dan815
  • dan815
calculate the probability of collision Suppose there are 2 circles of radius r, placed at the mid points along the lengths of a rectangle like the picture, What is the probability of collision between the circles before either circle hits a boundary?, the Particles can have any direction, and they must keep the direction until collision. Both particles travel at same speed
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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dan815
  • dan815
|dw:1441174469829:dw|
imqwerty
  • imqwerty
will we consider gravity ?
ganeshie8
  • ganeshie8
|dw:1441175324916:dw|

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ganeshie8
  • ganeshie8
I think the locus of intersection points is the perpendicular bisector of the line segment joining those two bubbles
dan815
  • dan815
yeah we need to consider that mid line only i think too, since this is case where the speed is equal
dan815
  • dan815
|dw:1441175465518:dw|
ganeshie8
  • ganeshie8
yeah that angle should be same
dan815
  • dan815
now with in this angle space itself we should find the prob and then see what the total prob is
dan815
  • dan815
i think we should narrow it down to like how many angle different is possible such that a collision is still possible
dan815
  • dan815
**how much angle difference**
dan815
  • dan815
sometimes i m surprised at my own english hehe
ganeshie8
  • ganeshie8
|dw:1441175544779:dw|
dan815
  • dan815
it doesnt have to be a perfect collision though, it can be off course as long as they touch
ganeshie8
  • ganeshie8
cant we treat the bubbles as point masses
ganeshie8
  • ganeshie8
oh they gave radius ok
dan815
  • dan815
this could be a useful one too like a circle, i think there might be some really neat way to solve this one
dan815
  • dan815
|dw:1441175998926:dw|
imqwerty
  • imqwerty
ummm....i am little confused but is this the answer- probability=L/[2(W+L)]
dan815
  • dan815
there is some modular form of solution for this problem if u limit the angles these circles can travel to 2pi*k/n , where 0
dan815
  • dan815
tbh i dont know yet qwerty id have to see your full work
dan815
  • dan815
here is one way of approaching it
dan815
  • dan815
|dw:1441176643347:dw|
dan815
  • dan815
now there is a perfect collision for every infinitessimal angle by the proportional gets smaller with some factor
dan815
  • dan815
as the perfect collision surface area thins out with the angle from the center
dan815
  • dan815
|dw:1441176750326:dw|
imqwerty
  • imqwerty
what i did was - like if we have a system like this -|dw:1441176400369:dw| and we are asked that what is the probability that a ball dropped from above falls in the smaller square....then the probability is - (ar. of small square)/(ar. of big square) umm wait i think my answer was not correct but can we follow such an approach?
dan815
  • dan815
ya it is definately something simliar to that
dan815
  • dan815
the only main thing to consider is how the difference in angle changes
dan815
  • dan815
as u move above the straight line
dan815
  • dan815
for imperfect collisions
ganeshie8
  • ganeshie8
thats really a good idea for simplicity maybe treat bubbles as point masses first define two random variables : \(\theta_1, \theta_2\)
dan815
  • dan815
|dw:1441176897902:dw|
dan815
  • dan815
yah that might actually still be equal if we just use point masses in the end
ganeshie8
  • ganeshie8
so the probability for perfection collision is \(0\) because we don't get any area |dw:1441177185300:dw|