1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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 and a chain

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    We have a chain of 10 blocks, all of them joined by a thin rope and placed in a straight line.
    Suddenly, other two blocks collide with [itex]v[/itex] speed at one end with the chain of the 10 blocks.
    It is assumed that the table is frictionless and the collision is elastic.
    The main question is: after the colission, which blocks are going to move and at what speed?.

    2. Relevant equations

    Since we have an elastic collision, we have to take into account the following equations: [tex] P_{i}=P_{f} \rightarrow m_{1}v_{1i}+...+m_{n}v_{ni}=m_{1}v_{1f}+...+m_{n}v_{nf} [/tex]
    Moreover, the colission is completely elastic, which means: [tex]E_{k(i)}=E_{k(f)}[/tex]

    3. The attempt at a solution

    Initially, the velocity of the blocks of the chain is zero, which means that the linear momentum is the following one: [tex]P_{i}=mv+mv=2mv [/tex].
    We also know that the collision is elastic, and, therefore, (all the blocks have the same weight), the final velocity of the whole system, should be exactly [itex]2v[/itex] (please, correct me if I am wrong)
    However, then, I have to calculate [itex]P_{f}=m(v_{1}+...+v_{12})[/itex], but I don't see how to calculate the relation between all the velocities... In other words, I have the following system:
    [itex]P_{i}=P_{f} \rightarrow 2v=v_{1} + ... + v_{12}[/itex]
    [itex]E_{k(i)}=E_{k(b)} \rightarrow 2v=v_{1} + ... +v_{12}[/itex]
    I post an image so that you can get an idea about how the collision is.
  2. jcsd
  3. Nov 16, 2013 #2
    first of all, since the collision is elastic, the momentum of the block will be transferred to the last block and the rest block will be at rest, so you know the momentum of last i.e MV and the rest 9 blocks will be at rest, i think tis will be enough for you...!!!!!!
  4. Nov 16, 2013 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If I understand the problem correctly, you have to assume the blocks are all the same mass and the ropes are are all the same length (or, rather, that none is longer than the one preceding). Just consider the first collision to begin with: how will the two blocks move after? As Kishlay says, you will be able quickly to move on to considering the point at which the last block has been hit. But now it gets interesting. We are not told whether the ropes are elastic. If not, what happens when the rightmost rope becomes taut?
  5. Nov 17, 2013 #4
    to whom are you telling this thing??
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted