Two cars collide at an intersection. car a, with a mass of 1900 kg, is going from west to east, while car b, of mass 1500 kg, is going from north to south at 17.0 m\s. as a result of this collision, the two cars become enmeshed and move as one afterwards. in your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60.0degrees south of east from the point of impact. part a was: how fast were the enmeshed cars moving just after the collision? i got 8.66 for velocity in part a which was correct but i can't figure out part b: how fast was car a going just before the collision

8.66 m/s and  7.75 m/s

Explanation:

from the question we were given

mass of A = 1900 kg

mass of B = 1500 kg

velocity of B = 17 m/s

direction of A and B after collision = 60 degrees south of east.

and we can get the momentum from the parameters above

momentum = mass x velocity

momentum of car B = 1900 x 17 = 25,500 kgm/s

NOTE: the first diagram shows the cars and the direction they move at, while the second diagram shows the cars and their directions arranged to form a triangle.

  The momentum of the cars is the parameter that would be used in calculations for the triangle.

therefore momentum of car (A+B) = (momentum of car B) / (cos θ)

momentum of car (A+B) = 25,500 / cos 30

                              = 29,444.87 kgm/s

recall that momentum = mass x velocity

therefore velocity of (A+B) = mass of (A+B) / momentum of (A+B)

                     velocity (A+B) = 3,400 / 29,444.87

                                              = 8.66 m/s

momentum of A =

  momentum of A = 14,722.44 kgm/s

recall that momentum = mass x velocity

therefore velocity of car A = momentum / mass

                = 14,722.44 / 1,900

                   = 7.75 m/s

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