Mechanics: Explosion of an Object Vector Diagram

[AN630078]

Don’t be sorry, you have been doing everything correctly so far.
Let’s step back to try to visualize what is physically happening, before we apply Math.

The original object was moving horizontally at constant speed (no acceleration).
If the thing desintegrates by itself in midair, there has not been any external force that changes the original dynamic situation.

For that reason, beyond the point of the explosion, we have a system of particles (three fragments), which center of mass must keep moving horizontally at constant speed, just like the original object was doing.

The act of separation among the fragments is an equilibrium dynamic process.
That means that, impulsed by a common expansive force at that center of mass, each is moving away from it at a speed that is inversely proportional to its mass.

Then, the velocity of each fragment must be a combination of its own separation velocity respect to the CM and the velocity of the CM itself.

Based on all the above plus your correct calculations, an observer on the ground would see:

Fragment #1 (of mass 8 Kg) moving horizontally at 25 m/s, carrying a momentum of 200 Kg-m/s.

Fragment #2 (of mass 5 Kg) moving up 50° from horizontal at 30 m/s, carrying a momentum of 150 Kg-m/s.

Fragment #3 (of mass 3 Kg) moving down 23.5° from horizontal at 96 m/s, carrying a momentum of 288 Kg-m/s.

Thank you that is kind to say. Ok, I think I understand what you are saying, thank you for explicating what is physically happening during the explosion.

"That means that, impulsed by a common expansive force at that center of mass, each is moving away from it at a speed that is inversely proportional to its mass."

So, is the speed of all of the fragments inversely proportional to their masses or just the original object?

"Then, the velocity of each fragment must be a combination of its own separation velocity respect to the CM and the velocity of the CM itself."
I apologise but I am a little confused by this statement.

Using the information you have provided I have tried to correct my vector diagram which I have attached.

Fragment #1 (of mass 8 Kg) moving horizontally at 25 m/s, carrying a momentum of 200 Kg-m/s.

Fragment #2 (of mass 5 Kg) moving up 50° from horizontal at 30 m/s, carrying a momentum of 150 Kg-m/s.

Fragment #3 (of mass 3 Kg) moving down 23.5° from horizontal at 96 m/s, carrying a momentum of 288 Kg-m/s.

Would this be correct? I feel that I am not drawing what the question is specifying, as it asks one to use a vector diagram to find the speed and direction of the third fragment, but I think I have done so somewhat in reverse, performing calculations and using these to construct the diagram?
Perhaps, if I had drawn the diagram as below, resolving that P3 must move downwards to return to the x-axis as you state the "summation of the momenta of all fragments in the y-axis must equal zero, as it was for the original object before the explosion". Then, I could have either;
-used a ruler and protractor to find the angle of the direction and magnitude of P3
-or used trigonometry of the right angled triangle formed having found the sides lengths by calculating its horizontal and vertical components to find the direction and magnitude ?

Then to find the velocity of the third fragment I would use the magnitude of the P3 (the momentum) divided by the mass of the fragement, which was found to be 3kg by the conservation of mass. Would this be the correct approach?

I can write it out in order to show my thought progression if that helps

 

[Lnewqban]

There is no error in your diagram in post #9, vector $P_1$ should begin at the same place as vector $P$ does and vector $P_3$ should end where vector $P$ ends.

Revise this new diagram of post #31 by relocating vectors $P_1$, $P_2$ and $P_3$ 560 units on x-axis towards the left.

 

[Lnewqban]

Thank you that is kind to say. Ok, I think I understand what you are saying, thank you for explicating what is physically happening during the explosion.

"That means that, impulsed by a common expansive force at that center of mass, each is moving away from it at a speed that is inversely proportional to its mass."

So, is the speed of all of the fragments inversely proportional to their masses or just the original object?

You are welcome, happy to be of some help

The product mass times velocity implies that, for conservation of the momentum to happen in elastic collisions or explosions, the lighter fragment will move faster after the occurrence.

Please, refer to the representation of the exploding bomb at rest represented in post #26 above.

"Then, the velocity of each fragment must be a combination of its own separation velocity respect to the CM and the velocity of the CM itself."
I apologise but I am a little confused by this statement.

Yes, sorry.
We will be having a different diagram of vectors if our exploding object were originally in repose.
Vector $P_1$ would be pointing horizontally to the left.
Vector $P_2$ would be pointing up, but tilted to the left some.

... I feel that I am not drawing what the question is specifying, as it asks one to use a vector diagram to find the speed and direction of the third fragment, but I think I have done so somewhat in reverse, performing calculations and using these to construct the diagram?
Perhaps, if I had drawn the diagram as below, resolving that P3 must move downwards to return to the x-axis as you state the "summation of the momenta of all fragments in the y-axis must equal zero, as it was for the original object before the explosion". Then, I could have either;
-used a ruler and protractor to find the angle of the direction and magnitude of P3
-or used trigonometry of the right angled triangle formed having found the sides lengths by calculating its horizontal and vertical components to find the direction and magnitude ?

Then to find the velocity of the third fragment I would use the magnitude of the P3 (the momentum) divided by the mass of the fragement, which was found to be 3kg by the conservation of mass. Would this be the correct approach?

I believe that you have been doing everything properly.
In order to calculate the direction and magnitude of $V_3$ vector, you needed to determine the balance of the momentum of the system of fragments first.

Copied from:
https://en.m.wikipedia.org/wiki/Momentum

“If the velocities of the particles are u1 and u2 before the interaction, and afterwards they are v1 and v2, then

This law holds no matter how complicated the force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds up to zero, so the total change in momentum is zero. This conservation law applies to all interactions, including collisions and separations caused by explosive forces.”

 

[AN630078]

There is no error in your diagram in post #9, vector $P_1$ should begin at the same place as vector $P$ does and vector $P_3$ should end where vector $P$ ends.

Revise this new diagram of post #31 by relocating vectors $P_1$, $P_2$ and $P_3$ 560 units on x-axis towards the left.

Thank you very much for your reply.
Ok, I have just tried to summarise my workings to comprehensively answer the question.
I have redrawn the diagram in post #31 by relocating the vectors 560 untis to the left (see attached)
Right, so if I need to determine the balance of the momentum of the system of fragments first in order to calculate the direction and magnitude of P3 vector;
using the formula you have provided;
m1u1+m2u2=m1v1+m2v2

How, initially then would I find the velocity of P3. I assume I would rearrange p=mv to v=pm.
However, how with the initial information given would I find the momentum of P3, as I earlier made the mistake of finding the momentum by adding the magnitudes of the fragments to equal the momentum of P. Would I find the horizontal and vertical components of P1, P2 and P and then substitute to find the missing value of the magnitude of the momentum of P3?

Piece	x-component	y-component
P1	200 cos 0=200	200 sin 0=0
P2	150 cos 50 =96.4	150 sin 50=115
P3	?	?
P	560 cos 0=560	560 sin 0 = 0

Px=mvx=m1vx1+m2vx2+m3vx3
560=200+96.4+m3vx3
560-296.4=m3vx3
263.6 kgms^-1=mvx3

Pxy=mvy=m1vy1+m2vy2+m3vy3
0=0+115+m3vy3
-115 kgms^1=m3vy3

Then either by constructing a right-angled triangle or using Pythagoras' Theorem the hypotenuse will be equal to the magnitude of the momentum of P3:
Momentum of P3= √-115^2 + 263.6^2
Momentum of P3~ 288 kg ms^-1 to 3.s.f

To find the direction of P3:
tan θ = 115/263.6
θ=tan^-1 (115/263.6)
θ~23.6 degrees to 3.s.f

This angle would be negative as the vector is traveling downwards towards the x-axis as "the summation of the momenta in the y-axis must equal zero, as it was before the explosion, while the summation of the momenta in the x-axis must equal the magnitude of the momentum before the explosion" as you earlier stated.

Then since we have now found the momentum to find the velocity of P3:
v=p/m
v=288/3
v=96 ms^-1

So the third fragment has a momentum of 288 kgms^-1 traveling at a velocity of 96ms^-1 at -23.6 degrees to the x-axis.

Would my vector diagram also now be correct and complete? Thank you again for all of your help I am very grateful

 

[Lnewqban]

Happy to help, I know that you are tenaciously trying.

Your summary is correct, you have found the requested magnitude and direction of vectors $V_3$ and $P_3$.

There is, however, a conceptual error in the last vector diagram.
Could you tell us what it is?

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

Hint:
Before the explosion, our mass of 16 Kg had a momentum of 560 N-m/s.
No external force to the system doing any work means that that momentum must be conserved; hence, the total momentum of the system of particles (total mass still is 16 Kg) must keep that value, even when each particle has its own individual velocity and momentum.

The vector momentum is slave of the vector velocity regarding direction, mass is just a multiplier.

 

[Lnewqban]

In the following diagram, consider vector R to represent the momentum before any explosion resulting in three fragments.
Consider vectors A, B and C to represent the moment of each fragment after explosion.

Note that the condition of conservation of momentum implies that vector R and the sequence of vectors A, B and C must have a common origin and end.
The momentum of the center of mass is the same before and after the explosion, whether or not the fragments are flying together or dispersing.

 

[AN630078]

In the following diagram, consider vector R to represent the momentum before any explosion resulting in three fragments.
Consider vectors A, B and C to represent the moment of each fragment after explosion.

Note that the condition of conservation of momentum implies that vector R and the sequence of vectors A, B and C must have a common origin and end.
The momentum of the center of mass is the same before and after the explosion, whether or not the fragments are flying together or dispersing.

View attachment 269092

Thank you for your reply. I read the hyper physics page on vectors and found it very informative and helpful thank you So would the error in my latest diagram be that although the vectors share a common origin they do not share a common end, as P3 ends at the origin? How could I amend this because I had previously drawn that P3 ends at the end of P (see post 9) was this correct? I will redraw this diagram more clearly to show what I mean.

 

[Lnewqban]

Thank you for your reply. I read the hyper physics page on vectors and found it very informative and helpful thank you So would the error in my latest diagram be that although the vectors share a common origin they do not share a common end, as P3 ends at the origin? How could I amend this because I had previously drawn that P3 ends at the end of P (see post 9) was this correct? I will redraw this diagram more clearly to show what I mean.

Exactly!
If the after-explosion vectors make a closed loop, that means that the combined or system momentum and velocity is zero, which is not what we have.

 

[AN630078]

Exactly!
If the after-explosion vectors make a closed loop, that means that the combined or system momentum and velocity is zero, which is not what we have.

Ok, yes that does make sense. Thank you very much for your reply and all of your help. I will post my amended diagram shortly