Collisions in two dimensions

  • Thread starter rvnt
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  • #1
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Homework Statement


Billiard ball A of mass mA= 0.400kg moving with speed vA=1.80m/s strikes ball B initially at erst, of mass mB=0.500kg.As a result of the collision,ball A is deflected off at an angle of 30.0degrees with a speed v'A=1.10m/s. Taking the x-axis as the original direction of motion of ball A, solve for the speed v'B and angle Θ'B of ball B. Do not assume the collision is elastic.


Homework Equations


px conserved: mAvA=mAv'ACosΘ'A+mBv'BCosΘ'B
py conserved: mAvA=mAv'AsinΘ'A+mBv'BSinΘ'B



The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
-1.5407=CosΘ'BSinΘ'B
 

Answers and Replies

  • #2
Doc Al
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To solve for the angle, use a double angle trig identity:
sin(2θ) = 2 sinθ cosθ
 
  • #3
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Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
 
  • #4
Doc Al
Mentor
45,084
1,394
Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
My bad. I didn't read your work carefully and just looked at the last line. You were on the right track until that point. That last line was an error.

The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
That looks OK.
-1.5407=CosΘ'BSinΘ'B
Not sure how you got this from the previous line. Note:
-0.44=(0.677897/CosΘ'B)SinΘ'B = (0.677897)(SinΘ'B/CosΘ'B) = (0.677897)(TanΘ'B)

That will give you the angle; then you can find the speed.
 
  • #5
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Oh right! k now I got it thank-you
 

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