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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?

1 Answer

7 votes

Answer:

The carriage speed B was 67.4 mph

Step-by-step 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 =
p_(fx)

m v₁ₓ + 0 = (m + M) vₓ

Y axis

poy =
p_(fy)

0 + M
v_(2y) = (m + M)
v_(y)

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

sin 66 =
v_(y) / v

cos 66 = vₓ / v


v_(y) = v sin 66

vₓ = v cos 66

We substitute and write the system of equations

m v₁ₓ = (m + M) v cos 66

M
v_(2y) = (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


v_(2y) = (m + M)/m v sin 66


v_(2y) = (1000 + 1500) /1000 29.5 sin 66


v_(2y) = 67.4 mph

The carriage speed B was 67.4 mph

User Harry Blue
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