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Lets try the blocks again

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    no one gave correct help the first time i sent this out. if you can't do it, dont worry.

    a 100g granite cube slides down a 40 degree frictionless ramp. at the bottom, just as it exits onto a horizontal table, it collides with a 200 g steel cube at rest, how high above the table should the granite cube be released to give the steel cube a speed of 150cm/s

    2. Relevant equations
    energy equations, momentum equations

    3. The attempt at a solution
    do not assume this is straight forward. i am a first year university student with alot of physics behind me and i cant get the solution. i tried posting it before, but no one could get the correct answer. before you help, please post your answer.
    i used regular equivalent momentums before and after then subbed in the remaining velocity into the energy equations to yield h. sounds simple huh? try it.
  2. jcsd
  3. Nov 28, 2008 #2
    Let's assume the collision is elastic, so that all the kinetic energy is transfered to the steel cube.
    We can use the definition of momentum to find the final momentum of the steel cube:

    Psteel = mv

    Just plug in the given constants to find the momentum.
    In order for our granite cube to have that momentum, it must satisfy:

    Pgranite = Psteel
    mgranitevgranite = Psteel
    vgranite = Psteel / mgranite

    Since mgranite is given to us, we can find vgranite.

    Now, that's just the setup. Did you get that far?
    The tricky part is finding the needed table height.
    How might you go about finding the height that would result in a final velocity found above?
    I'll give you a hint: Take a look at Newton's equations. Which one is the most useful here?
    Last edited: Nov 29, 2008
  4. Nov 29, 2008 #3
    ok so, that was one of my many numerous methods. try getting the answer, it will be wrong. you will probably get 40 something. also, you can't assume the collision is inelastic. in fact, my first post of this question had numerous people say i have eto assume it is elastic!

  5. Nov 29, 2008 #4
    I am quite sure this can be considered an elastic collision. I got 25.8 cm. But I did it pretty quickly too. Do you know the answer?
    You were doing correctly in your other post. I do not know why you stopped!
    Last edited: Nov 29, 2008
  6. Nov 29, 2008 #5
    can you give me your steps please? general steps nothing perfuse. thanks! you got the correct answer!

  7. Nov 29, 2008 #6
    You're quite right. I meant to type elastic rather than inelastic. A typo. Sorry about that.
  8. Nov 29, 2008 #7
    Yes. Break the problem into two parts. That may be easier than trying to do it all at once like in your other post.

    Part 1 should be to determine at what speed does the granite clock need to be traveling at when it makes contact with the steel block. This can be done by noting that both linear momentum and kinetic energy are conserved assuming an elastic collision.

    That gives you two equations/two unknowns-->solve for the initial velocity of the granite block.

    Part 2) Now using that velocity as the final velocity of the granite block in a conservation of mechanical energy scenario involving only the granite block, find out what height the granite block must be released from to achieve that velocity.

    After doing this, perhaps take a good look at the problem and note how those two parts could have been merged into one.
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