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!

Newton's Second Law for a System of Particles

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?


    2. Relevant equations
    Equation 1: R=((v0^2)/g)*sin(2θ0)


    3. The attempt at a solution
    R gives the horizontal range of the projectile, however, due to the change in mass and velocity of the shell, I divided the R by 1/2 to get the distance traveled by the shell before its explosion. After that, I'm not sure what to do.
     
  2. jcsd
  3. Feb 20, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi peaceandlove! :smile:

    What is the speed of the shell just before it explodes?

    What equation can you use to find the speed of the other fragment just after the explosion?
     
  4. Feb 20, 2009 #3
    Why hello tiny-tim!

    I have no clue what the speed of the shell is right before it explodes, so I'm just gonna say 29 m/3.

    I dunno if this would be the right equation to use, but according to (m1+m2)(v0)(cos (θ0))=(m1v1)+(m2v2) where v1=0 and (m1+m2)=1, the velocity of the fragment that continues to travel through the air along the y-axis is 32.4332 m/s.
     
  5. Feb 20, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ok, let's do that first …

    what is its vertical component of speed right before it explodes?

    and what is its horizontal component of speed right before it explodes? :wink:
     
  6. Feb 20, 2009 #5
    I figured it out! I found the coordinates of the point where the shell explodes and the velocity of the fragment that did not fall straight down. Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. I then used this information to determine where the fragment lands by analyzing a projectile launched horizontally at time t=0 and got 119.35124 m.
     
  7. Feb 20, 2009 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Excellent!

    :biggrin: Woohoo! :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Newton's Second Law for a System of Particles
  1. Newton's Second Law (Replies: 5)

Loading...