!!!!!!!!!. that number schould really be larger because i aslo rounded.
    well, here you go
     
    Last edited: Jul 12, 2008
  6. Jul 12, 2008 #5
    I looked over that method. Thanks! Using the law of conservation of energy is more straightforward than how I approached the problem. That will be helpful for future problems.
     
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