Question

Железнодорожный вагон массой 35 тонн подъезжает к стоящему на том же пути неподвижному вагону массой 28 тонн и автоматически сцепляется с ним после сцепки вагоны движутся прямолинейно со скоростью 0,5 метров в секунду какова была скорость вагона массой 35 тонн перед сцепкой?

Answer 1

 по ЗСИ: p1+p2=p; m1v1+m2v2=(m1+m2)v;  35000v+0=63000 *0,5; v=31,5/35; v=0,9м/с

Answer 2

Calculation of the Speed of the 35-ton Wagon Before Coupling

To calculate the speed of the 35-ton wagon before coupling, 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 calculated by multiplying its mass by its velocity. In this case, we have two wagons: one with a mass of 35 tons and an unknown velocity, and the other with a mass of 28 tons and a velocity of 0.5 meters per second.

Let's denote the velocity of the 35-ton wagon before coupling as v.

The total momentum before the coupling is given by the sum of the momentum of the 35-ton wagon and the momentum of the 28-ton wagon:

Total momentum before coupling = (mass of 35-ton wagon) × (velocity of 35-ton wagon) + (mass of 28-ton wagon) × (velocity of 28-ton wagon)

Using the given values, we can write the equation as:

35 × v + 28 × 0.5 = (mass of 35-ton wagon + mass of 28-ton wagon) × (velocity after coupling)

Simplifying the equation, we have:

35v + 14 = 63 × (velocity after coupling)

Since the velocity after coupling is given as 0.5 meters per second, we can substitute this value into the equation:

35v + 14 = 63 × 0.5

Solving for v, we get:

35v + 14 = 31.5

35v = 31.5 - 14

35v = 17.5

v = 17.5 / 35

v = 0.5 meters per second

Therefore, the speed of the 35-ton wagon before coupling was 0.5 meters per second.