# The rocket body weighing 200 grams contains 300 grams of gunpowder.

**The rocket body weighing 200 grams contains 300 grams of gunpowder. Determine the speed of the gas release if the speed of the rocket is 400m / s. consider the combustion of gunpowder instantaneous**

mk = 200 g = 0.2 kg.

mp = 300 g = 0.3 kg.

Vk “= 400 m / s.

Vп “-?

Let us write down the law of conservation of momentum for a closed boy-boat system: mk * Vk + mp * Vp = mk * Vk “+ mp * Vp”, where mk, mp are the mass of the hull and gunpowder, respectively, Vk, Vk “- the speed of the hull before and after combustion of gunpowder, Vp, Vp “- the speed of the powder before and after combustion.

Since at the initial moment of time the rocket is at rest Vk = Vp = 0 m / s, then 0 = ml * Vl “+ mm * Vm”.

mk * Vk “= – mp * Vp”.

Vp “= – mk * Vk” / mp.

The sign “-” means that the speed of the rocket is directed in the opposite direction of the speed of emission of gases.

Vp “= 0.2 kg * 400 m / s / 0.3 kg = 266.6 m / s.

Answer: the speed of the rocket is Vk “= 266.6 m / s.