A car weighing 20 tons moves at a speed of 0.5 m / s catching up with a car weighing 30

A car weighing 20 tons moves at a speed of 0.5 m / s catching up with a car weighing 30 tons moving at a speed of 0.2 m / s What is the speed of the cars after elastic interaction.

By the condition of the problem:

m1 = 20 tons = 20,000 kg – mass of the first car;

v1 = 0.5 m / s – speed of the first car;

m2 = 30 tons = 30,000 kg – the mass of the second car;

v2 = 0.2 m / s – speed of the second car.

It is required to find the speed v3 of the cars after elastic interaction.

According to the law of conservation of momentum (momentum):

m1 * v1 + m2 * v2 = (m1 + m2) * v3, hence

v3 = (m1 * v1 + m2 * v2) / (m1 + m2) = (20,000 * 0.5 + 30,000 * 0.2) / (20,000 + 30,000) =

16,000 / 50,000 = 0.32 m / s.

Answer: the speed of the cars after elastic interaction is 0.32 m / s.