Question

Железнодорожный вагон массой 36 т, движущийся со скоростью 1 м/с, подъезжает к стоящей на том же пути платформе массой 24 т и автоматически сцепляется с ней. Определить скорость движения платформы и вагона после их сцепки.

Answer 1

Закон сохранения импульса:

m1V1 = (m1+m2)V = > V = (m1V1)/(m1+m2)

V = 3,6 * 10^4/6 * 10^4 = 0,6 м/c.

Answer 2

Calculation of the Speed of the Platform and the Wagon after Coupling

To determine the speed of the platform and the wagon after they are coupled, we can apply 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 a wagon with a mass of 36 tons (36,000 kg) moving at a velocity of 1 m/s, and a platform with a mass of 24 tons (24,000 kg) initially at rest.

Let's denote the velocity of the platform after the coupling as v and the velocity of the wagon after the coupling as V.

The total momentum before the coupling is given by:

Total momentum before = (mass of wagon) x (velocity of wagon) + (mass of platform) x (velocity of platform before)

Total momentum before = (36,000 kg) x (1 m/s) + (24,000 kg) x (0 m/s)

The total momentum after the coupling is given by:

Total momentum after = (mass of wagon + mass of platform) x (velocity of wagon after coupling)

Total momentum after = (36,000 kg + 24,000 kg) x (V)

According to the principle of conservation of momentum, the total momentum before the coupling is equal to the total momentum after the coupling:

(36,000 kg) x (1 m/s) + (24,000 kg) x (0 m/s) = (36,000 kg + 24,000 kg) x (V)

Simplifying the equation:

(36,000 kg) + 0 = (60,000 kg) x (V)

V = (36,000 kg) / (60,000 kg) = 0.6 m/s

Therefore, the speed of the platform and the wagon after their coupling is 0.6 m/s.

Please note that the above calculation assumes an ideal scenario without any external forces or friction.