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Velocity of the wedge

  1. Mar 27, 2003 #1
    I think this may be correct I just need a second set of eyes on it to look it over. Thanks in advance.

    The question is:

    A small block of mass m1 =0.500kg is released from rest at the top of a curve shaped frictionless wedge of mass m2=3.00kg, which sits on a frictionless horizontal surface. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right

    a) what is the velocity of the wedge after the block reaches the horizontal surface

    b) what is the height h of the wedge



    Vwedge=(-mBlockVblock)/mWedge = -(0.500)*(4)/3
    Vwedge= 0.667 m/s


    (0+Mblock*g*h=1/2*Mblock*V^2block + 1/2*Mwedge*v^2wedge+0)

    h= (1/2*Mblock*V^2block + 1/2*Mwedge*v^2wedge)/gMblock

    h=0.953 m

    any input?
  2. jcsd
  3. Mar 27, 2003 #2
    Looks OK to me. Better use 2/3 than .668
  4. Mar 27, 2003 #3
    naa, this is a physics class, not a Calculus class. Exact numbers rounded to 3 places is best.

    Thanks for the check.

    Peace out bro

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