Conservation of mom. explosion

1. Mar 5, 2013

burton95

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. Mar 6, 2013

ehild

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.

ehild

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3. Mar 6, 2013

burton95

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?

4. Mar 6, 2013

burton95

Then
-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??????????

5. Mar 6, 2013

burton95

Nevermind i got it. I was setting up v2 for the wrong equation. Thx