How do I find the direction of C?

[lioric]

Homework Statement

A shell of mass 12 kg moving horizontally at 320 ms^ -1 explodes into three fragments A,B and C, which continue to move in a common vertical plane.

Fragment A: mass 2 kg continues at 450ms^-1,45 degree angle above the horizontal.

Fragment B: mass 6kg continues at 400ms^-1 horizontally.

Fragment C: mass m continues at speed v, in a direction theta to the horizontal.

Homework Equations

Momentum + Velocity x mass
Conservation of momentum : total initial momentum = Total final momentum

3. The Attempt at a Solution
So i tried to find the direction of the C by using this
Initial momentum before explosion is simply in x axis
which is 12kg x 320m/s = 3840kgm/sMomentum after explosion has three parts
Fragment A: 2x450=900kgm/s
Fragment B: 6x400 =2400kgm/s
Fragment C (since A is in NE direction and B is in E I suppose C is in somewhat SW direction ) so I guess since momentum is conserved 3840-900-2400 = 540 in SW some angle direction

since all the momentum in an explosion in a particular axis is also conservered and the initial momentum in y-axis is 0
so I tried with
final momentum of fragment A in y-axis + final momentum of fragment C in axis = 0
900sin45 + -540sinX = 0
but I get math error

 

[Merlin3189]

Firstly, you need to remember momentum is a vector quantity - the direction is important.
So we look at momentum in a particular direction.
You can do a calculation for more than one direction, but each is a separate calculation.

 

[lioric]

Firstly, you need to remember momentum is a vector quantity - the direction is important.
So we look at momentum in a particular direction.
You can do a calculation for more than one direction, but each is a separate calculation.

Well since the fragment C is in SW direction I did take the Y axis for is in the negative and the fragment A as a positive

 

[Merlin3189]

Fragment C (since A is in NE direction and B is in E I suppose C is in somewhat SW direction ) so I guess since momentum is conserved 3840-900-2400 = 540 in SW some angle direction

This is what I thought wrong. These numbers (in red) are not in the same direction, so you can't add them.

I didn't look too hard at the latter part, because you said it was an error. I'll look at that more carefully now.

 

[lioric]

This is what I thought wrong. These numbers (in red) are not in the same direction, so you can't add them.

I didn't look too hard at the latter part, because you said it was an error. I'll look at that more carefully now.

Ok
So I need to add them in the particular axis

Total Y axis
900sin45 -Asinx =0
Asinx= -900sin45 = 636.396

Total x axis
900cos45 + 2400 - Acosx =3840
Acosx=3840-3036.369=803.60

this is sufficient to make a right angle triangle
and using pythogorus theorum
√636.396² + 803.60²
=1025kgm/s
now I can use tan
which gives me 38.38 degrees from the horizon in the negative direction

Is this OK?

 

[Merlin3189]

Yes. And this approach:-

final momentum of fragment A in y-axis + final momentum of fragment C in axis = 0
900sin45 + -540sinX = 0

is ok. It's just your 540 is not correct.

 

[lioric]

Yes. And this approach:-

is ok. It's just your 540 is not correct.

I actually don't know the correct answer so is it possible for someone to check it for me

 

[Merlin3189]

Ok
So I need to add them in the particular axis

Total Y axis
900sin45 -Asinx =0
Asinx= -900sin45 = 636.396

Total x axis
900cos45 + 2400 - Acosx =3840 Why -Acosx ? But it depends which direction you draw it and which angle A is. Perhaps you need a diagram?
Acosx=3840-3036.369=803.60 Then, here you change it to + Acosx and thus get the right answer (if A and x are what I think they are.)
So now you go on to get a correct answer.
this is sufficient to make a right angle triangle
and using pythogorus theorum
√636.396² + 803.60²
=1025kgm/s
now I can use tan
which gives me 38.38 degrees from the horizon in the negative direction

Is this OK?

 

[lioric]

The A is not an angle it is the resultant momentum of the fragment C
x is the angle
since i took right side is positive and left side is negative and fragment C is going left in x-axis I gave it a negative
but when I solved the equation as -Acosx =
I took the negative to the other side and the answer is a -803.60
The same for the y-axis too the answer is negative
I did draw a diagram and i knew the direction as SW
and when we use Pythagoras theorem the signs doesn't matter since the squares get rid of the negeaives
I did find the direction which will compensate for the lack of negative signs

 

[Merlin3189]

My mistake saying A instead of x. I understood what you meant by A.

From your comments you obviously understand the situation clearly, so that's fine.
IMO the two lines of your calculation are inconsistent and that sort of thing can lead to mistakes when you are less sure about where things are going.
My diagram had the initial mass going W to E and A going N $5 E, B going E and C going E θ S. So all x component are positive and the y components as you said.

 

[lioric]

Thanks a lot
Your guidance helped