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Collisions and Energy

  1. Nov 7, 2015 #1
    Hello everyone, here's the problem
    1. The problem statement, all variables and given/known data
    A pole with mass m2 = 2 Kg is planted on the ground with a consecutive hammer hits. The hammer ha a mass m1 = 4 kg and is dropped from 1 meter upon the pole. With a single hit the pole penetrate the ground of 2 cm.
    Find:
    a) the total resistence R to the penetration, supposing constant that R is costant for each hit and the collision is perfectly inelastic.
    b)The dissipation of energy after the collision

    2. Relevant equations
    Momentutm
    Conservation of energy

    3. The attempt at a solution
    I've found the velocity of the hammer just before the collision using the conservation of energy, that is 4.4 m/s.
    Using the momentum consrvation and knowing that it is an perfectly inelastic collision I've found that the velocity of the sistem after the hammer hit is 2.9 m/s.
    I think till now there is anything to say (If I made any mistakes please tell me).
    After that I've though to find the acceleration of the pole knowing that the penetration is 0.02 meters and that the final velocity is 0, using this formula [tex]a=(V_0)^2/2S[/tex] where [tex]V_0 = 2.9[/tex].
    Knowing that the [tex]F=ma[/tex] I've thougt to find the resistance R but the result is wrong.
    Can you help me please? (The result of R should be 84.9 N)
     
  2. jcsd
  3. Nov 7, 2015 #2

    SammyS

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    That looks like the correct approach.

    You don't show your results for the last two parts.

    What did you get for acceleration?

    What did you use for the mass in the last step? What did you get for the resisting force?
     
  4. Nov 7, 2015 #3
    To find the acceleration I've used the equations for decelerated motion:
    [tex]v = v_0 - at[/tex] and [tex]s=v_0t - \frac{1}{2}a t^2[/tex]
    expressing the time t from the first equation, the second equation becomes:
    [tex]s=\frac{v_0^2}{2a}[/tex]
    with a inverse function the acceleration is equal to
    [tex]a=\frac{v_0^2}{2s}[/tex]
    The value computed is 217.78 m/s^2
    Using the second principle of dynamics [tex]F=ma[/tex] and using only the mass of the pole, the force is about F = 434 N
    I thought the resistence should be equal to this force but as I told the result is 84.9 N.

    Do you think I did any other mistakes?
     
  5. Nov 7, 2015 #4

    gneill

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    I suspect that the given answer is incorrect. I can't see how a force of about 85 N could stop the masses in such a short distance. The force of gravity alone would eat up nearly 60 N of it, leaving only about 25 N to slow them. Your acceleration of 217.78 m/s2 is about 22g's, so expect something around 22 x 60 N as a ballpark figure.
     
  6. Nov 8, 2015 #5
    Well, that value for the acceleration is the value I got, but I'm not completely sure It's correct... what do you think about my resoning?
     
  7. Nov 8, 2015 #6

    gneill

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    Your acceleration value looks reasonable to me. I think you should draw a free body diagram for the situation after the collision, while the masses are decelerating. As I mentioned, the force due to gravity plays a role in how much force needs to be applied in order to accomplish the required deceleration.
     
  8. Nov 8, 2015 #7

    haruspex

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    Why only the mass of the pole?
     
  9. Nov 9, 2015 #8
    Because I thought that the penetration is due only to the mass of the pole... What do you think about that?
     
  10. Nov 9, 2015 #9

    gneill

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    The problem states that the collision is perfectly inelastic. What does that tell you about the masses after collision?
     
  11. Nov 9, 2015 #10

    haruspex

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    What will happen to the hammer after impact? Does it disappear? Move more slowly than the pole?
     
  12. Nov 9, 2015 #11
    The masses after the collision move with the same velocity, but I don't know the direction.
    I supposed the direction was opposite one to each other and I decide to use just the mass of the pole to compute the force.
     
  13. Nov 9, 2015 #12

    gneill

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    I think you need to review the details of what constitutes a perfectly inelastic collision. Look it up.
     
  14. Nov 9, 2015 #13
    Ok! You're right! The objects stick together after a perfectly inelastic collision. So the mass to consider is 6 kg istead of 2.
    This leads to a greater force of 1302 N... I'm gonna draw a free body diagram and post it as soon as I can because I'm not figuring it out :-(
     
  15. Nov 9, 2015 #14

    haruspex

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    That would violate the conservation of momentum equation you used. Momentum is a vector, so the direction of travel matters.
     
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