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!

Collision of two carts

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A cart of mass m moves with a speed v on a frictionless air track and collides with an identical cart that is stationary. If the two carts stick together after the collision, what is the final kinetic of the system in terms of m and v?


    2. Relevant equations

    pi=pf

    3. The attempt at a solution

    mivi=mfvf
    0=mvf
    so m=vf
    is that right?
     
  2. jcsd
  3. Mar 22, 2012 #2

    BruceW

    User Avatar
    Homework Helper

    your first line is correct (mivi=mfvf), but I don't understand the rest of your working. One step at a time, what is the initial momentum?

    EDIT: remember, the momentum of the system is conserved, so you should work out the total momentum.
     
  4. Mar 22, 2012 #3
    the initial momentum is 0 because it says the cart starts at rest
     
  5. Mar 22, 2012 #4

    BruceW

    User Avatar
    Homework Helper

    But there are two carts to begin with. You need to find the total momentum, because it is the momentum of the system which is conserved.
     
  6. Mar 22, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The initial momentum of one cart is 0 but it is total momentum that is conserved. BruceW was asking for the total momentum.
     
  7. Mar 22, 2012 #6
    so instead of 0=mvf i should do
    m1iv1i+m2iv2i=m1fv1f+m2fv2f
    which is the same as
    mv1i=mv1f+mv2f
    mv1i=2mvf
    so vf=mv1i/2m
    is that right?
     
  8. Mar 22, 2012 #7

    BruceW

    User Avatar
    Homework Helper

    Exactly. Well written, too :) The examiner will like it when you set out your steps nicely, as you have done here.

    So you've now got vf (and you can further simplify the fraction). And the next step is to get "the final kinetic of the system in terms of m and v" (I'm guessing this means the final kinetic energy).
     
  9. Mar 23, 2012 #8
    so if i simplify i get vf=v/2
    is that the correct simplification?
     
  10. Mar 23, 2012 #9

    BruceW

    User Avatar
    Homework Helper

    Yep, that's right. next step is to find the final kinetic energy.
     
  11. Mar 23, 2012 #10
    oh right.
    so i can use Ei=Ef
    and then say
    Ki+Ui=Kf+Uf
    do i have to do this twice, once for each cart?
     
  12. Mar 23, 2012 #11
    so Kf=1/2mvf
    which means
    Kf=1/2m (v/2)
    kf=mv/4
    but it said this is wrong!
     
  13. Mar 23, 2012 #12

    BruceW

    User Avatar
    Homework Helper

    energy is not necessarily conserved in this collision. In problems like this, if you're unsure, then a good rule of thumb is to use momentum conservation. You have already used conservation of momentum to get the final velocity of the two carts stuck together, so what is the final kinetic energy?
     
  14. Mar 23, 2012 #13
    but i didnt think final kinetic energy was in the momentum conservation formula
     
  15. Mar 23, 2012 #14
    Hello eagles12,

    See the question states that cart 1 was moving initially and cart 2 was at rest.After collision, both start moving with a common velocity v.

    Using momentum conservation you found out the new velocity as u/2 (where u is initial velocity of cart1).

    Now you need to find final kinetic energy of the two carts.

    You know their masses m and their velocites.So just apply the formula of kinetic energy.
     
  16. Mar 23, 2012 #15
    i applied the formula of kinetic energy Kf=1/2m (v/2)
    and then kf=mv/4
    but it said that this was incorrect
     
  17. Mar 23, 2012 #16
    Also it is v important to realise that during head on collisions (which you will deal with at school level)
    its the TOTAL momentum which is conserved .

    That is
    m1u1 +m2u2=m1v1+m2v2

    Later in higher classes you wil learn about oblique collisions in which individual components are also conserved.(But lets just move past that right now!).

    Also during collisions although total energy is conserved,
    using conservation of energy formula doesn't help.(except if the collision is elastic)

    This is because some of the energy is lost as sound or heat (etc) and you cannot acount how much amount was lost.


    So in your formula,
    K1 +U1=k2 + U2

    you never know about the values of U2 and U1 (which include terms like sound energy, heat, potential energy, )
    so using conservation of energy gives you no extra information.

    However, by using conservation of momentum you can find v2 and v1 and thus obtain the final kinetic energies (as you know m1 and m2).

    After this you can use K1-K2 =U1-U2 and thus find how much energy was lost as sound or heat or to elasticity :-)
     
    Last edited: Mar 23, 2012
  18. Mar 23, 2012 #17

    Formula of kinetic energy for a body having velocity u and mass is mu^2/2 not mu/2
     
  19. Mar 23, 2012 #18
    if i don't know U1 and U2 how do I figure out K1-K2=U1-U2
     
  20. Mar 23, 2012 #19

    BruceW

    User Avatar
    Homework Helper

    You must find the final kinetic energy without using the principle of energy conservation. emailanmol was saying that once you do find the kinetic energy, then you can calculate U2-U1. But this isn't in the question, so you don't need to worry about doing this for this question.

    You got pretty close here:
    As emailanmol was saying, it is this bit which was incorrect. Remember that the two carts get stuck together, so what is their mass? And it looks like you've also forgotten to square the speed.
     
  21. Mar 23, 2012 #20
    First lets focus on finding K2
    (We are drifting away from the main question here).

    Read my last post.

    You know velocity of each cart is v/2. Mass is m for each cart

    What is the kinetic energy of one cart?
    What is the total kinetic energy of both carts?


    Remember kinetic emergy is mu^2/2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Collision of two carts
  1. Cart-Cart Collision (Replies: 3)

Loading...