What Are the Speeds of the Fragments After an Explosion?

[StephenDoty]

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]

Assume that the speed of the container before the explosion was zero.

 

[Snazzy]

If there is no starting momentum before the explosion, then there will be no NET final momentum.

 

[mikelepore]

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.

 

[StephenDoty]

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?

 

[Snazzy]

Draw a picture, break the momentums into x and y vectors as mikelepore said, and use trigonometry.

 

[StephenDoty]

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?

 

[StephenDoty]

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?

 

[StephenDoty]

any help would be appreciated

 

[Doc Al]

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.

 

[StephenDoty]

Are the two equations:

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

 

[Doc Al]

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.

 

[StephenDoty]

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]

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.

 

[StephenDoty]

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?

 

[Doc Al]

OK so:
3/2 +V=W

Good.

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

Simplify that expression for V.

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

Sure. You already have the value for V. Plug that into get the value for W.