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Final energy

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data

    A 2kg mass with KE=80 J strikes and sticks to an initially stationary 8kg mass. Calculate the final energy of the stuck-together masses.

    2. Relevant equations
    KE=1/2 x m x v squared


    3. The attempt at a solution
    i have no idea, any help would be appreciated
     
  2. jcsd
  3. Dec 16, 2008 #2

    rl.bhat

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    Homework Helper

    Find the velocity of the 2 kg mass. then apply the law of conservation of momentum. Find the velocity of combined mass and hence the energy.
     
  4. Dec 16, 2008 #3
    found the velocity, but wat is the law of conservation of momentum? p=mv? if so, i then used K=(p squared)/(2m or m+m) and got 32 J

    is that right?
     
    Last edited: Dec 16, 2008
  5. Dec 16, 2008 #4
    p=mv where p is the linear momentum

    the conservation of linear momentum says
    [​IMG]
    where
    u signifies vector velocity before the collision
    v signifies vector velocity after the collision.

    For an inelastic collision
    [​IMG]
     
  6. Dec 16, 2008 #5
    so since i have no epsillon it is inelastic? so i use the formula for inelastic collisions or p=mv?
     
    Last edited: Dec 16, 2008
  7. Dec 16, 2008 #6
    whats your reasoning behind doing that?

    Think about it like this, the first mass has kinetic energy initially.
    [​IMG]
    now it strikes the second mass and they "stick together"... so in this process of sticking energy is lost to heat, bonding etc... but the momentum is conserved since there isnt a net external force on the system.
    After they're are stuck they(m1+m2) start moving at a new velocity.
    The question asks for the ENERGY in the final mass (m1+m2) after the collision

    IF they stick together what kind of a collision is it? elastic or inelastic?
     
  8. Dec 16, 2008 #7
    inelastic, so i used that formula, and got final velocity of the entire mass to be 1.6886, then i put it into K=1/2 x m(total mass) x v(velocity of total mass) squared and fot 14.2568, am i even close?
     
  9. Dec 16, 2008 #8
    you're close... can you write down exactly what you're doing so I know where you're going wrong
     
  10. Dec 16, 2008 #9
    K = 1/2 x m x v squared
    80=.5 x 2 x v squared
    v = 8.9443 m/s

    m1 x v1i + m2 x v2i = (m1+m2)vf
    2 x 8.443 + 8 x 0 = 2 + 8 x vf
    16.886 = 10 x vf
    vf = 1.6886 m/s

    K = 1/2 x m x v squared
    K = .5 x 10 x (1.6886) squared
    K = 14.2568 J
     
  11. Dec 16, 2008 #10
    v = 8.9443 m/s right... so
    2 x 8.443? + 8 x 0 = 2 + 8 x vf

    But everything else seems good.
     
  12. Dec 16, 2008 #11
    got 16.0001 J

    Thank you so much!!!
     
  13. Dec 16, 2008 #12
    No problem... anytime
     
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