# Answer to Question #6918 in Mechanics | Relativity for Caley

Question #6918

A car with a mass of 2 895kg is standing still when it is hit by a car with a mass of 2269kg from behind after the collision car 1 notices that they move forward at a speed of 36km/h. Car 1 claims that car 2 must have been driving faster than 60km/h spped limit. Is the driver of car 1 correct? Prove by calculations

Expert's answer

If after collision the driver of 1 car noticed that they move forward at the constant speed (36 km/h), then they move without a friction.

So, we consider a case when after the collision, two cars are move forward without a

friction.

Lets denote:

the mass of the first car is m1 = 2895 kgthe velocity of the first car before collision is v1 = 0 km/hthe mass of the second car is m2 = 2269 kgthe velocity of the second car is v2 and it's unknown the velocity of the two cars after collision is v3 = 36 km/hand the mass of two cars that move forward together after the collision is m3 = m1+m2

Due to a Law of conservation of momentum we have:

m1*v1 + m2*v2 = m3*v3

v1 = 0, so

m2*v2 = m3*v3

v2 = (m3/m2)*v3

because of m3 = m1+m2, we can write:

v2 = (m1/m2 + 1)*v3

v2 ~ 82 km/h

So, the driver of car 1

So, we consider a case when after the collision, two cars are move forward without a

friction.

Lets denote:

the mass of the first car is m1 = 2895 kgthe velocity of the first car before collision is v1 = 0 km/hthe mass of the second car is m2 = 2269 kgthe velocity of the second car is v2 and it's unknown the velocity of the two cars after collision is v3 = 36 km/hand the mass of two cars that move forward together after the collision is m3 = m1+m2

Due to a Law of conservation of momentum we have:

m1*v1 + m2*v2 = m3*v3

v1 = 0, so

m2*v2 = m3*v3

v2 = (m3/m2)*v3

because of m3 = m1+m2, we can write:

v2 = (m1/m2 + 1)*v3

v2 ~ 82 km/h

So, the driver of car 1

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