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Combination of conservation laws, and older things. Need Help.

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A small wooden block, of mass M, lies in the middle of a horizontal table of length L
    and height h above the floor. The coefficient of kinetic friction between this block
    and the surface of the table is μ. A bullet, of mass m, is shot with a horizontal
    velocity into the block. As a result, the block (with the embedded bullet) starts
    moving across the table and finally lands on the floor a horizontal distance D from
    the edge of the table. Use this information to find the speed of the bullet.

    2. Relevant equations

    ΔP=0 , f = N.μ, ΔE = Woth (1/2mV2 or mgh or f.x), x=V.t, h=1/2gt2

    3. The attempt at a solution
    Let Vi= speed of the bullet

    Well, first i found the final velocity of the block on the table with the bullet inside, then applied energy law to the block BTWN the moments 1 to 2. (1. just off the table 2. on the ground & stopped.) Then i found t, to calculate D from projectile motion relations, and used all i found in the energy law; thus i found :

    Vi= √(μ.L+(D2g)/4h)*√2*((M+m)/m)

    Is that answer true, I could not find it anywhere else so I need verification please ...
    Thanks! :)
     
  2. jcsd
  3. Dec 4, 2011 #2
    Vi= √(μ.L+(D2g)/4h)*√2*((M+m)/m)

    How can you add L + D2g/4h? Something is wrong.

    mu*L has dimensions of length
    2g*D/4h does not have dimension of length

    You also have an inelastic collision here.
     
  4. Dec 4, 2011 #3
    It is not 2g, it is D^2 times g. But still the dimensions are not alike.

    So how should i do it? Give me some advice please.
     
  5. Dec 4, 2011 #4

    BruceW

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    I got a slightly different answer to you yttuncel. I think you should go through it once again. You can tell it is not right because this: μ.L doesn't have the correct units.

    Your answer is close, so maybe you accidentally wrote something in the wrong place while doing the calculation?
     
  6. Dec 4, 2011 #5
    By the way LawrenceC, I did not applied the energy law directly before the collision, I found the velocity after collision then applied energy law. So as energy after collision is conserved, that would be no problem I think.

    BruceW, is it Vi= √(μ.g.L+(D2g)/4h)*√2*((M+m)/m) ?

    I forgot putting g in Normal force :/
     
  7. Dec 4, 2011 #6

    BruceW

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    You are even closer, but still not quite right. Did you take into consideration that the block only travelled a distance L/2 along the table? (Since it was originally placed in the middle)
     
  8. Dec 4, 2011 #7
    Then delta P = 0 is conservation of momentum. Fine.....I did not realize that's what it meant upon first looking at post.
     
    Last edited: Dec 4, 2011
  9. Dec 4, 2011 #8
    Didnt even see that mentioned in the question :/ Thank you both for your comments and help :)

    Edit: I think I did it wrong again. Because the final velocity on the table is not equal to the velocity just after the collision

    Edit2: Ok got it. Lol. Thanks again!
     
    Last edited: Dec 4, 2011
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