First, find the magnitude of v , that is, the speed v of the two-car unit after the collision.?

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.

Answer 1

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)

Answer 2

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.

Answer 3

The bottom one is the correct one for mastering chemistry
v = √[(m1v1)² + (m2v2)²] / (m1+m2)

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