1. Limited time only! Sign up for a free 30min personal 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!

Relationship between distance traveled and momentum

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Two bodies, A and B, arrive at the base of a frictionless inclined plane with equal momenta of magnitude p. If massb = 3massa, the ratio of the maximum distance that B travels up the plane to the maximum distance A travels up the plane is...
    with choices
    A) db = d a
    B) db = 9da
    C) db = 3da
    D) db = (1/3)da
    E) db = (1/9)da

    2. Relevant equations
    mv = p
    not sure what else :frown:

    3. The attempt at a solution
    Knowing they arrive at the base of this inclined plane with the same momentum, I'm thinking that massa is traveling at a faster velocity than massb. With this, I am assuming that massa is going to travel up the inclined plane farther than massb, but I do not really know where to start in terms of finding the actual ratio of the distance traveled between the two masses. Please and thank you for helping >_<
     
  2. jcsd
  3. Nov 28, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi abpandanguyen! Welcome to PF! :smile:

    What principle would you use to find the maximum height, given the initial speed? :wink:
     
  4. Nov 28, 2009 #3
    Alright, so I'm not exactly sure what the name of the principle is, but I used the equation

    (vf)2 = (vi)2 + 2as(sf-si)

    and plugged in a random but reasonable incline to get an acceleration. With this and other initial velocities I plugged in just for the sake of using numbers to get a better idea of what I was doing, I found that E is the best answer.

    The numbers I used were:
    massa = 2 kg
    massb = 6 kg
    so respectively I gave them initial velocities of 6 m/s and 2 m/s and the incline I used was 5o and I got db = 2.34m vs da = 21.07m

    Yes? No? D:<
     
  5. Nov 28, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    You're missing the point of this exercise.

    Basically, you're using the equation v2 = 2as (or v2 = 2gh).

    a is the same for both masses, so you don't need to know what a is, nor m.

    you can find s1/s2 just from v1/v2
     
  6. Nov 28, 2009 #5
    Okaaay, that makes sense to me now. Thank you very much :D
     
  7. Nov 29, 2009 #6
    I guess I still don't see the answer, yet, I think it is Db=1/3Da, can anyone confirm this?

    thanks
     
  8. Nov 29, 2009 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi samsorz! :smile:

    (are you the same person as abpandanguyen?)

    Write the equation for Db, and divide it by the equation for Da

    what do you get? :smile:
     
  9. Nov 29, 2009 #8
    no, i am not the same person, but probably same class :)

    assuming that s1/s2 is the same thing as D1b/D2b, then... since you said the velocity is just the v1/v2, and velocity isnt affected by a change in mass, are they equal then?
     
  10. Nov 29, 2009 #9
    no, this is abpandanguyen
    I got E as my answer. I am pretty sure that is right, unless I just screwed up my equation somewhere. Thanks again btw Tiny Tim.
     
  11. Nov 29, 2009 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi everyone! Welcome to PF, samsorz! :smile:
    Sorry, I'm not following you :confused: … they start with different velocities, but with equal momenta …

    write the equation for Db, and divide it by the equation for Da

    what do you get? :smile:
     
  12. Nov 29, 2009 #11
    ok, see if this is better

    Pb= mv m=2, v=6, p = 12
    Pa=3mv m=1, v=4, p = 12

    6/4, 3/2?, im getting more and more lost :(

    EDIT: V^2=2d?, so 36 and 16?
     
  13. Nov 29, 2009 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi samsorz! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)

    If you wrote proper equations, you wouldn't get into this mess! :rolleyes:

    Write it all out in full (and show us) …

    what is the equation relating pa and va?

    what is the equation relating pb and vb?

    so what is the equation relating va and vb?

    what is the equation relating da and va?

    what is the equation relating db and vb?

    so what is the equation relating da and db?
     
  14. Nov 29, 2009 #13
    what is the equation relating pa and va? pa=mava

    what is the equation relating pb and vb? pb=3mavb

    so what is the equation relating va and vb? pa/va=pb/3vb

    EDIT: the rest im having trouble finding
     
  15. Nov 29, 2009 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No, you haven't finished that part yet …

    what is va/vb ?

    The next step is the relation between v and d.

    You wrote v2 = 2d, which is the right idea, but you've left something out.​
     
  16. Nov 29, 2009 #15
    so what is the equation relating va and vb? Va/Vb=[Pa/ma]/[Pb/3ma]

    I'm not sure what im leaving out
     
  17. Nov 29, 2009 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Well, the question tells you the momenta (Pa and Pb) are equal,

    so Va/Vb=[Pa/ma]/[Pb/3ma = 1/3 :wink:

    Carry on. :smile:
     
  18. Nov 29, 2009 #17
    By carry on do you mean theres more parts? or is that it lol

    TY
     
  19. Nov 29, 2009 #18

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    I meant …
    hmm … i think misunderstood :redface: when you wrote …
    you meant in v2 = 2d, didn't you?

    That was supposed to be one of the standard constant acceleration equations …

    so where's the acceleration? :wink:
     
  20. Nov 29, 2009 #19
    You said that earlier so I thought we neglected it
     
  21. Nov 29, 2009 #20

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes, we don't need to know what a is …

    but that's no excuse for writing an equation that's wrong!! :rolleyes:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relationship between distance traveled and momentum
Loading...