A spacecraft with a mass of 10 tons moves at a speed of 9 km / s. During braking, 1450 kg of combustion products

A spacecraft with a mass of 10 tons moves at a speed of 9 km / s. During braking, 1450 kg of combustion products were thrown out of it with a speed of 3 km / s relative to the hull in the direction of motion. Determine the speed of the ship 100 m / s, determine the speed of the rocket.

mk = 10 t = 10000 kg.

Vк = 9 km / s.

mt = 1450 kg.

Vt / k = 3 km / s.

Vк “-?

Let us find the speed of the fuel Vt relative to the stationary frame of reference: Vt = Vt / k + Vk, since the fuel flies out in the direction of movement of the ship, it decelerates.

Vt = 3 km / s + 9 km / s = 12 km / s.

For a closed system: rocket-fuel, the law of conservation of momentum is valid.

mk * Vk – impulse before the release of gases from the ship.

(mk – mt) * Vk “- the momentum of the ship after the release of gases.

mk * Vk = (mk – mt) * Vk “+ mt * Vt.

Vk “= (mk * Vk – mt * Vt) / (mk – mt).

Vк “= (10000 kg * 9 km / s – 1450 kg * 12 km / s) / (10000 kg – 1450 kg) = 8.5 km / s.

Answer: the speed of the rocket when braking is Vk “= 8.5 km / s.