Billiard ball A of mass. Ma=0.400kg moving with speed Va=1.80m/s strikes ball B,
ID: 1976058 • Letter: B
Question
Billiard ball A of mass. Ma=0.400kg moving with speed Va=1.80m/s strikes ball B, initially at rest, of mass Mb=0.500kg. As a result of the collision, ball A is deflected off at an angle of 30.0(degree) with a speed V'a=1.10m/s. (a) Taking the x-axis to be the original direction of motion of ball A, write down the equations expression the conservation of momentum for the components in the x and y directions separately. (b) Solve these equations for the speed V'b and angle Theta'b of ball B. Do not assume the collision is elastic
Explanation / Answer
Relevant 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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