1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic momentum/loss of energy question

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Two skaters collide and grab onto each other on frictionless ice. one of them of mass "m1" is moving to the right at "v1", while the other, of mass "m2", is moving to left at "v2"

    given m1, m2, v1, v2, determine:

    the magnitude and direction of the velocity just after the collision
    the loss in kinetic energy of the system


    2. Relevant equations

    p=mv


    3. The attempt at a solution


    m1*v1 - m2*v2 = (m1+m2)v3

    v3 = (m1*v1-m2*v2) / (m1+m2)

    firstly I should know if that was right.. then:

    1/2*m1*v1^2 + 1/2*m2*v2^2 = 1/2(m1+m2)(v3)^2

    m1*v1^2 + m2*v2^2 = (m1+m2)(m1*v1-m2*v2)^2 /(m1+m2)^2

    m1*v1^2 + m2*v2^2 = (m1*v1-m2*v2)^2 /(m1+m2)

    (m1*v1^2 + m2*v2^2) (m1+m2) = (m1*v1-m2*v2) (m1*v1-m2*v2)

    m1^2*v1^2 + m1*m2*v1^2 + m1*m2*v2^2 + m2*v2^2 = m1^2*v1^2 - 2*m1*m2*v1*v2 + m2^2*v2^2

    m1*m2*v1^2 + m1*m2*v2^2 = -2*m1*m2*v1*v2
    v1^2 + v2^2 = -2*m1*m2*v1*v2 / ( m1*m2)

    2*v1*v2 + v1^2 - v2^2


    theres a chance I have a clue I knew what I'm doing on the first part.. the second part clearly not?
     
  2. jcsd
  3. Oct 25, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi oreosama! :smile:
    yes :smile:
    the question asks for the difference in KE …

    you should have a minus in the middle of those lines! :rolleyes:

    now just gather the v12 terms, the v22 terms, and the v1v2 terms​
     
  4. Oct 25, 2012 #3
    thanks for that :bugeye:

    peddling along through this assignment i've come across:

    A bomb of mass "m" at rest explodes. Half of the mass is thrown off in the +x-direction at a speed "v". A quarter of the mass is thrown off in the +y-direction at a speed "3v".

    given[m,v]

    determine the magnitude and direction of the remaining piece

    determine the energy of the explosion

    i have no idea what I'm supposed to do, but I would think everything should cancel out?

    (1/2*m*v) i + (1/4*m*3v)i + something = 0

    -(1/2*m*v) i - (3/4*m*v) j = something ?

    everything begins at rest and blows up..

    (1/2*1/2*m*v^2 + 1/2*1/4*m*(3v)^2) * 2 (the third piece being exact opposite of the first two means energy should be double the first 2 added up?)

    ...

    11/4*m*v^2
     
  5. Oct 26, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi oreosama! :smile:

    (just got up :zzz:)
    (try using the X2 button just above the Reply box :wink:)

    yes, momentum (and angular momentum) is conserved in every collision

    but your equation should be (1/2*m*v) i + (1/4*m*3v)j + something = 0, shouldn't it? :wink:

    try again :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic momentum/loss of energy question
  1. Energy Loss Question (Replies: 3)

Loading...