# Homework Help: Nasty Momentum problem

1. Mar 26, 2005

### 2sonian

I'm running into trouble how to conceptualize and determine my reference frames to make the solving of this problem easy:

It's a variation on the "exploding shell" problem, except that an intial velocity of 300 m/s is given to the particle. Here's the setup:
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A plane has exploded into three pieces. At the instant of the explosion, the plane was flying at an altitude of 1808 ft (551.079 m) at 671.08 mph (300.000 m/s) heading due north.

Two pieces of the plane were recovered.
Half the plane was recovered 5000 meters due north of the explosion point
One quarter of the plane was recovered 3000 meters away from the explosion point, at a compass heading of 240° (i.e. 60° south of due east from the explosion point)
Each piece was found at an elevation of approx. 200ft, due to the topography of the area, you can assume that the last piece will be found at a similar elevation.

It is believed that the plane broke up into only three pieces, so only one piece is left. Using the information given, can you find the last piece?

Based on the reports at the time, it was a calm, clear windless day and due to the special nature of the plane, wind resistane is not a factor, and all three pieces impacted simultaneously.

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So I figured that it took 10.0012 seconds for the pieces to impact the ground, but am not sure why I would need to know this at all.
I generally am thinking that this is a conservation of linear momentum problem. I that I could take the two momentum vectors (the one due north and the southeasterly one) and sum them together and then add a third one so I get the overall momentum to be zero. The magnitude and the direction of this third vector, would seem to me to necessarily be symmetrical to the southeasterly momentum vector since the first vector lands along the same line of flight as before the explosion.
Something in the back of my mind says this is too simple an approach and that I'm missing something here.

Any ideas on how to get this set up properly and determine the final piece location?

Thanks
2sonian

2. Mar 26, 2005

you have the right idea. The only problem you will run into is the velocity of the plane just befoure explosion. The third piece will be tricky to locate since the plane was not stationary befoure explosion.

Regards,

3. Mar 27, 2005

### whozum

That right, since momentum is conserved in all directions, the third piece must have landed in a western direction. Sum all 3 momentum vectors so the net momentum is 0, and that should give you the direction of the 3rd piece. If you know the momentum of the third piece you can figure out where it landed using the time of free fall

4. Mar 27, 2005

### 2sonian

So I'm still not clear on this or just being vastly stupid. If I sum the vectors to zero, then I get a vector for the missing piece that is of the same magnitude as the second piece and in a direction that is 60° south of due west. How does freefall time matter at that point?

--2sonian--

5. Mar 28, 2005

### plusaf

not stupid at all.....

but, like many other entries i've read here, it looks like you're trying to find something complicated inside something simple....

looking at the original proglem....
=============================
A plane has exploded into three pieces.
At the instant of the explosion, the plane was flying due north.
Two pieces of the plane were recovered.
Half the plane was recovered 5000 meters due north of the explosion point
One quarter of the plane was recovered 3000 meters away from the explosion point, at a compass heading of 240° (i.e. 60° south of due east from the explosion point)
Each piece was found at an elevation of approx. 200ft, due to the topography of the area, you can assume that the last piece will be found at a similar elevation.
It is believed that the plane broke up into only three pieces, so only one piece is left. Using the information given, can you find the last piece?
It was a windless day and due to the special nature of the plane, wind resistance is not a factor, and all three pieces impacted simultaneously.
=============================

simply by conservation of momentum, if the plane breaks into three pieces and one piece (the half) is recovered due north of the explosion when the plane had been flying due north, the other two halves must have taken equal amounts of any sideways momentum away from the explosion.

if the first piece was half the plane and the second piece was 1/4 of the plane, the last piece must also have been 1/4 of the plane.

if the first "fourth" went 60 degrees south of due east, the other "fourth" had doggone well must have gone 60 degrees south of due west.

and it's nice that the plane was flying due north over perfectly flat land and that none of the pieces were affected by wind or air resistance as they fell. that's the part that makes the rest easy.

try picturing it before you pull out the equations.
some people are visual, some are auditory, some see equations in their heads.... none is better or worse or good or bad; just go with the technique that works best for you.

me? very visual, as you might glean from most of my suggestions for solutions here.....

6. Mar 30, 2005

### lektor

Hi, i gave this one a go in my spare time, would i be able to get a confirmation that the distance is 3000m 60 degrees south of due west?

7. Mar 31, 2005

### 2sonian

8. Mar 31, 2005

### BobG

What a cool idea!!

Your first problem is this: North is 0 degrees, East is 90 degrees, South is 180 degrees, West is 270 degrees. In otherwords, 240 degrees is not 60 degrees South of East.

Each piece has a velocity of 300 m/s due North from the plane's initial velocity prior to the explosion. The 10 sec to reach the ground allows you to calculate how much momentum was added by the explosion. In other words, how far would the 1/2 piece have travelled without the explosion? How far and what direction did it actually travel? The difference is due to the momentum added by the explosion. You have to do the same for known quarter piece.

The sum of the momentum due to the explosion has to equal zero, which allows you to find the direction and velocity the explosion sent the missing piece. Then you have to add the piece's initial velocity prior to the explosion back in to figure out the actual path of the final piece.

Edit: Is it no surprise that the guy who did find it used a slide rule? I'm going to have to check to see if anyone does that around where I live - we certainly have the right type of area for it.

Last edited: Mar 31, 2005
9. Apr 5, 2005

### plusaf

ah, help me understand this......

so, the explosion blows the half of the airplane forward along the same path it was traveling, north. it also blows 1/4 of the plane off to one side, slowing it down [reducing its velocity], so that that piece lands 3000 meters away, 60 degrees south of due east of the explosion point.

are you saying that the final fourth doesn't land 3000 meters away and 60 degrees south of due west???

i never used the heading of 240 degrees in my solution statement. if 60 degrees south of due east is where one of the "fourths" was found, you should find the "final fourth" piece one due west, by conservation of momentum! yes, 240 degrees heading isn't 60 degrees south of due east, so either the 240 is misstated or the 60 south of east is misstated. i assumed the 60 south of east was correct. if 240 degrees is correct (60 degrees east of south, not south of east, then the final piece would be found 60 degrees west of south, not south of west. (120 degrees heading, right?)

+af

10. Apr 5, 2005

### whozum

3 pieces break.

Ignoring the half that went due north, we have momentum change in the direction perpendicular to the path of travel:

Piece one went 60 degrees south of due east, and fell 3000m away.

Piece two, assuming it has the same mass MUST have gone the same distance, in an angle 60 degrees south of due west.