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??...Part B:How fast was car A going just before the collision

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.

from the second diagram, to find the momentum of car (A + B) we make use of the trigonometric formula cos θ = (momentum of car B) / (momentum of car (A +B))

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

To find out how fast car A was moving we have to find its momentum from the second diagram using Pythagoras theorem.

momentum of A = [tex]\sqrt{momentum of (A+B)^{2} - momentum of B\\^{2} }

momentum of A = [tex][tex]\sqrt{29,444.87^{2}- 25500^{2}}[/tex]

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