Question #157299

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.


1
Expert's answer
2021-01-25T14:05:20-0500

Given,

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

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

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

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


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


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


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


dmm=dvv\frac{dm}{m}=-\frac{dv}{v}


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

but here gravity is also working in downward direction,

m1mdmdmm=v1vdvv\int_{m_{1}}^{m-dm} \frac{dm}{m}=\int_{v_1}^v \frac{dv}{v}


[lnm]m1m1dm=[lnv]v1v\Rightarrow [\ln m]_{m_1}^{m_1-dm}=[\ln{v}]_{v_1}^v

ln[m1dm]ln[m1]=ln[v]ln[v1]\Rightarrow \ln[m_1-dm]-\ln[m_1]=\ln[v]-\ln[v_1]

m1dmm1=vv1\Rightarrow \frac{m_1-dm}{m_1}=\frac{v}{v_1}

c)m1dm=vv1m1m_1-dm=\frac{v}{v_1}m_1

dm=m1vm1v1dm=m_1-\frac{vm_1}{v_1}

if dm/dt = constant

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


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