Mechanics: Explosion of an Object Vector Diagram

In summary: The same is true of the other vectors in the equation.In summary, the conversation discusses the conservation of momentum and how it applies to a system of objects in an isolated system. It is explained that momentum is a vector and cannot be added by simply adding their magnitudes. Different methods, such as using components or graphically, can be used to add momentum vectors. However, the equation given in the conversation is incorrect as it does not specify the direction of the vectors.
  • #1
AN630078
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Homework Statement
I have a mechanics question concerning an explosion of an object which I am truly struggling with. I think perhaps I have been thinking about it for too long to reach a credible solution.

An object of mass 16 kg is moving in the x direction at 35 m/s. The object explodes and breaks into three pieces. One piece, of mass 8 kg, continues moving in the x direction at 25 m/s. Another piece, of mass 5 kg, moves on at 30 m/s in a direction at 50o to the x direction.
Using a vector diagram find the speed and direction of the third fragment.
Relevant Equations
p=mv
Well, I understand that according to the conservation of momentum the total momentum of a system is conserved for objects in an isolated system, that is the sum of total momenta before the collison is equal to the sum of momenta after the collision.

In this case, the momentum of the object before the collison is equal to p=mv=16*35=560 kg ms^-1
The first piece has a momentum of; P1=8*25=200kgms^-1
The second piece has a momentum of; P2=5*30= 150 kg ms^-1
Since p before = p after
560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3

Therefore, the momentum of the third piece; P3=560-350=210 kgms^1
The mass is also conserved so, 16kg = 8kg + 5g + ?
So the mass of P3=16-13= 3kg

Rearranging the equation for momentum in terms of velocity;
p=mv
v=p/m
v=210/3=70 ms^-1 (the speed of the third fragment)

What I am struggling with is actually constructing the vector diagram to show this.
I have attached a rough sketch of my preliminary approach. However, I am rather stuck on how to find the angle of the third fragement. Moreover, would the vector diagram exhibit the momentum of the pieces or rather their velocity? I have shown it with momentum but in hindesight I am uncertain.
I have not had approached a problem where I have to find the missing vector in this way before, typically if I am using a vector diagram for say three vectors I am doing so to find the resultant force and angle.
In this case, would I do so by resolving the momentum vectors into their horizontal and vertical components (given to 3.s.f);
Piecex-componenty-component
P1200 cos 0 = 200200 sin 0 = 0
P2150 cos 50 ~ 96.4150 sin 50 ~ 115
P3210 cos θ210 sin θ
Total 200 + 96.4 + ?0+115 +?
The resultant momentum (560kgms^-1) would be equal to the square root of the total x-component squared + total y-component squared
R =√x^2+y^2
560=√x^2+y^2

I just cannot fathom how to find the angle of the third piece, nor am I certain whether my inital train of thought would be correct. I think I am perhaps overthinking the problem which has caused me some confusion. I would be very grateful of any help or guidance here 😁
 
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  • #2
AN630078 said:
The first piece has a momentum of; P1=8*25=200kgms^-1
The second piece has a momentum of; P2=5*30= 150 kg ms^-1
Since p before = p after
560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3
Momentum is a vector. How do you add vectors?
 
  • #3
jbriggs444 said:
Momentum is a vector. How do you add vectors?
Do you mean algebraically or graphically? If you mean graphically then to add vectors, lay the first one on a set of axes with its tail at the origin. Place the next vector beginning at the previous vector’s end. When there are no more vectors, draw a straight line from the origin to the end of the last vector. This line is the sum of the vectors.
However, would it be that if I drew the vector of P1 and placed vector P2 starting from the end of P1 that then P3 would start at the end of P2. The resultant vector of the momentum would be from the beginning of P1 to the end of P3?
 
  • #4
AN630078 said:
Do you mean algebraically or graphically? If you mean graphically then to add vectors, lay the first one on a set of axes with its tail at the origin. Place the next vector beginning at the previous vector’s end. When there are no more vectors, draw a straight line from the origin to the end of the last vector. This line is the sum of the vectors.
However, would it be that if I drew the vector of P1 and placed vector P2 starting from the end of P1 that then P3 would start at the end of P2. The resultant vector of the momentum would be from the beginning of P1 to the end of P3?
The point is that you do not add them by adding their magnitudes.

You can add them graphically if you have a protractor and a ruler. You can add them using trigonometry. Or you can add them by using their components.
 
  • #5
So would 560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3 be incorrect then?

In which case which method would be most suitable to use here? Would it be adding them by their components or rather graphically from the vector diagram the question is asking you to draw?
 
  • #6
AN630078 said:
So would 560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3 be incorrect then?

In which case which method would be most suitable to use here? Would it be adding them by their components or rather graphically from the vector diagram the question is asking you to draw?
The equation as written is wrong because it is not a vector equation.

##\vec{P_1}## is not correctly specified by a scalar value of 200 kg m/sec. It is a vector. It needs a direction. ##\vec{P_2}## is not correctly specified by a scalar value of 150 kg m/sec. It is a vector. It needs a direction. The original momentum (##\vec{P}##) is not correctly specified by a scalar value of 560 kg m/sec. It is a vector. It needs a direction.

If you had written:$$\vec{P}=\vec{P_1} + \vec{P_2} + \vec{P_3}$$then that would be a correct equation. But now you have to do the vector diagram and solve it.
 
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  • #7
jbriggs444 said:
The equation as written is wrong because it is not a vector equation.

##\vec{P_1}## is not correctly specified by a scalar value of 200 kg m/sec. It is a vector. It needs a direction. ##\vec{P_2}## is not correctly specified by a scalar value of 150 kg m/sec. It is a vector. It needs a direction. The original momentum (##\vec{P}##) is not correctly specified by a scalar value of 560 kg m/sec. It is a vector. It needs a direction.

If you had written:$$\vec{P}=\vec{P_1} + \vec{P_2} + \vec{P_3}$$then that would be a correct equation. But now you have to do the vector diagram and solve it.
Thank you very much for your reply, oh I see my error. Ok I can understand that.
However, I am still uncertain of how to construct the vector diagram and of how to find the direction of P3?
 
  • #8
AN630078 said:
Thank you very much for your reply, oh I see my error. Ok I can understand that.
However, I am still uncertain of how to construct the vector diagram and of how to find the direction of P3?
Let us start from the vector equation given above:$$\vec{P}=\vec{P_1} + \vec{P_2} + \vec{P_3}$$We can start by drawing in the vector ##\vec{P}##. That is the initial momentum of the unbroken piece.

Draw its momentum vector on the middle of a sheet of paper. You know the direction of this vector and you can calculate its magnitude.

Now we want to draw in a representation of the right hand side. Recall that the vector sum of three vectors can be represented by drawing those vectors head to tail, one after another.

So on another sheet of paper, draw the vectors ##\vec{P_1}##, ##\vec{P_2}## and ##\vec{P_3}##. You know the magnitudes and directions for the first two, so those should be straightforward. But you do not know the third. So maybe just put in a short little dotted line and a question mark.

You know by conservation of momentum that the vector ##\vec{P}## that you drew on your first sheet of paper has to match the sum of ##\vec{P_1}##, ##\vec{P_2}## and ##\vec{P_3}## from your second sheet of paper.

So copy that chain of vectors onto your first sheet of paper, starting the chain at same place where ##\vec{P}## starts.

Now you know which way that dotted line for ##\vec{P_3}## has to point. It has to end up at the tip of ##\vec{P}##. That is what it means for ##\vec{P}## to be the sum of the three new vectors.
 
  • #9
jbriggs444 said:
Let us start from the vector equation given above:$$\vec{P}=\vec{P_1} + \vec{P_2} + \vec{P_3}$$We can start by drawing in the vector ##\vec{P}##. That is the initial momentum of the unbroken piece.

Draw its momentum vector on the middle of a sheet of paper. You know the direction of this vector and you can calculate its magnitude.

Now we want to draw in a representation of the right hand side. Recall that the vector sum of three vectors can be represented by drawing those vectors head to tail, one after another.

So on another sheet of paper, draw the vectors ##\vec{P_1}##, ##\vec{P_2}## and ##\vec{P_3}##. You know the magnitudes and directions for the first two, so those should be straightforward. But you do not know the third. So maybe just put in a short little dotted line and a question mark.

You know by conservation of momentum that the vector ##\vec{P}## that you drew on your first sheet of paper has to match the sum of ##\vec{P_1}##, ##\vec{P_2}## and ##\vec{P_3}## from your second sheet of paper.

So copy that chain of vectors onto your first sheet of paper, starting the chain at same place where ##\vec{P}## starts.

Now you know which way that dotted line for ##\vec{P_3}## has to point. It has to end up at the tip of ##\vec{P}##. That is what it means for ##\vec{P}## to be the sum of the three new vectors.
Thank you very much for your reply. I will go and draw this diagram now and then upload a picture...
I am still a little confused, should the vector P3 follow the direction of it as the dotted line, in which case the angle it makes would be between this line and the x-axis? Or would it be from the end of P2 to the end of P (the solid black line marked P3 in my diagram?)
Thank you very much again for your help I very much appreciate it!👍
 

Attachments

  • Vector diagram.JPG
    Vector diagram.JPG
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  • #10
AN630078 said:
Thank you very much for your reply. I will go and draw this diagram now and then upload a picture...
I am still a little confused, should the vector P3 follow the direction of it as the dotted line, in which case the angle it makes would be between this line and the x-axis? Or would it be from the end of P2 to the end of P (the solid black line marked P3 in my diagram?)
Thank you very much again for your help I very much appreciate it!👍
Good work. I would erase the dotted line from that diagram. The solid P1, P2 and P3 are drawn where they should be.

So now you are at the point where you will need to use trigonometry, coordinates or ruler and protractor to figure out ##\vec{P_3}##
 
  • #11
jbriggs444 said:
Good work. I would erase the dotted line from that diagram. The solid P1, P2 and P3 are drawn where they should be.

So now you are at the point where you will need to use trigonometry, coordinates or ruler and protractor to figure out ##\vec{P_3}##
Thank you for your reply. Which method do you think would be most suitable to use here?
If I am using trigonometry wiuld this be the cosine rule?
 
  • #12
AN630078 said:
Thank you for your reply. Which method do you think would be most suitable to use here?
If I am using trigonometry wiuld this be the cosine rule?
I always go with coordinates. Here, you have two vectors with very easy coordinate representations to start with.
 
  • #13
jbriggs444 said:
I always go with coordinates. Here, you have two vectors with very easy coordinate representations to start with.
Ok, so how should I approach doing so? Would it be the components of P1 and P2 ( as below)
Piecex-componenty-component
P1200 cos 0 = 200200 sin 0 = 0
P2150 cos 50 ~ 96.4150 sin 50 ~ 115
P3
Total 200 + 96.4 + ?0+115 +?

or would it be of the traingle formed by P2, P3 and the horizontal side 560-200=360kgms^-1?
 
  • #14
AN630078 said:
Ok, so how should I approach doing so? Would it be the components of P1 and P2 ( as below)
Piecex-componenty-component
P1200 cos 0 = 200200 sin 0 = 0
P2150 cos 50 ~ 96.4150 sin 50 ~ 115
P3
Total200 + 96.4 + ?0+115 +?

or would it be of the traingle formed by P2, P3 and the horizontal side 560-200=360kgms^-1?
You know that the x components of P1, P2 and P3 have to add up to the x component of P.
You know that the y components of P1, P2 and P3 have to add up to the y component of P.

In other words, you already know the Total. You do not have to put a question mark there.
 
  • #15
jbriggs444 said:
You know that the x components of P1, P2 and P3 have to add up to the x component of P.
You know that the y components of P1, P2 and P3 have to add up to the y component of P.

In other words, you already know the Total. You do not have to put a question mark there.
Piecex-componenty-component
P1200 cos 0 = 200200 sin 0 = 0
P2150 cos 50 ~ 96.4150 sin 50 ~ 115
P3
Total560 cos 0 = 560560 sin 0 = 0

x-component of P = 560
P1, P2 and P3 have to add up to the x component of P, therefore;
200+96.4+P3=560
P3=263.6

y-component of P=0
P1, P2 and P3 have to add up to the y component of P, therefore;
0+115+P3=0
P3=-115

How would I proceed to find the angle from this knowledge?
Would the x-componet of P3 be;
210 cos θ = 263.6
And the y-component of P3;
210 sin θ = -115
 
  • #16
AN630078 said:
How would I proceed to find the angle from this knowledge?
Would the x-componet of P3 be;
210 cos θ = 263.6
And the y-component of P3;
210 sin θ = -115
Where did the 210 come from? You never calculated that.

You are going to have a VERY hard time coming up with a value of ##\theta## such that ##210 \cos \theta = 263.6##

Edit: I see it now. You took the magnitude of P and subtracted the magnitudes of P2 and P1: 210 = 560 - 200 - 150.

No. That is wrong. That is what you started out doing and what I started out telling you was wrong. You do not add or subtract vectors that way. The magnitude of the sum is not the sum of the magnitudes.

Let us use trigonometry to find the angle of vector ##\vec{P_3}##. We know its components: ##\vec{P_3}=263.6x - 115y##

Draw that as a right triangle A B C with the point A at the upper left, point B 263.6 units to the right and point C 115 units down from there. The hypotenuse of that triangle is your vector ##\vec{P_3}##.

How long is the hypotenuse? That is the magnitude of ##\vec{P_3}##.
Look at angle A. Can you figure out how to compute it now?
 
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  • #17
jbriggs444 said:
Where did the 210 come from? You never calculated that.

You are going to have a VERY hard time coming up with a value of ##\theta## such that ##210 \cos \theta = 263.6##

Edit: I see it now. You took the magnitude of P and subtracted the magnitudes of P2 and P1: 210 = 560 - 200 - 150.

No. That is wrong. That is what you started out doing and what I started out telling you was wrong. You do not add or subtract vectors that way. The magnitude of the sum is not the sum of the magnitudes.

Let us use trigonometry to find the angle of vector ##\vec{P_3}##. We know its components: ##\vec{P_3}=263.6x - 115y##

Draw that as a right triangle A B C with the point A at the upper left, point B 263.6 units to the right and point C 115 units down from there. The hypotenuse of that triangle is your vector ##\vec{P_3}##.

How long is the hypotenuse? That is the magnitude of ##\vec{P_3}##.
Look at angle A. Can you figure out how to compute it now?
210 was earlier from if the three pieces add to equal the momentum of P:
560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3
P3=210 kg ms^-1

Oh I see that I was wrong. I have tried to draw the triangle ABC, I am not sure if I have done so correctly.
So to find P3, the hypotenuse;
P3=√236.4^2+115^2
P3=√82709.96
P3=287.5933...~288 kgms^-1 to 3.s.f

Then the angle at A;
tan θ = 263.6/115
θ=tan^-1 (263.6/115)
θ=66.4299...~66.4 degrees
 

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  • #18
AN630078 said:
210 was earlier from if the three pieces add to equal the momentum of P:
560 kg ms^-1 = 200kgms^-1 + 150 kg ms^-1 + P3
P3=210 kg ms^-1

Oh I see that I was wrong. I have tried to draw the triangle ABC, I am not sure if I have done so correctly.
So to find P3, the hypotenuse;
P3=√236.4^2+115^2
P3=√82709.96
P3=287.5933...~288 kgms^-1 to 3.s.f

Then the angle at A;
tan θ = 263.6/115
θ=tan^-1 (263.6/115)
θ=66.4299...~66.4 degrees
The angle in the problem statement was an angle from the horizontal. The angle you just computed is an angle from the <what>?
 
  • #19
jbriggs444 said:
The angle in the problem statement was an angle from the horizontal. The angle you just computed is an angle from the <what>?
Sorry, right so would the angle be;

tan θ = 115/263.6
θ=tan^-1 (115/263.6)
θ=23.5700...~23.6 degrees (shown as pink in my diagram)
Would this be the angle we are trying to find or since the horizontal is measured anticlockwise from the x-axis would this be the ange I have shown as white in my diagram, ie. 180-23.6=156.4 degrees?
Also, was the magnitude of P3 I calculated earlier correct?
 

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  • #20
AN630078 said:
Sorry, right so would the angle be;

tan θ = 115/263.6
θ=tan^-1 (115/263.6)
θ=23.5700...~23.6 degrees (shown as pink in my diagram)
Would this be the angle we are trying to find or since the horizontal is measured anticlockwise from the x-axis would this be the ange I have shown as white in my diagram, ie. 180-23.6=156.4 degrees?
Also, was the magnitude of P3 I calculated earlier correct?
Yes, I agree with the magnitude. You correctly used the Pythagorean theorem.

I do not agree with the angle. In which direction does vector ##\vec{P_3}## point?

Personally I would have drawn the triangle with the sharp angle on the upper left, the right angle in the upper right and the less sharp angle on the lower right. That is how I suggested that you draw the triangle in post #16 above.
 
  • #21
jbriggs444 said:
Yes, I agree with the magnitude. You correctly used the Pythagorean theorem.

I do not agree with the angle. In which direction does vector ##\vec{P_3}## point?

Personally I would have drawn the triangle with the sharp angle on the upper left, the right angle in the upper right and the less sharp angle on the lower right. That is how I suggested that you draw the triangle in post #16 above.
Thank you for your reply. The vector points downwards in a SE direction. Ok, sorry I am having a little trouble visualising this, could you perhaps draw the triangle as you mean, only if it is not too much trouble.
 
  • #22
AN630078 said:
Thank you for your reply. The vector points downwards in a SE direction. Ok, sorry I am having a little trouble visualising this, could you perhaps draw the triangle as you mean, only if it is not too much trouble.
You have a vector ##\vec{P_3}##. You know its coordinate value: (263.6, -115).

Here is how I would draw the triangle:

1599331106642.png
 
  • #23
jbriggs444 said:
You have a vector ##\vec{P_3}##. You know its coordinate value: (263.6, -115).

Here is how I would draw the triangle:

View attachment 268913
Oh oh thank you that has helped tremendously. So we would be trying to find the angle between the x-axis;
So, tan θ = 115/263.6
θ=tan^-1 (115/263.6)
θ=23.5700...~23.6 degrees
Would this be measured anticlockwise to the x-axis;
360-23.6 =336.4 degrees?
 

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  • #24
You could call it -23.6 or +336.4 or "23.6 degrees below the horizontal".
 
  • #25
jbriggs444 said:
You could call it -23.6 or +336.4 or "23.6 degrees below the horizontal".
@jbriggs444 Ok, thank you for your help. I have tried to answer the question fully on one piece of paper, which I have attached to show my workings. Would the vector diagram showing P1,P2,P3 and P be correct because having found the magnitude of P3 to be 288kgms^-1 this does not correspond to the length of P3 in the diagram, which I believe is significantly greater.
Would my diagram and solutions now be correct?
 

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  • #26
AN630078 said:
...
Would my diagram and solutions now be correct?

It seems to me that the vector diagram you show in this last post can’t be correct.
Sorry, I can’t read the rest of the text and calculations in the picture.

The sequence of after-explosion momentum vectors should start at the tail of the before-explosion momentum vector and end up at its head.

The reason is that 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.

momentum3.jpg


momentum21-1024x445.png
 
  • #27
Lnewqban said:
It seems to me that the vector diagram you show in this last post can’t be correct.
Sorry, I can’t read the rest of the text and calculations in the picture.

The sequence of after-explosion momentum vectors should start at the tail of the before-explosion momentum vector and end up at its head.

The reason is that 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.

View attachment 268942

View attachment 268943
Thank you for your reply. Right, so if the sequence of after-explosion momentum vectors should begin at the end of the before-explosion momentum vector and finish up at its head would it appear as in the diagram attached?

Also, would you like me to rewrite my previous text and calculations in a clearer format?
 

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  • vector diagram before-after collision.JPG
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  • #28
This last diagram (post 27) shows two fragments moving up, which contradicts the natural fact that 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.

What is wrong with your diagram of post #9?
I don’t know about the magnitudes and directions of the vectors, but the sequence seems to be correct.

For the magnitude and direction of the requested velocity of fragment #3, I would solve these equations for ##V_{x_3}## and for ##V_{y_3}##

##\sum P_x=mV_{x~Before~explosion}=(m_1V_{x_1})+(m_2V_{x_2})+(m_3V_{x_3})##

##\sum P_y=mV_{y~Before~explosion}=0=(m_1V_{y_1})+(m_2V_{y_2})+(m_3V_{y_3})##
 
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  • #29
Lnewqban said:
This last diagram (post 27) shows two fragments moving up, which contradicts the natural fact that 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.

What is wrong with your diagram of post #9?
I don’t know about the magnitudes and directions of the vectors, but the sequence seems to be correct.

For the magnitude and direction of the requested velocity of fragment #3, I would solve these equations for Vx3 and for Vy3

∑Px=mVx Before explosion=(m1Vx1)+(m2Vx2)+(m3Vx3)

∑Py=mVy Before explosion=0=(m1Vy1)+(m2Vy2)+(m3Vy3)
Thank you for your reply. Would the error in my diagram in post 9 be that P1 begins at the same place as P (the Origin)?

Oh oh, if it is incorrect that two fragments move up how would I show that P2 is at an angle of 50 degrees as stated in the question?

Thank you for providing those equations, how would I find the velocity of P3 then? Would I use v=p/m?Or would I proceed as follows?
Px before explosion = m*Vx = 16*cos 35=13.1 kg ms^-1
Px before explosion = (m1Vx1)+(m2Vx2)+(m3Vx3)
Px before explosion =(8*cos0)+(5*cos50)+(3cos θ)
Px before explosion =8+3.21+3cos θ
13.1=11.21+3cos θ
1.89=3cos θ
cos θ =1.89/3
θ=cos^1 (1.89/3)
θ=50.9 degrees to 3.s.f

I do not think this is correct though, sorry I am a little confused.
 
  • #30
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.

Simultaneously, an observer flying in formation with the CM would see three fragments flying away from that CM and from each other at different speeds and directions.
 
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  • #31
Lnewqban said:
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 😁👍
 

Attachments

  • vector diagram p1,p2,p3.JPG
    vector diagram p1,p2,p3.JPG
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  • #32
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.
 
  • #33
AN630078 said:
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 :smile:

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.
AN630078 said:
"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.

AN630078 said:
... 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
m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}.

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.”
 
Last edited:
  • #34
Lnewqban said:
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?

Piecex-componenty-component
P1200 cos 0=200200 sin 0=0
P2150 cos 50 =96.4150 sin 50=115
P3??
P560 cos 0=560560 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 😊👍
 

Attachments

  • Vector Diagram new.JPG
    Vector Diagram new.JPG
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  • #35
Happy to help, I know that you are tenaciously trying. :smile:

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.
 

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