The stoplight had just changed and a 1800 kg cadillac had entered the intersection, heading north at 4.0 m/s , when it was struck by a 1200 kg eastbound volkswagen. the cars stuck together and slid to a halt, leaving skid marks angled 35∘ north of east. how fast was the volkswagen going just before the impact?

8.6 m/s

Explanation:

Assuming no external forces present during the collision, total momentum must be conserved.

As momentum is a vector, if we decompose it in two perpendicular components (one N-S and the other W-E), total momentum must be conserved for both directions.

Prior to collision, the only component in the eastbound direction was due to the Volkswagen:

px = mvolk. Vvolk = 1,200 kg. Vvolk

After the collision, we have a component along this axis, which is the projection of the  momentum vector due to both cars stuck together, 35º north of the east:

        pxf = (mvolk + mcad) . vf . cos 35º (1)

With this same reasoning, we can write the following equation for the momentum on the N-S axis:

py1 = 1,800 kg. 4.0 m/s = (1,800 + 1,200) kg. Vf. sin 35º (2)  

Dividing both sides in (2) and (1), we get:

1,800. 4.0 / 1,200. Vvolk = tg 35º

Solving for Vvolk:

Vvolk= 7,200 / 1,200. tg 35º = 8.6 m/s

15.3 cos 65 degrees trig

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