Answer to Question #157299 in Classical Mechanics for mehrab hossain

Question

A spaceship of mass m has velocity v in the positive x direction of an inertial reference

frame. A mass dm is fired out the rear of the ship with constant exhaust velocity (-v0) with

respect to the spaceship. a) using conservation of momentum, show that dv/v0 = dm/m, b)

By integration, find the dependence of v on m if v1 and m1 are the initial values. c) Can the

acceleration be constant if dm/dt, the burning rate is constant.

Answer 1

Given,

The mass of the spaceship be m and velocity be v.

a) So, initial momentum of the spaceship be $(p)=mv$

Now, as the dm mass fired from the spaceship, then also momentum always be conserve

$\frac{dp}{dt}=0$

$\Rightarrow \frac{d}{dt}(mv)=0$

$\Rightarrow \frac{dm}{dt}v+m\frac{dv}{dt}=0$

$\Rightarrow \frac{dm}{dt}v=-\frac{dv}{dt}m$

$\frac{dm}{m}=-\frac{dv}{v}$

b) $\frac{dm}{dt}v+m\frac{dv}{dt}$

but here gravity is also working in downward direction,

$\int_{m_{1}}^{m-dm} \frac{dm}{m}=\int_{v_1}^v \frac{dv}{v}$

$\Rightarrow [\ln m]_{m_1}^{m_1-dm}=[\ln{v}]_{v_1}^v$

$\Rightarrow \ln[m_1-dm]-\ln[m_1]=\ln[v]-\ln[v_1]$

$\Rightarrow \frac{m_1-dm}{m_1}=\frac{v}{v_1}$

c)$m_1-dm=\frac{v}{v_1}m_1$

$dm=m_1-\frac{vm_1}{v_1}$

if dm/dt = constant

then $\dfrac{dv}{dt}=-g+k\frac{v'}{m}$