The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 43 mph straight into a 3 ton SUV traveling at 13mph in the other direction. The car and the SUV crumple from the collision, and stick together. What is the final velocity, in MPH (you don't need to enter MPH in your answer), of the SUV/car entanglement (the positive direction is the direction the car was initially going) ?

Question

Physics

Courtney

12 April, 10:52

The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 43 mph straight into a 3 ton SUV traveling at 13mph in the other direction. The car and the SUV crumple from the collision, and stick together. What is the final velocity, in MPH (you don't need to enter MPH in your answer), of the SUV/car entanglement (the positive direction is the direction the car was initially going) ?

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Answers (1)

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Kenna Garrett · Answer 1

vcs = 1 mph (in the direction the car was initially going)

Explanation:

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

As the movement is along one dimension, we can treat to momentum as a scalar, looking only to the magnitude, taking into account that both vehicles were moving in opposite directions.

So, we can put the following:

p₀ = p₁ ⇒ mc * vc - ms*vs = (mc + ms) * vcs

Replacing by the values of the masses and the speeds of both vehicles, and solving for the final speed of the SUV/car entanglement, we have:

vcs = (1 ton*43 mph - 3ton*13 mph) / 4 ton

⇒ vcs = (43 ton*mph - 39 ton*mph) / 4 ton = 1 mph