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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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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 ...

Solution Summary

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

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