How Do You Determine the Velocity and Angle of a Billiard Ball After Collision?

AI Thread Summary
The discussion focuses on determining the velocity and angle of billiard ball B after it collides with ball A, using conservation of momentum equations. The initial conditions include ball A's mass and speed, and its deflection angle and speed post-collision. Participants analyze the equations derived from momentum conservation in both x and y directions, identifying an error in the calculations related to the angle of deflection. The correct approach involves using the relationship between sine and cosine to solve for the angle and subsequently the speed of ball B. Clarifications and corrections lead to a better understanding of the problem-solving process.
rvnt
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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
 
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To solve for the angle, use a double angle trig identity:
sin(2θ) = 2 sinθ cosθ
 
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?
 
rvnt said:
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.

rvnt said:

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.
 
Oh right! k now I got it thank-you
 
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