Physics hw help please?

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