A vessel at rest explodes, breaking into three pieces

In summary, the explosion is going to be symmetric and result in the sum of the vectors being zero. The magnitude and angle of the final speed will be determined by the angle and magnitude of the vectors when the explosion started.
  • #1
Deleote
2
0
1. Ok, I've been working on this problem for ever and it seems easy but I know that there is something wrong.
A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 90 m/s. One goes along the negative x-axis and the other along the negative y-axis. The third piece has three times the mass of each other piece. What are the direction and magnitude of its velocity immediately after the explosion? a) What is the angle with respect to the x-axis?
b)What is the magnitude of its velocity?
for A) I understand that the value is 45 degrees however I do not exactly understand why it is 45 degrees. and for B) I have no idea where to start...I know that momentum is involved but that's about it.


2. P=mv ? not sure where to go from there.
 
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  • #2
Leave the formulae aside for a moment and try to intuit the problem. Then the formulae should make more sense.

The explosion is gonig to be symmetric. Since it started with 0 momentum, the sum of the pieces will also have to be 0 momentum. i.e. everything will have to cancel out. The vector sum of the known pieces m(a) and m(b) has to be equal and opposite to the unknown third piece's m(c) vector to maintain 0 momentum. This means it has to be opposite in both angle AND magnitude.


Code:
[FONT="Courier New"]       [COLOR="SeaGreen"][B]O[/B][/COLOR] m(c)
m(a)[COLOR="Red"]  /[/COLOR]
  90 [COLOR="Red"]/[/COLOR]
[COLOR="seagreen"][B].[/B][/COLOR]---*
[COLOR="silver"]|[/COLOR]  [COLOR="Blue"]/[/COLOR]|
[COLOR="silver"]|[/COLOR] [COLOR="Blue"]/[/COLOR] | 90  
[COLOR="silver"]|[/COLOR][COLOR="Blue"]/[/COLOR]  |
 [COLOR="silver"]---[/COLOR][COLOR="seagreen"][B].[/B][/COLOR]  m(b)


[COLOR="Blue"]/[/COLOR] = sum of vectors of known m(a) and m(b)
[COLOR="Red"]/[/COLOR] = vector of unknown m(c)

angle of [COLOR="Red"]/[/COLOR] must be equal & opposite angle of [COLOR="Blue"]/[/COLOR]
magnitude of [COLOR="Red"]/[/COLOR] must be equal to magnitude of [COLOR="Blue"]/[/COLOR] [B]times the ratio of masses[/B]
[/FONT]

Momentum comes in when we calculate the magnitude of the unknown vector. m(c)'s momentum will have to exactly balance the sum of the momenta of the m(a) and m(b). So m(c)v(c) = m(a)v(a)+m(b)v(b). Tripling the object's mass while maintaining the same momentum will require the object to go one third as far.
 
Last edited:
  • #3
At least, that seems right. I only studied this in High School - 30 years ago...

How'm I doin? :rolleyes:
 
  • #4
Thats what I believed to be true also but that would yield an answer of 60 m/s for the magnitude of the final speed which is incorrect...
I attempted to solve using that method...3m(c)v(c)=m(a)*90 m/s + m(b)*90 m/s.
 
  • #5
You need to incorporate the angles. Again intuitively: if A and B flew directly apart from each other, C's velocity would have to be zero. (A and B already satisfy the conservation law, any C velocity would violate it.) Conversely, if A and B flew off in exactly the same direction, then we would have DaveC426913's equation above, and C's speed would be 60 m/s.

Since neither of the above is exactly the case here, you're probably going to have a sine or cosine somewhere in the mix. Make sense?
 

1. What causes a vessel at rest to explode?

There are several potential causes for a vessel at rest to explode, including buildup of combustible gases, structural weaknesses, and external sources of heat or pressure. It is important to properly maintain and monitor vessels to prevent these potential hazards.

2. Can a vessel at rest spontaneously explode?

While it is possible for a vessel to spontaneously explode due to extreme pressure, this is rare and typically only occurs in very specific circumstances. In most cases, there is a specific cause for the explosion.

3. How much force is generated when a vessel at rest explodes?

The force generated during an explosion can vary greatly depending on the size and type of vessel, as well as the specific circumstances leading up to the explosion. In general, the force can be quite significant and can cause extensive damage.

4. Can a vessel at rest explode without any warning signs?

In some cases, there may be warning signs such as unusual noises or vibrations, changes in temperature or pressure readings, or visible leaks or cracks. However, in other cases, a vessel may explode without any warning signs, making it important to regularly inspect and maintain vessels to prevent accidents.

5. What are the potential risks and consequences of a vessel at rest exploding?

The risks and consequences of a vessel exploding can be severe, including injury or death to nearby individuals, damage to surrounding property, and environmental hazards. It is essential to follow proper safety protocols and regulations when working with vessels to minimize these risks.

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