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Change in rocket kinetic energy

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data
    During the ascent from a height R to a height 2R, the speed of the rocket changes
    from 7600 m s–1 to 7320 m s–1. Show that, in SI units, the change ΔEK in the kinetic
    energy of the rocket is given by the expression
    ΔEk = (2.09 × 106)m.

    m : mass of rocket
    2. Relevant equations
    KE = 1/2 m.v^2


    3. The attempt at a solution
    ΔEK= final ke - initial ke
    = 1/2 m.7320^2 - 1/2 m. 7600^2
    = - (2.09 × 10^6)m


    But the answer in the marking scheme is + (2.09 × 10^6)m.

    Plz help and clarify the concept
     
  2. jcsd
  3. Feb 12, 2014 #2

    NascentOxygen

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    Staff: Mentor

    hi Miraj! I'm afraid the marking scheme is wrong. :frown:
     
    Last edited: Feb 12, 2014
  4. Feb 12, 2014 #3
    But the question is from CIE A level M/J 2007 Paper 4.
    How can Cambridge university be wrong?
     
  5. Feb 12, 2014 #4

    NascentOxygen

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    Staff: Mentor

    A simple typo maybe?

    You have quoted the problem correctly? It doesn't ask to show that "the magnitude of the change ΔEK in the kinetic
    energy of the rocket is given by ..." ?
     
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