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!

Homework Help: Conservation of mom. explosion

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    (2) A fuel tank with a total mass of 6 kg is moving with a speed of vi = 0.5 m/s when it explodes into three pieces. The three pieces fly away from the explosion in the directions shown in the figure; N.B., vi and v3 point in the same direction. The masses of the pieces are: m1 = 1 kg, m2 = 2 kg and m3 = 3 kg. What is the speed (in m/s) of the m1 piece if v3 = 6 m/s?

    v1 and v2 point in negative x direction(left). v1 points positive Y(up) and v2 points negative Y

    θ1 between v3 and v1 = 127 degrees
    θ2 between v1 and v2 = 90
    θ3 between v3 and v2 = 143

    2. Relevant equations

    Pf = Pi

    3. The attempt at a solution

    I set Pi and v3 to be on the x axis moving right as positive. There is no Piy

    Pix = 6 kg * .5 m/s = 3 kg m/s
    Piy = 0

    Pfx = 3 kg m/s = 3 kg * 6 m/s - m1v1sinθ - m2v2sinθ

    Pfy = 0 = m1v1cosθ - m2v2cosθ

    If I can get one of the remaining velocities I'm home free. I tried getting the hypotenuse using tan X/6 = 127 and got x to = 536.7...or some such nonsense.
    Last edited: Mar 5, 2013
  2. jcsd
  3. Mar 6, 2013 #2


    User Avatar
    Homework Helper

    Is the scenario that shown in the picture?

    Are you sure you used sine and cosine correctly in the components of momentums?
    Plug in the data into the equations, isolate one of the speeds and substitute into the other equation.
    Show your work in detail.


    Attached Files:

  4. Mar 6, 2013 #3
    Yeah thats it. What did u use to draw it?

    Originally i tried to take sin127 to get the hypo between v3 and v1. That gave me .79. Not right so i broke up theta1 now = 37° between y axis and v1. To get the value in x direction would be sine right?
  5. Mar 6, 2013 #4
    -v1m1sin37 - m2v2sin53 = -15
    V2 = -15 + m1v1sin37 / (-m2sin53)
    Insert into original
    3 = 18 - m1v1sin37 - (m2(-15+m1v1sin37)/-m2sin53)(sin53)

    Eventually i end up at
    -15 = -m1sin37v1 -15 + m1v1sin37 = 0??????????
  6. Mar 6, 2013 #5
    Nevermind i got it. I was setting up v2 for the wrong equation. Thx
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted