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A body consists of equal masses M of inflammable and non-inflammable material. The body descends freely under gravity from rest. The combustible part burns at a constant rate of kM per second where k is a constant. The burning material is ejected vertically upwards with constant speed u relative to the body, and air resistance may be neglected. Show, using momentum considerations, that d/dt[(2-kt)v]=k(u-v)+g(2-kt) where v is the speed of the body at time t. Hence show that the body descends a distance (g/2k^2)+(u/k)(1-ln2) before all the inflammable material is burnt.

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The equal masses of inflammable and non-inflammable materials are determined.

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Q1. A body consists of equal masses M of inflammable and non-inflammable material. The body descends freely under gravity from rest. The combustible part burns at a constant rate of kM per second where k is a constant. The burning material is ejected vertically upwards with constant speed u relative to the body, and air resistance may be neglected. Show, using momentum considerations, that d/dt[(2-kt)v]=k(u-v)+g(2-kt) where v is the speed of the body at time t. Hence show that the body descends a distance (g/2k^2)+(u/k)(1-ln2) before all the inflammable material is burnt.

Solution : Total initial mass of the body = 2M
Rate of burning of the combustible part = kM
Mass of the body at any time t = 2M - kMt

As the combustible part of the body burns, total mass of the body decreases gradually and its velocity increases. During an ...

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