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Block hits a rod with axle in middle

  1. Apr 30, 2009 #1
    A cube of mass m slides without friction at speed v0 and undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass 2m. The rod is pinned through its center through a frictionless axle. And initially hangs straight down at rest. What is the cubes velocity both speed and direction after?

    Im kinda lost on this one on where to start.

    So i tried momentum equations
    mv0 =mv1+2mv1

    but the answer is v1=(1/5)v0

    Im guessing both objects do not receive the same speed so this would be where my equation fails...
  2. jcsd
  3. Apr 30, 2009 #2
    So this just popped in my head but i'd like to check if my idea is correct
    M/V is a constant. Thus the velocity for the rod must be 2v1

    Go back to the momentum equation
    mv0= mv1 +2m*2v1
    solve for v1
    v1=1/5 v0

  4. May 1, 2009 #3
    Why are you using a common value for the velocities after the collision? (v1)
    That's generally for an inelastic collision, where both objects 'stick together' and move off with a common v.
    Here the collision is perfectly elastic.
    What do you know about perfectly elastic collisions? What quantities are preserved?
  5. May 1, 2009 #4
    I tried pairing momentum and energy equations together and solving in terms of v0 and v1. but the answer was incorrect
    .5m v02 = .5mv12+.5Iw2
    where I=(1/12)MR^2
    and w= v2/R
    thus mv02=mv12+(1/6)mv22
    With the R left i was unsure of what i could do here
    Last edited: May 1, 2009
  6. May 1, 2009 #5

    Doc Al

    User Avatar

    Staff: Mentor

    No need to introduce v2; stick with ω.
    You want angular momentum, not just momentum. Fix two of those terms.
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