Conservation of Momentum of a bomb shell

[heycoa]

I must be thinking way to hard about this problem. Can someone please give me a nudge in the right direction?

Homework Statement

A shell traveling with speed v0 exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0. What is the velocity of the other fragment?

Homework Equations

Let M=m1+m2 (the mass of the shell before explosion equals the sum of the masses of the fragments)

Conservation of momentum: v0*M=v0*m1+V2*m2 (where V2 is the vector pointing in the x direction and v0*m1 is the momentum of the fragment traveling directly upwards)

The Attempt at a Solution

When I work this out, everything pretty much cancels out and V2 = v0. Except I think this is too easy. When I try to shift this problem so that the initial velocity of the shell is traveling in the x direction, and the fragments both exit with an x and y component with magnitudes at a 90 degree angle to each other, I get a completely different answer (V2=(2*sqrt(2)-1)*v0). And this doesn't seem right either.

Please give me a hand

 

[ehild]

Homework Statement

A shell traveling with speed v0 exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0. What is the velocity of the other fragment?

Homework Equations

Let M=m1+m2 (the mass of the shell before explosion equals the sum of the masses of the fragments)

Conservation of momentum: v0*M=v0*m1+V2*m2 (where V2 is the vector pointing in the x direction and v0*m1 is the momentum of the fragment traveling directly upwards)

The Attempt at a Solution

Momentum is vector, and conservation of momentum holds for each component. Write an equation for both the horizontal and vertical components of the shell and its pieces.


ehild

 

[PhysicsLover21]

Your first attempt was correct. It does cancel out each other...what is your doubt?

 

[ehild]

Your first attempt was correct. It does cancel out each other...what is your doubt?

Once vo is the horizontal velocity the shell travels before explosion. After explosion, one piece travels vertically up with speed vo. You can not add horizontal and vertical components of vectors.

ehild

 

[PhysicsLover21]

Ah..now I get you...ya, echild is right...gotta use vector or cartesian way to add them...

 

[heycoa]

ok so I got V2=vo*cos(45) in the x direction. Does that sound reasonable?

 

[SammyS]

I must be thinking way to hard about this problem. Can someone please give me a nudge in the right direction?

Homework Statement

A shell traveling with speed v0 exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0. What is the velocity of the other fragment?

Homework Equations

Let M=m1+m2 (the mass of the shell before explosion equals the sum of the masses of the fragments)

Conservation of momentum: v0*M=v0*m1+V2*m2 (where V2 is the vector pointing in the x direction and v0*m1 is the momentum of the fragment traveling directly upwards)

...

What do you mean by the x direction?

Usually, x is horizontal and y is vertical. For this problem that would mean that the xy-plane is vertical with the x-axis pointing North (or South).

Please define your coordinate system.

 

[heycoa]

Yes, I am using the general cartesian coordinate system where Y is north and X is east. My original velocity vector (before explosion) is pointing 45 degrees from horizontal. The mass m1 has a velocity exactly north, and I have extrapolated that the mass m2 has a velocity pointing exactly east.

 

[SammyS]

Yes, I am using the general cartesian coordinate system where Y is north and X is east. My original velocity vector (before explosion) is pointing 45 degrees from horizontal. The mass m1 has a velocity exactly north, and I have extrapolated that the mass m2 has a velocity pointing exactly east.

If the y-axis points North and the x-axis points East, then what is up, the vertical direction? That must be the z-axis.

The initial velocity vector, v₀, is due North and horizontal, according to the given information.For the set of coordinate axes you have chosen here, that means that v₀ is purely in the y direction.

If v₁ is the velocity of the fragment that goes vertically up, (magnitude of vector, v₁, is v₀, the magnitude of vector, v₀) then v₁ is purely in the z direction.

Neither of these two velocities has any component Eastward, i.e. neither has an x component.


You may want to rethink your choice of coordinates.

The coordinate system usually used to describe this situation has the x-axis pointing North, and the y-axis pointing vertically.

 

[heycoa]

forget the coordinate system. I can work in any conventional or non conventional set of coordinates. But that is not the issue right now since V1 and V2 are exactly perpendicular to each other. All I want to do is find the value of V2 if the vector V1 is 90 degrees perpendicular to V2 and the original (Pre explosion) velocity vector has these components (which are perpendicular to each other).

 

[heycoa]

This is a 2 dimensional problem not 3 dimensional.

 

[SammyS]

This is a 2 dimensional problem not 3 dimensional.

Yes. That's is consistent with what I said in post #7, part of which I "Quote" below.

...

Neither of these two velocities has any component Eastward, i.e. neither has an x component.


You may want to rethink your choice of coordinates.

The coordinate system usually used to describe this situation has the x-axis pointing North, and the y-axis pointing vertically.

So, as I suggested, choosing the x-axis to point North, and the y-axis to point up (vertically) will express the problem in two dimensions. Everything occurs in the xy plane.

 

[heycoa]

Ok. Now that that's out of the way, I have already solved the problem using vector notation in the x-y-plane, as I stated in my original post. If you know weather or not V2=(2*sqrt(2)-1)*v0, is correct or not, it would actually help me.

 

[SammyS]

Ok. Now that that's out of the way, I have already solved the problem using vector notation in the x-y-plane, as I stated in my original post. If you know weather or not V2=(2*sqrt(2)-1)*v0, is correct or not, it would actually help me.

I can't comment on that answer until I see how you got that answer.


Up to this point, your choice of axes has been unclear at best, and perhaps inconsistent. At some point in the above discussion, you have a fragment heading east and the x-axis pointing east. (None of the velocities or momenta have any components in the eastward direction.)

I'm not trying to be a pain or be condescending, it's just that until your coordinate system is well-defined and you give components of the velocities and/or momenta in terms of that coordinate system, it makes no sense to deal with any other details of the problem. Once a relevant coordinate system is established, the problem you are trying to solve is not too difficult.

 

[heycoa]

i will post a picture. I re-did it this morning and it should be more clear.

 

[heycoa]

here it is. Please ignore the center of mass problem on the bottom (problem 3.15). As you can see, I chose the original velocity vector to be in the x-y-plane, with x and y components. The resulting fragments each have one of the components.

 

[heycoa]

I guess that didnt work. The picture is here: http://imgur.com/0WNwbtA .
-Thx

 

[SammyS]

here it is. Please ignore the center of mass problem on the bottom (problem 3.15). As you can see, I chose the original velocity vector to be in the x-y-plane, with x and y components. The resulting fragments each have one of the components.

I guess that didnt work. The picture is here: http://imgur.com/0WNwbtA .
-Thx

Yes, that image is way to large to display in a post.

It turns out that the coordinate system is the issue here.

You show that the initial velocity vector, v₀, has equal x and y coordinates.

According to the problem statement

The "... shell [is] traveling with speed v0 exactly horizontally and due north ... ".​

Therefore, v₀ will have only one non-zero component if the x-axis points due North.

Please define your coordinate system.

 

[heycoa]

I shouldn't have to. I can just work with vector components.

 

[SammyS]

I shouldn't have to. I can just work with vector components.

It's impossible to give components of v₀ without specifying a coordinate system.

If you would like, I can specify a coordinate system.



... Come to think of it I've already given two systems, one of which I said I preferred.

 

[heycoa]

youre right. I need a coordinate system to have components. However, it should be evident what my coordinate system is based on the Vx and Vy components of the vector V

 

[SammyS]

you're right. I need a coordinate system to have components. However, it should be evident what my coordinate system is based on the Vx and Vy components of the vector V

All that is evident is that there is a serious error.

In the problem, the initial velocity, v₀ of the shell (mass of 2m) is horizontal and due North.

Immediately after exploding, one of the fragments (mass of m) has velocity (You call it v₁ which is fine. ) which is vertical.

Thus the velocity vector v₀ is perpendicular to the velocity vector v₁ . After all, one is vertical, the other horizontal.The problem I see is that you have an angle of 45° between the directions of these two angles.

Choosing a specific coordinate system will go a long ways towards resolving this problem.

 

[heycoa]

so you are saying that this is a 3 dimensional problem? That horizontal and due north does not have a vertical component in it?

 

[SammyS]

so you are saying that this is a 3 dimensional problem? That horizontal and due north does not have a vertical component in it?

No. It's readily done in two dimensions.

Let the y-axis be vertical, pointing upward, i.e. perpendicular to level ground.

Let the x-axis be perpendicular to the y-axis and point due North. (It is then horizontal.)

v₀ is horizontal and due North. Thus it has a vertical component of zero. It points solely in the x direction.



No vector needed to solve this problem has any component in the East-West direction.

 

[heycoa]

So the shell is traveling in the x-direction, and when it explodes it has a fragment traveling "vertically up" (i.e. y-direction) and another fragment traveling in an undefined direction?

Please let me know what this undefined direction is, I am terribly confused.

 

[SammyS]

So the shell is traveling in the x-direction, and when it explodes it has a fragment traveling "vertically up" (i.e. y-direction) and another fragment traveling in an undefined direction?

Please let me know what this undefined direction is, I am terribly confused.

The main point of the problem is

"What is the velocity of the other fragment?"​

Velocity is a vector, so it's asking to find that "undefined direction", as you refer to it.

You solve this by using Conservation of Momentum, as suggested by the title you chose for this thread.

 

[SammyS]

More specifically:

If velocity, v₀ is in the x direction, then $\ {\vec{v}}_0=v_0\hat{x}\,,\$ right?

What's the initial momentum? (Call it something like p₀ .)For the fragment which goes vertically, what is its velocity?

What is its momentum? (Call it something like p₁ .)​

Call the momentum of the other fragment something like p₂ .)

How are p₀, p₁, and p₂ related?

 

[heycoa]

p0=p1+p2 i believe

 

[SammyS]

p0=p1+p2 i believe

Yes, if added as vectors.

What are p₀ and p₁ ?

 

[heycoa]

p0=2m*v0
p1=m*v0

 

[SammyS]

p0=2m*v0
p1=m*v0

Don't forget they have directions associated with them.

Now, find p₂ ?

 

[heycoa]

so the direction of p0 is in the x-direction, and p1 is in the y-direction?

 

[SammyS]

so the direction of p0 is in the x-direction, and p1 is in the y-direction?

Yes ...

 

[heycoa]

well p2=v2*m

and when i solve for v2 i get 2*v0(x hat) + v0(y hat)

 

[SammyS]

well p2=v2*m

and when i solve for v2 i get 2*v0(x hat) + v0(y hat)

From that I get $\ \vec{p_2}=2\,m\,v_0\hat{x}+m\,v_0\hat{y}\ .$

Adding p₁ and p₂ should give p₀, right?