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Three piece explosion question

Question: If a container explodes and breaks into three fragments that fly off 120 degrees apart from each other, with mass ratios 1 : 4: 2. If the first piece flies off with a speed of 6m/s what is the speed of the other two fragments. (All fragments are in the plane.)


I do not even know how to set up this problem. I know that the momentum before the explosion has to equal the momentum after the explosion. But there is no starting velocity of the container before the explosion.

I just need some help setting the problem up.

Thank you

Stephen
 

Doc Al

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Assume that the speed of the container before the explosion was zero.
 
458
0
If there is no starting momentum before the explosion, then there will be no NET final momentum.
 
Maybe Stephen needs to be reminded that to utter the phrase "conservation of momentum" means that you have two equations: one equation consisting of all x-components, and one equation consisting of all y-components.
 
ok
so
0= m*6cos(theata) + 4m*Vxcos(theata) + 2m*Vxcos(theata)
0= m*6sin(theata) + 4m*Vysin(theata) + 2m*Vysin(theata)

But what would theata equal?
 
458
0
Draw a picture, break the momentums into x and y vectors as mikelepore said, and use trigonometry.
 
right
so the first mass would be at 120 degrees thus x=-6cos60 and y=6sin60 since 180 - 120 = 60
the second mass would be at 240 degrees thus x=-Vcos60 and y=-Vsin60 since 240-180= 60
the third mass would be at 360 degrees thus x=Vcos0 and y=Vsin0 since 360 is at zero.

Is this right?
 
thus the two equations are:
7mVfx= -3m-2Vxm + 2Vxm
7mVfy= 3sqrt(3)* m - 2*sqrt(3)*Vy*m + 0

Right?
The velocities could be:
0.9m/s and 3m/s
or
1m/s and 3m/s
or
1.5m/s and 09m/s
or
1.5m/s and 3m/s


would the velocities be 1.5m/s and 3m/s?
 
Last edited:
any help would be appreciated
 

Doc Al

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44,725
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so the first mass would be at 120 degrees thus x=-6cos60 and y=6sin60 since 180 - 120 = 60
the second mass would be at 240 degrees thus x=-Vcos60 and y=-Vsin60 since 240-180= 60
the third mass would be at 360 degrees thus x=Vcos0 and y=Vsin0 since 360 is at zero.
Good, but do not use the same letter (V) to stand for two different speeds. Call the speed of the second mass V and the speed of the third mass W (or whatever).

Now write your conservation of momentum equations and solve for V and W.
 
Are the two equations:

0= -3m-2Vxm + 2Vxm
0= 3sqrt(3)* m - 2*sqrt(3)*Vy*m + 0
 

Doc Al

Mentor
44,725
1,021
Are the two equations:

0= -3m-2Vxm + 2Vxm
0= 3sqrt(3)* m - 2*sqrt(3)*Vy*m + 0
Almost. Looks like you used Vx to stand for the speed of both mass 2 and mass 3 in that first equation. (Probably just a typo.) Correct that and you're good to go.
 
ok so the equations are:
0= -3m-2Vxm + 2Wxm
0= 3sqrt(3)* m - 2*sqrt(3)*Vy*m + 0
so:
3m+2Vxm=2Wxm
or
3/2 +Vx=Wx

and

2sqrt(3)*m*Vy=3*sqrt(3)*m

Vy=(3*sqrt(3))/(2sqrt(3))


Now what?
 

Doc Al

Mentor
44,725
1,021
ok so the equations are:
0= -3m-2Vxm + 2Wxm
0= 3sqrt(3)* m - 2*sqrt(3)*Vy*m + 0
There are only two speeds; call them V (mass 2) and W (mass 3). I'd write the equations as:

0= -3m -2Vm + 2Wm
0= 3sqrt(3)*m - 2*sqrt(3)*V*m + 0

Work from there and solve for V & W.
 
OK so:
3/2 +V=W

V=(3*sqrt(3))/(2sqrt(3))


Now do I replace V in 3/2 + V = W so I would get a value for V and W?????
 

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