- #1

as2528

- 40

- 9

- Homework Statement
- Two automobiles of equal mass approach an intersection. One vehicle is traveling with a velocity of 13.0 m/s toward the east and the other is traveling north with a speed of v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0° north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth?

- Relevant Equations
- arctan(fy/fx)=theta

m1viy+m2v2iy=(m1+m2)vfy

I calculated:arctan(fy/13.0)=55=>fy=18.566 m/s

Then I calculated, using the momentum equation:

m1viy+m2v2iy=(m1+m2)vfy=>

mv2i=2*m*18.566=>v2=37.132 m/s

I thought that because the cars were stuck together, the kinetic energy from the northbound car would be lost. So, the speed would have lowered after the accident. However, the answer is 18.566 m/s. Why do we only have to calculate the arctan? Would the speed not drop after the strike due to the mass being larger since the cars are stuck, and since momentum is conserved, the car would have been traveling at 37.132 m/s before the strike?

Then I calculated, using the momentum equation:

m1viy+m2v2iy=(m1+m2)vfy=>

mv2i=2*m*18.566=>v2=37.132 m/s

I thought that because the cars were stuck together, the kinetic energy from the northbound car would be lost. So, the speed would have lowered after the accident. However, the answer is 18.566 m/s. Why do we only have to calculate the arctan? Would the speed not drop after the strike due to the mass being larger since the cars are stuck, and since momentum is conserved, the car would have been traveling at 37.132 m/s before the strike?