In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic.
initial p = final p
√((m1v1)² + (m2v2)²) = (m1+m2)v
So your answer looks good except for some paren issues. I’d write
v = sqroot( (m1*v1/(m1+m2))^2+(m2*v2/(m1+m2))^2 )
or just v = √((m1v1)² + (m2v2)²) / (m1+m2)
The collided vehicles circulate off at some attitude ø (enable’s say measured from the vertical) Momentum is conserved in each course. The east-west momentum (after collision) is then (m1+m2)*v*sinø and the north-south momentum is (m1+M2)*v*cosø. The east-west momentum of the finished equipment until now collision is purely that of the eastward vehicle, p1 = m1*v1, and in addition for the north-south momentum p2 = m2*v2; summing until now and after momenta for each course: m1*v1 = (m1*m2)*v*sinø m2*v2 = (m1+m2)*v*cosø you have 2 equations in 2 unknowns, v and ø; you may resolve.
The bottom one is the correct one for mastering chemistry
v = √[(m1v1)² + (m2v2)²] / (m1+m2)