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Conservation of Energy Question

  1. Dec 30, 2005 #1
    Question is:
    ------------

    Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring loaded gun that is mounted on a table. The target box is 2.20 m horizontallyl from the edge of the table. Bobby compresses the spring 1.10 cm, but the marble falls 27.0 cm short. How far shuld Rhoda compress the spring to score a hit?

    Answer:
    --------

    Diagram:

    _______ projectile |
    _|_spring_|-------------___ v
    | | marble<--- ---O_
    | | --__
    | | --__
    | |__________________________|box|_
    |<------------2.20 m ---------->|

    Given
    -----
    dx = 2.20 m
    Sc (spring compressed) = .011 m
    (falls .27 m short)

    Sc=?
    (hits the targe)

    so basically what should be the compression of the spring before it can hit the target?

    -----------------
    I didn't know how to start. I wrote down F=-ks. I figured that the initial velocity is 0. PE at start is 0. Total change in energy is 0. But I do not know what else to do. Please help!!!
     
  2. jcsd
  3. Dec 30, 2005 #2
    Clear diagram of previous post!

    Attached is a clearer diagram of the previous posted diagram! :smile:
     

    Attached Files:

  4. Dec 30, 2005 #3

    LeonhardEuler

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    Gold Member

    Are you sure that the PE at the start is 0? What about the compression in the spring or the hieght of the object?
     
  5. Dec 30, 2005 #4
    with kinematics you should be able to find the horizontal velocity that got the projectile to go 220 - 27 cm. Knowing that velocity should let you solve for the spring constant. With this knowledge, find out what horizontal velocity will get the projectile to go where you want it to go. Then using spring equations again, figure out how much spring compression will make that initial velocity.

    Does this seem to help?
     
  6. Dec 30, 2005 #5

    LeonhardEuler

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    Gold Member

    However, since he does not know the hieght of the table, he will not be able to find the spring constant. It doesn't matter though. Solve for the height in terms of the spring constant and both terms will cancel out in the end.
     
  7. Dec 30, 2005 #6
    Ahh, this is a classic one.

    You will be needing :
    1) split up the problem in two parts (one part with the spring on tha table, the second part when the object has left the table and fall down because of gravity)

    2) Nexton's second law (determin all the forces in the horizontal and vertical direction when the object is falling down : gravity)

    3) conservation of total energy when the particle just leaves the table. This will be the "connection" between the particle just leaving the spring (calculate the energy it gets from the spring) and the motion under gravity.

    good luck

    marlon

    edit : Both k and m will not be important for your calculation. You do not need to know them, because they will cancel out
     
    Last edited: Dec 30, 2005
  8. Dec 30, 2005 #7
    Oh yeah, I missed that detail. That makes it much more tricky. Build a model.
     
  9. Dec 30, 2005 #8
    Guys, just use k and m for springconstant and mass. You will see that both quantities will cancel out (if you solve this question correctly :) ). You don't need them...

    marlon
     
  10. Dec 30, 2005 #9
    Please show work


    Hi, do you think you can show me the process of how you started the problem. I would greatly appreciate it.

    Thanks
     
  11. Dec 30, 2005 #10

    LeonhardEuler

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    Gold Member

    The way I did the problem was to write equations for the x- and y-coordinates as functions of time. Then I solved for time in the x equation, and plugged the expression I got for t into the y equation. This gives y as a function of x. The expression for x has an initial velocity, v_0 in it. Use conservation of energy to solve for v_0 in terms of k, the spring constant, m, the mass, and s, the displacement of the spring. Then it is just a matter of using the information given to solve for h in terms of k, m, and s. k and m cancel away in the end.
     
  12. Dec 30, 2005 #11
    First forget about the 1.73 cm. That's the solution of another question. Sorry...

    well, start with conservation of energy : [tex]\frac{kx^2}{2} = \frac{mv^2}{2}[/tex] ; x is the distance the spring is compressed.

    From this, you get a velocity (just bring on k and m...don't mind them) when x = 0.011m

    When leaving the table, the object moves under the influence of gravity.
    Applying Newton's second law gives you

    [tex]y = \frac{gt^2}{2}[/tex]
    [tex]x' = vt[/tex] ; x' is the horizontal distance travelled by the object. You know that x' = 1.93 m if the spring is compressed 0.011 m.

    From the above two equations, write y as a function of x' and v. Calculate the y-value with the 0.011m and 1.93m data.

    Then, using this y-value (k and m are still not known) you can get the x-value (distance that the spring is compressed) when you know that x' = 2.2m

    marlon
     
  13. Dec 30, 2005 #12
    Correctamundo

    marlon
     
  14. Dec 30, 2005 #13
    Where are you getting the 2 from and what is the t for?

    is 2 part of the equation and t is for time and if t is time then how would you get time or 1.93 m.

    Thank you
     
  15. Dec 30, 2005 #14
    does anyone know how marlon got [tex]y = \frac{gt^2}{2}[/tex]

    any help appreciated
     
  16. Dec 31, 2005 #15
    Don't you know the equation for the vertical position of an object moving under the influence of gravity ?

    In general, the formula is :

    [tex]y = y_0 + v_{0y}t + \frac{a_y t^2}{2}[/tex]

    [tex]y_0 [/tex] is the initial position ; for you it is 0

    [tex]v_{0y}[/tex] is the initial velocity component in the y direction ; for you this is 0 because the v after energy conservation only has a component along the x direction.

    [tex]a_y[/tex] is the acceleration in the y direction ; for gravity this is g


    regards
    marlon
     
  17. Dec 31, 2005 #16

    Thank you marlon. I still do not know what to do after I get the Y.

    I got Y=301686.116/(k*m)

    Help would be appreciated.
     
  18. Jan 2, 2006 #17
    read my first post carefully. it's all in there

    marlon
     
  19. Jan 2, 2006 #18
    trust me I have been looking at it for days, but I can't seem to get an answer. I literally have five pages of work and to no avail. I need a solution. As always, appreciated.
     
  20. Jan 2, 2006 #19
    conservation of energy yields (for the spring pressed over 0.011m)
    [tex](1/2)k(0.011)^2 = (1/2)mv^2[/tex]
    [tex]v^2= (k/m)(0.011)^2[/tex]

    conservation of energy yields (for the spring pressed over x' m)
    [tex](1/2)k(x')^2 = (1/2)mv'^2[/tex]
    [tex]v'^2= (k/m)(x')^2[/tex]

    Newton's second law yields:
    [tex]h = gt^2/2[/tex]
    [tex]x = vt[/tex]

    thus

    [tex]h = \frac{g(1.93)^2}{2v^2}[/tex]

    and ofcourse also

    [tex]h = \frac{g(2.2)^2}{2v'^2}[/tex]

    Calculate this h for x = 1.93m and replace the v by the first equation for the velocity coming from conservation of energy. This gives you height h.

    Use this h in

    [tex]v'^2 = \frac{g(2.2)^2}{2h}[/tex]

    Above formula will give you v' and use this v' in

    [tex]v'^2= (k/m)(x')^2[/tex]

    to solve for x'

    marlon

    edit : g=9.81 m/s²
     
    Last edited: Jan 2, 2006
  21. Jan 2, 2006 #20
    Thank you!

    Just out of curiousity, is the coefficient we get for V' the answer to X'.

    I got [tex]V' = (.017) (sqrt (k)/sqrt(m))[/tex]

    and when I plug in for [tex]v'^2= (k/m)(x')^2[/tex]

    I get x=.017. Is it suppose to be that way, I am really confused. Well if it is thank you Marlon and everyone else who has posted on this topic. I couldn't have done this problem alone. I hope to back for more help. I need help in succeeding in physics. I have a passion for physics, but I am not good at it right now, but I hope to be from this forum.

    Thank you again. :smile:
     
    Last edited: Jan 2, 2006
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