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!

Instantaneous velocity and a catapult

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone of mass 5 g is projected with a rubber catapult. If the catapult is streched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released.

    2. Relevant equations
    work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone.


    (a) 1/2 fe = 1/2 mv2


    Or

    (b) 1/2 ke2 = 1/2 mv2

    3. The attempt at a solution

    mass m = 5g = 0.005kg; extension e = 7cm = 0.07m;
    force f = 70 N; velocity = ?;

    using:
    1/2 fe = 1/2 mv2

    v = square root ( fe/m)

    v = (70 * 0.07/0.005)
    = 31. 304 m/s

    Using either of the equations gives the same result. Am not sure wether I have arrived because the book am using gave a different solution for the problem. So you watching at this thread, what do you think?
     
  2. jcsd
  3. Apr 11, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks right. Did the book give a different answer?
     
  4. Apr 12, 2014 #3
    Yes it did. It gave 44.27 ms-1 as the answer.
     
  5. Apr 12, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    We made the same mistake. The 70N is the average force, not the peak force.
     
  6. Apr 12, 2014 #5
    So how do we go about it? By the way, what is the difference between average force and peak force?
     
  7. Apr 12, 2014 #6
    Average force is the definite integral of F(x)dx/definite integral of dx

    (kx^2)/x = kx/2

    k*.07[M]/2 = 70[N] so k = 2000 [N/M]

    V = SQRT(ke^2/m) = 44.27 [M/s]
     
  8. Apr 12, 2014 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That is clearly how it is intended in the question statement, but it's not right. (It's a common textbook and teacher blunder - much too common:mad:.)
    We define average acceleration as Δv/Δt, which is the same as ∫a.dt/∫dt. Consequently, the reasonable definition of average force is ∫F.dt/∫dt. This is not in general the same as ∫F.dx/∫dx.

    If we assume the catapult is stretched at a constant rate, it won't matter which definition is used. The force increases linearly from 0 to peak, so the average is half the peak.
     
  9. Apr 12, 2014 #8
    But they give the distance stretched and not the time to stretch.
    So it is quite clear that they mean average over distance and not over time.
     
  10. Apr 13, 2014 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The intention is clear, but it does not make the usage correct.
    If I were to tell you a particle accelerated in a straight line over a distance of 1m with an average acceleration of 2m/s2, would you conclude the final velocity is 2m/s?
     
  11. Apr 13, 2014 #10
    It depends on what the independent variable is. In the problem, it was displacement. In acceleration it is almost always time. But if acceleration was given as a function of position, the average would be found by integrating over distance.
     
  12. Apr 13, 2014 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't accept there can be two definitions of average acceleration leading to different answers. I have certainly never seen it defined as Δv2/(2Δs).
     
  13. Apr 13, 2014 #12
    It may be any number of definitions. I don't see how is this a matter of acceptance.
    They don't give different answers as long as the one relevant for the specific problem is used.
    As long as the author specifies which definition is using, it should be no problem.

    As you say, it's a matter of use not of correctness or acceptance.
    For velocity and acceleration the use is overwhelmingly in favor of time averages.
    This may be because the integrals over time of velocity and acceleration have clear and useful meanings whereas the integrals over distance not so.

    In the case of force both integrals have useful, common meanings so I don't think the case for "use" is so strong as for velocity and acceleration. In collision problems average force is the average over time, so it may incline the scale towards time average. But this does not make average over distance incorrect or even unusual. But in problems with kinetic energy and work, average over time is less useful, isn't it?
    Is there a "standard" that establishes the "acceptable" definitions of averages? In respect to what variable can we average a given quantity?

    In this problem the author did not specify which average he means. This is his mistake.
    I suppose he assumed too that there is only one average that is "acceptable". :smile:
    Just that he assumed on the "wrong" side.
     
  14. Apr 13, 2014 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I agree that it is always possible to define an average as being over a specific parameter. I'm only concerned with the case where no specific base has been given. I don't think it is acceptable for the base to be inferred from the available information. Suppose you were given both time and distance.
    I think it's more definite than that. Acceleration is defined as change in velocity per unit time, so average acceleration should mean Δv/Δt, just as average velocity means Δx/Δt.
    To me, that's "lamppost logic". If given only energy and distance, the only kind of average force that can be calculated is an average with respect to distance. But that doesn't make it reasonable to call this "average force" without qualification. Yes, exactly that happens in many posed problems, and it leads students to believe that this is a valid way to compute average force, not realising that it may give a different answer compared to the usual average over time. On this very forum I've had to point out to other homework helpers that it can give a different answer.
    All I'm saying is that where an average force or acceleration is to be taken with respect to distance it should say so. An unqualified average implies wrt time.
     
  15. Apr 14, 2014 #14
    I agree with what you say. It's just that I think an unqualified average should not imply anything, in general. There is no special status of the time average, again in general.
    You seemed to imply that there is. :smile:
     
  16. Apr 14, 2014 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, I believe "average force", unqualified, means average over time.
    I just Googled average force. Of all the hits in the first page, none of these qualified the term and all took it to be an average over time:
    http://formulas.tutorvista.com/physics/average-force-formula.html
    http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html
    http://www.phys.ufl.edu/courses/phy2053/fall09/lecture14.pdf
    http://www.acs.psu.edu/drussell/bats/impulse.htm
    http://www.dummies.com/how-to/content/how-to-calculate-force-from-impulse-and-momentum.html
    https://curricula2.mit.edu/pivot/book/ph1101.html?acode=0x0200
    http://www.mcasco.com/Answers/qa_avgf.html
    Just one hit, an answer posted on another forum, used average over distance:
    https://answers.yahoo.com/question/index?qid=20140411190559AAwPwNq
    and one other (in biomechanics) appears to be concerned with average static force over an area (so really should be average pressure).
     
  17. Apr 14, 2014 #16

    You did not define some variables in the equation:
    (kx^2)/x = kx/2
    How come that equation? I believe k is constant. But what is x?
     
  18. Apr 15, 2014 #17
    I can easily accept that. :)
    Yes, the Google statistics supports the usage as you see it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Instantaneous velocity and a catapult
  1. Catapult and velocity (Replies: 2)

  2. Instantaneous Velocity (Replies: 1)

  3. Instantaneous velocity (Replies: 16)

  4. Instantaneous Velocity (Replies: 2)

  5. Instantaneous velocity (Replies: 7)

Loading...