2D elastic collisions with equal masses

  • Thread starter blader324
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  • #1
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Hi everyone, i was just wondering, when two objects of the same mass collide (in an elastic collision) the angles that they bounce off from should combine to 90 degrees. right? so i'm trying to prove this, but i'm a little stuck. i used vector component equations, but i arrived at a little problem. i was just wondering if anyone knew that if the velocities of two objects after the collision multiplied to equal zero...what does that mean?
 

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  • #2
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when two objects of the same mass collide (in an elastic collision) the angles that they bounce off from should combine to 90 degrees. right?

i couldnt understand. could you show a diagram or something


if the velocities of two objects after the collision multiplied to equal zero...

this too
 
  • #3
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i don't know how to draw a diagram on this thing.
 
  • #4
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Hi everyone, i was just wondering, when two objects of the same mass collide (in an elastic collision) the angles that they bounce off from should combine to 90 degrees. right?

Really? Let's say two the two objects (of same mass) head for each other. After an elastic collision, they retrace their paths. How would you calculate the angles and what do they add up to in my example?


Do you have a more general problem at hand? If so, post that question, we will try to help.
 
  • #5
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it's a two dimensional collision so they deflect at an angle off of each other...and i don't have a general problem...i have to prove that they bounce off at a combined angle of 90 degrees
 
  • #6
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i havent heard of that, it depends upon the path of approach.
if it was so as you said, people wouldnt be playing a game called snooker. try watching a game and see if the combined angle is 90
 
  • #7
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okay...here's the situation. object a is moving with an intial velocity of whatever...object b is at rest. they collide elastically and in 2 dimensions and bounce off of each other...the combined angles that they bounce off each other should be 90 degrees...maybe that'll help clear things up?
 
  • #8
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duh.. why wud it be 90 degrees???
assuming that the path of approach of object a passes through the centre of object b, object b will go in the direction of motion of object a and object a will retrace its path(provided the masses of the objects allow this condition), change in the direction of object a is 180 degrees and object b is... duh what was the initial direction of motion of b???
 
  • #9
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okay...i get what you're saying...but they're not colliding in a straight line...cause otherwise what you're saying would make sense. object b is at rest initially (before it's hit)
 
  • #10
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Okay, I didn't quite notice the 2D-part. Sorry.

It's an elastic collision. What is its definition? With the given information can you arrive at a simple relation between initial and final velocities of the objects?

Next, use the conservation of momentum and the above result to show what you have been asked for. (It's just algebra)
 
  • #11
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i did all that already...and then i get the final velocity of A and B equal to zero
 
  • #12
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Could you show your work? We could make corrections at places where you may have done something wrong.
 
  • #13
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ugh...nevermind...it's really hard for me to put the equations up...but thank you so much for your time and effort anyways...i have to get going to my class right now...hopefully a classmate can help me figure it out!

thanks so much again!!!! THANK YOU THANK YOU THANK YOU THANK YOU!
 
  • #14
Doc Al
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i did all that already...and then i get the final velocity of A and B equal to zero
I presume you mean that you get that the dot (scalar) product of the final velocity vectors equal zero: [itex]\vec{V}_a \cdot \vec{V}_b = 0[/itex]. If so, you're done since that implies that the angle between the vectors is 90 degrees.
 
  • #15
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Are You Serious!!!!
 
  • #16
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but how does that work?
 
  • #17
Doc Al
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That's a fact, jack. :smile:
 
  • #18
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this is my first time taking physics, can you explain why the angle between those two vectors is 90 degrees, with the fact that vector quantities multiply to be zero
 
  • #19
Doc Al
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but how does that work?
The scalar product of the vectors [itex]\vec{A} \cdot \vec{B} = AB\cos\theta[/itex]. That can only be zero if theta = 90 (or A or B equals zero, which is not the case here).
 
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  • #20
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thank you so much sooooooooooooooooooooooooooo much Doc Al...you just helped me sooo omuch....take care and once again...THANK YOU!!!!!
 
  • #21
Doc Al
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My pleasure. (Next time, stick to one thread! :wink:)
 
  • #22
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Psst,doc... theta = pi/2. ;)
 
  • #23
Doc Al
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Psst,doc... theta = pi/2. ;)
D'oh! Thanks, neutrino. I'll fix that! :redface: (I left out the 9 in 90.)
 
  • #24
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First all, I think this result applies only if one of the balls is initially at rest, right? You can then prove the result by applying the conservation laws of energy - since it's an elastic collision - and of momentum.

Conserving energy gives you one equation in terms of the magnitudes of the three velocities (initial velocity of the one moving ball and two final velocities), and conserving momentum gives you a vector equation relating the three velocities. Combine the two equations and do a little algebra and you get the result you're looking for, i.e. the dot product of the two final velocities is zero.

And by the way, for those posters who were wondering about the collision headed directly for the center of the stationary ball, you can work out what final velocities must be in that case, and you'll see that the ball that was initially moving will be at rest and the initially stationary ball will move with the initial velocity. The dot product is still zero, so this result is consistent with the statement that the velocities are perpendicular.
 
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