Answer to Question #6918 in Mechanics | Relativity for Caley
So, we consider a case when after the collision, two cars are move forward without a
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 correct.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!