# 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.

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