oubaas
m8*V'8*cos 28+m6*V'6*cos 29.5 = m8*V8 (*)
m8*V'8*sin 28= m6*V'6*sin 29.5 (**)
V'6 = m8*V'8*sin 28/(m6*sin 29.5)
since m8 = m6
V'6 = V'8*sin 28/sin 29.5 = 0.953V'8
substituting V'6 in equation (*) and cancelling masses (since m8 = m6)
V'8*0.883+0.953V'8*0.870 = 3.7
V'8(0.883+0.870) = 3.7
V'8 = 3.7/(0.883+0.870) = 2.1 m/sec
V'6 = 2.11*0.953 = 2.0 m/sec
P'8 = 2.1*0.5 = 1.05 kg*m/sec
P'6 = 2.0*0.5 = 1.0 kg*m/sec
?
(a) conserving vertical momentum yields
0 = m*U*sinΘ - m*V*sinΦ → where V is the 6-ball's velocity
and U is the 8-ball's velocity (post-collision)
so U = V*sinΦ / sinΘ
conserving horizontally yields
m*v = m*U*cosΘ + m*V*cosΦ
cancel mass and substitute for U --
v = (V*sinΦ / sinΘ)*cosΘ + V*cosΦ
v = V*[sinΦ/tanΘ + cosΦ]
V = v / [sinΦ/tanΘ + cosΦ]
(b) V = 3.7m/s / [sin29.5º/tan28º + cos29.5º] = 2.06 m/s
round to 2.1 m/s
(c) U = 2.06m/s * sin29.5º/sin28º = 2.16 m/s
round to 2.2 m/s