Two cars collide at an intersection. Car A, with a mass of 2000{\ m kg} , is goi

Question

Two cars collide at an intersection. Car A, with a mass of 2000{ m kg} , is going from west to east, while car B, of mass 1300{ m kg} , is going from north to south at 16.0{ m m}/{ 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 60.0^circ south of east from the point of impact.

Part A-How fast were the enmeshed cars moving just after the collision?

Explanation / Answer

Anyway, define our coordinate system: East is the +x direction North is the +y direction I am more familiar with numeric labeling, so car A is car 1...and car B is car 2. Initial velocity of each car in terms of speed: v1i = V1* v2i = V2* Total initial momentum: p = m1*v1i + m2*v2i The cars have an inelastic collision, where conservation of momentum applies, and end moving at a final velocity vf: p = (m1 + m2)*vf Thus: m1*v1i + m2*v2i = (m1 + m2)*vf Solve for vf: vf = (m1*v1i + m2*v2i)/(m1 + m2) Remember this is a vector equation: vf = In terms of the given angle, it implies that: tan(theta) = -vfy/vfx Which means that: tan(theta) = (m2*V2/(m1 + m2))/(m1*V1/(m1 + m2)) And it simplifies: tan(theta) = m2*V2/(m1*V1) So, V2 = m1*V1*tan(theta)/m2 And we can substitute: vf = And combine with Pythagorean theorem to find Vf, the post-impact speed: Vf = sqrt(vfx^2 + vfy^2) Vf = sqrt((m1*V1)^2 + (m1*V1*tan(theta))^2)/(m1 + m2) Simplifies: Vf = m1*V1*sqrt(1 + tan(theta)^2)/(m1+m2) Summary: Vf = m1*V1*sqrt(1 + tan(theta)^2)/(m1+m2) V2 = m1*V1*tan(theta)/m2 Data: m1:=2000 kg; m2:=1300 kg; V1:=16.0 m/s; theta:=60 deg; Substitute the values in above equation :)

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