# Collisions in two dimensions

1. Aug 7, 2010

### rvnt

1. The problem statement, all variables and given/known data
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.

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

3. 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

2. Aug 8, 2010

### Staff: Mentor

To solve for the angle, use a double angle trig identity:
sin(2θ) = 2 sinθ cosθ

3. Aug 8, 2010

### rvnt

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. Aug 8, 2010

### Staff: Mentor

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.

That looks OK.
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. Aug 8, 2010

### rvnt

Oh right! k now I got it thank-you