Billiard ball A of mass mA = 0.115 kg moving with speed vA = 2.80 m/s strikes ball B , initially at rest, of mass mB = 0.144 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s , and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.
Part A
Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the x direction.
Taking the x axis to be the original direction of motion of ball A, choose the correct equation expressing the conservation of momentum for the components in the x direction.
0=mAv′Asinθ′A−mBv′Bsinθ′B | |||||||||
mAvA=mAv′Acosθ′A+mBv′Bcosθ′B | |||||||||
mAvA=mAv′Acosθ′A−mBv′Bsinθ′B | |||||||||
0=(mAvA+mBv′B)sinθ′B Part B Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the y direction.
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mAvA=mAv′Acosθ′A+mBv′Bcosθ′B
part b )
0=mAv′Asinθ′A−mBv′Bsinθ′B
because initally there is no y component
ma*vai = ma*vaf + mb*vbf
vbf = ma/mb*vai - ma/mb*vaf
vbf = ma/mb*(vai - vaf)
vbf = ma/mb*(vai - vaf)
vai = 2.8i + 0j m/s
vaf = 2.10 m/s * (cos(30)i + sin(30)j) = 1.819i + 1.05j m/s
vbf = 0.115 kg / 0.144 kg * ((2.8i + 0j) m/s - (1.819i + 1.05j) m/s)
vbf = 0.115/0.144*(0.981i - 1.05j m/s)
vbf = 0.783i - 0.838 j
part c )
theta = tan^-1(0.838/0.783) = 46.94 degree = 47 degree
part d )
vb = sqrt(vx^2 + vy^2)
vb = sqrt(.783^2 + .838^2)
vb = 1.147 m/s
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