1. Not finding help here? Sign up for a free 30min 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!

Law of conservation of linear momentum and energy

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A body of 2M(KG) mass at velocity V collided with a body of mass M(KG) at rest.
    The first body after collision has a velocity ao 1/3V , and the second moves at 4/3V.
    Verify both laws of conservation of linear momentum and energy with explanation.

    2. Relevant equations
    m1v1f + m2v2f= m1v1i+ m2v2i
    KE = (1/2)mv^2
    PE=mgh

    3. The attempt at a solution
    KE = (1/2)mv^2
    KE before collision = 1/2 *2MV^2+ 1/2* M(0)=MV^2
    KE after collision =1/2*2M(1/3V)^2+1/2*m(4/3V)^2=1/9MV62+8/9MV^2=MV^2
    KE before collision=KE after collision
    there is no height then there is no potential energy
    (KE+PE) before collision + (KE+PE) after collision
    then mechanical energy is constant at any point
     
  2. jcsd
  3. Apr 6, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    I will presume that you figured the momentum was conserved. But that said it should be enough to suggest that since the kinetic energy before equals the kinetic after that no energy was lost to other sources, like potential, or friction, or maybe even sound.
     
  4. Apr 8, 2009 #3
    I found it :)
    m1v1f + m2v2f= m1v1i+ m2v2i
    momentum before collision=2mv+zero=2mv
    momentum after collision=2m(1/3v)+m(4/3v)=2/3mv+4/3mv=6/3mv=2mv
    then Momentum before collision =Momentum after collision
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook