Why do we not have to calculate momentum here?

Question

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 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?

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?

Answer 1

The calculated 18.566 m/s is not the final velocity in the y-direction,

but the initial velocity in the y-direction, i.e., v_2i.

And you correctly concluded that the final velocity in the y-direction must

be half the initial velocity in the y-direction.

Let

v_1i = 13 m/s be the velocity of the East-bound car,

v_2i (to be calculated) be the velocity of the North-bound car,

v_f be the final velocity,

m be the common mass,

α = 55° be the angle of v_f.

Conservation of momentum in the x- and y- direction are

mv_1i = 2mv_fcos(α)

mv_2i = 2mv_fsin(α)

Divide the second equation by the first to get

v_2i/v_1i = tan(α)

That is why

v_2i = v_1itan(α)