1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring gun problem can't find start

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    spring-loaded gun on tabletop. Target box 2.20m horizontally from edge of table on floor.
    Shot 1 - spring compressed 1.10 cm - falls 27 cm short.
    How far should the spring be compressed to hit the target with shot 2?


    2. Relevant equations
    GPE1 + SPE1 + KE1 = GPE2 + SPE2 + KE2<br>
    0 + k(x2)/2 + 0 = mgh + 0 + 0


    3. The attempt at a solution
    I've been beating myself up trying to solve this one and the next one I'll post and I don't know where to start. <p>
    I tried:<br>
    to use a range solution (V2Sin(2Theta)/g) and cut it in half to get the initial velocity so I could attempt to figure out KE (mV2/2 - but don't have an angle<p>
    to figure out time of flight - but don't have the height of the table<p>
    I don't have a spring constant for it. And I just don't think I should be using proportions to determine the answer, not this late in a physics (w/calculus) class.<p>
    I don't know where to start, and btw it didn't give me mass either.<P?
    Thanks in advance
    o
     
  2. jcsd
  3. Nov 11, 2008 #2
    See "Spring and projectile problem" on Nov 30. Same problem, already discussed on the forum.
     
  4. Nov 11, 2008 #3

    LowlyPion

    User Avatar
    Homework Helper



    Whatever the height of the table you know that the horizontal velocity times the time to fall is short i.e. too slow. So to get to the proper distance the horizontal speed needs to be increased to 220/193 of whatever speed it comes off the top of the table.

    Working backward, the v will come from the kinetic energy that came from the potential energy of the spring in the gun. Since the mass and the spring constant are invariant that means that the x is proportional to the v.

    So ... that means that if the x is 220/193 of the 1.1 cm you might have a better chance?
     
  5. Nov 11, 2008 #4
    Thank you both, I understand LowlyPion's answer better, but I think I learned more from figuring out how to get where nasu's answer pointed. Thanks again.
     
  6. Nov 12, 2008 #5

    LowlyPion

    User Avatar
    Homework Helper

    I just read the other thread. It can be found here. (Not Nov 30, Oct 30)
    https://www.physicsforums.com/showthread.php?t=268009

    Interesting ... but a long way around the barn I think.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?