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Energy/ i stuck on the last question, i need help, appreciate

  1. Nov 7, 2007 #1
    A quarter-circle of radius R, block A mass M is release from the top of the quarter-circle, slide down the curve section. And collide inelastically with identical block point B. The two blocks move together to the right and stop with the distance L. The coefficient of Kinetic energy between the block and horizontal is Uk.
    a)find speed Block A before it hits block B
    b)find speed of combined blocks after collision
    c)find the amount of kinetic energy lost

    a)1/2mv^2=mgh
    v=Squaroot 2gL

    b)M(Squaroot 2gL) + M ( 0 )= (2M)V
    v= (Squaroot 2gL)/2

    c) i know the equation for this one but i dont know how to find the KE lost
    1/2(2M)[(Squaroot 2gL)/2]=Uk*mgL
    (mgR)/2=Uk*mgL
     
  2. jcsd
  3. Nov 7, 2007 #2
    You have a typo; L for R in a) and b)

    Parts b and c are ambiguous. At what time(s) are the speed and KE required?

    Your answer for b is correct immediately after collision. Trivially, the speed after travelling distance L is zero.

    In your proposed solution for c you assume the question requires the energy lost after travelling distance L. The velocity then is zero so all the initial GPE has been lost.

    Alternatively you might calculate the energy loss immediately after collision.

    Either way the inclusion of L and Uk in the question seems designed to mislead; they are not required.
     
  4. Nov 7, 2007 #3
    In an inelastic collision kinetic energy is not conserved, you're looking for the change in kinetic energy of the system from before the object hits (from friction) to after the collision (amount lost to do the actual collision).
     
  5. Nov 7, 2007 #4
    sorry im so confuse, can u give which equation im going to use?
     
  6. Nov 7, 2007 #5
    KE before mgR
    KE after collide 1/2(2M)[(Squaroot 2gR)/2]^2 => 1/2mgR
    So it lose 1/2 of it KE?
     
  7. Nov 7, 2007 #6
    Is there any friction before the collision? The question says "The coefficient of Kinetic energy between the block and horizontal is Uk". There is no Uk specified for the curved section so no choice but to assume no friction.
     
  8. Nov 7, 2007 #7
    yes, its happen at the horizontal not the curve
     
  9. Nov 7, 2007 #8
    Check your working for the KE after collision
     
  10. Nov 7, 2007 #9
    1/2(2M)[(Squaroot 2gR)/2]^2
    M*[ (2gR)/4 ]
    it is 1/2mgR
     
  11. Nov 7, 2007 #10
    Oops! Sorry! You were right the first time. It is mgR/2
     
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