# Collisions in two dimensions

## 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

Doc Al
Mentor
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?

Doc Al
Mentor
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.

Oh right! k now I got it thank-you