Question

Вагон массой 30 т, движущийся по горизонтальному питии со скоростью 1,5 м/с, автоматически на ходу сцепляется с неподвижным вагоном массой 20 т. С какой скоростью движется сцепка.

Answer 1

Задача на закон сохранения импульса.
До сцепки импульс равен  MV, после сцепки (M+m)V₁
MV = (M+m)V₁ ⇒ V₁ = MV/(M+m) = 1,5·30/(30+20) = 0,9 м/сек.

Ответ: 0.9 м/сек.

Answer 2

Calculation of the Final Speed of the Coupled Wagons

To determine the final speed of the coupled wagons, we can use the principle of conservation of momentum. According to this principle, the total momentum before the coupling is equal to the total momentum after the coupling.

The momentum of an object is given by the product of its mass and velocity. Therefore, we can calculate the initial momentum of the first wagon and the final momentum of the coupled wagons and equate them to find the final velocity.

Let's denote the mass of the first wagon as m1 (30 tons) and its initial velocity as v1 (1.5 m/s). The mass of the second wagon is m2 (20 tons), and its initial velocity is 0 m/s since it is stationary.

Using the principle of conservation of momentum, we have:

m1 * v1 = (m1 + m2) * vf

where vf is the final velocity of the coupled wagons.

Substituting the given values, we get:

30 tons * 1.5 m/s = (30 tons + 20 tons) * vf

Simplifying the equation:

45 tons * m/s = 50 tons * vf

Dividing both sides by 50 tons:

0.9 m/s = vf

Therefore, the final velocity of the coupled wagons is 0.9 m/s.

Please note that the calculation assumes an ideal scenario without considering factors such as friction or air resistance.