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!

Projectile question

  1. Dec 10, 2013 #1
    I have another question for you guys:)! I have solved something and I just want to know if my method is ok!


    A ball of mass m is ejected from a mortar at speed v and angle θ to the horizontal, at the top of its trajectory it explodes into two equal pieces of mass m/2 and one drops vertically what is the velocity of the other right after the explosion?

    The first thing I done was components of velocity

    v(x)= (vcosθ)
    v(y)=(vsinθ)

    Then calculated momentum before and made it equal to momentum after as there was no external force so momentum is conserved

    Although I used (vcosθ)i+(vsinθ)j

    And when cancelling out I got my final answer as

    v=gt/2sinθ

    Is my methodology correct or should I go about this problem in a different way?:)


    Cheers guys!
     
  2. jcsd
  3. Dec 10, 2013 #2

    Mentz114

    User Avatar
    Gold Member

    There's not enough information. What does 'drop vertically' mean ? Also your answer should be a vector.
     
  4. Dec 10, 2013 #3

    Doc Al

    User Avatar

    Staff: Mentor

    OK, so what was the momentum just before the explosion?

    And what was the initial momentum of the piece that "drops" vertically?

    How did you get this? Why does your answer depend on time?
     
  5. Dec 10, 2013 #4
    Hold on I've noticed a flaw in my work! I shall get back with an update ASAP:)
     
  6. Dec 10, 2013 #5
    Ok so the object is falling at g?

    So v=g*t

    So p(falling)=mgt/2

    M/2 as the mass has halfed?
     
  7. Dec 10, 2013 #6

    Doc Al

    User Avatar

    Staff: Mentor

    No.

    They want the speed immediately after the explosion, not after some time later when it has fallen.
     
  8. Dec 10, 2013 #7
    Ok so where would I go from the fact that momentum is conserved?
     
  9. Dec 10, 2013 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Start by answering the two questions I asked in my first post.
     
  10. Dec 10, 2013 #9
    Ok so I had the momentum before :)

    So momentum after would be....

    p=mv/2 as would the momentum of the other piece!

    So

    p(falling)=m(vsinθ) as it has no horizontal component

    p(other)=m(vsinθ)+(vcosθ)/2

    I missed the unit vectors but I did mean to put them in
     
  11. Dec 10, 2013 #10

    Doc Al

    User Avatar

    Staff: Mentor

    What is it? Give magnitude and direction.
     
  12. Dec 10, 2013 #11
    p(before)= m((vcosθ)i+(vsinθ)j))
     
  13. Dec 10, 2013 #12

    Doc Al

    User Avatar

    Staff: Mentor

    No. Note that the moment in question is when the ball is at the top of its trajectory, not when it is first launched.
     
  14. Dec 10, 2013 #13
    Ahh ok so then we would just have


    p(before)=m(vcosθ)i

    As at the top of the trajectory y component of velocity is zero?
     
  15. Dec 10, 2013 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Right!

    And after the explosion, what is the momentum of the piece that "drops"? (I would assume the "usual" meaning of "drop", though it is somewhat ambiguous.)
     
  16. Dec 10, 2013 #15
    v=tanθ/2
     
  17. Dec 10, 2013 #16
  18. Dec 10, 2013 #17
    So as a vector


    v= (vcosθ)i-tanθ/2 j

    Although I have not considered the loss of mass for the x component in the above! I know horizontal velocity is constant with constant mass but mass here changes?
     
  19. Dec 10, 2013 #18

    Doc Al

    User Avatar

    Staff: Mentor

    Not sure where this comes from. For one thing, the right hand side has no units! So it cannot equal a velocity.

    What's the initial velocity of something that is dropped vertically?
     
  20. Dec 10, 2013 #19
    It is zero
     
  21. Dec 10, 2013 #20

    Doc Al

    User Avatar

    Staff: Mentor

    Right! (That's how I interpret the given information.) So what is its momentum?

    And what must be the momentum of the second piece? And thus its velocity?
     
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: Projectile question
  1. Projectile question (Replies: 1)

  2. Projectile Questions (Replies: 41)

  3. Projectiles questions (Replies: 1)

  4. Projectiles Question (Replies: 14)

  5. Projectile Question (Replies: 6)

Loading...