Time to be a crash investigator! two cars are in an accident: car a was driving due east when car b, driving due north, ran a stop sign and collided with his car, causing an inelastic collision (the cars stuck together). the skid marks from the accident go off at a 66 degree angle above the horizontal (east). a traffic camera witnessed car a going 30 mph prior to the crash, and both cars traveling 29.5 mph after the crash. unfortunately, the camera was angled so it did not see car b prior to the crash.
car a has a mass of 1000 kg, and car b has a mass of 1500 kg.
the driver of car b claims that he was driving the 30 mph speed limit, and didn’t see the stop sign because of foggy weather. driver a claims driver b was speeding and that is why he missed the stop sign.
how fast was car b traveling prior to the crash?

The carriage speed B was 67.4 mph

Explanation:

This is an exercise for the moment, that as a vector we must look for the solution of each axis (x, y). We define a system formed by the two cars, for this system the forces during the crash are internal, so the moment is preserved.

The data they give is the car more A m = 1000kg and its speed is v1₁₀ = 30 mph i^ and the mass of the car B M = 1500 kg

Let's write the moment for each axis

X axis

     p₀ₓ =

     m v₁ₓ + 0 = (m + M) vₓ

Y axis

     poy =

      0 + M = (m + M)

Let's look for the components of the final velocity with trigonometry

     sin 66 =  / v

     cos 66 = vₓ / v

       = v sin 66

     vₓ = v cos 66

We substitute and write the system of equations

     m v₁ₓ = (m + M) v cos 66

     M  = (m + M) v sin66

From the first equation

    v = m / (m + M) v₁ₓ / cos 66

    v = 1000 / (1000 + 1500) 30 / cos 66

    v = 29.5 mph

From the second equation

     = (m + M)/m   v sin 66

   = (1000 + 1500) /1000     29.5 sin 66

     = 67.4 mph

The carriage speed B was 67.4 mph

this would be about 43%