Question

Вагон массой 40 тонн цепляется с неподвижным вагоном массой 60 тонн. С какой скоростью движется сцепка по горизонтальному участку дороги, если до удара первый вагон имел скорость 0,6 м/с

Answer 1

По закону сохранения импульса m1*V1=9m1+m2)*V
V=m1*V1/(m1+m2)=40*0,6/(040+60)=0,24 м/с

Answer 2

Calculation of the Speed of the Coupling

To calculate the speed of the coupling between a wagon with a mass of 40 tons and a stationary wagon with a mass of 60 tons, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The equation for conservation of momentum is:

m1 * V1 = (m1 + m2) * V

Where: - m1 is the mass of the first wagon (40 tons) - V1 is the initial velocity of the first wagon (0.6 m/s) - m2 is the mass of the second wagon (60 tons) - V is the final velocity of the coupling

By substituting the given values into the equation, we can solve for V:

V = (m1 * V1) / (m1 + m2)

V = (40 * 0.6) / (40 + 60)

V = 24 / 100

V = 0.24 m/s

Therefore, the coupling moves with a speed of 0.24 m/s on the horizontal section of the track.