A body of mass `m_(0)` is placed on a smooth horizontal surface . The mass of the body is decreasing exponentially with disintegration constant , `lambda` . Assuming that the mass is ejected backwards with a relative velocity u . If initially the body was at rest , the speed of body at time t is

Answer

`v = v_(0)lambdat`

Explanation

`N_(1)=M_(0)e^(-lambdat)`
Thus momentum of this mass
`=Mv_(1) v=` velocity of mass at time `t=M_(0)e^(-lambdat) v`
Let at time `dt` defect mass is elected with relative velocity `v_(0) : v_() -v_(0)+v`
Linear momentum of M at time `(t+dt)`
`=(M-dm)(v+dv)+dm (v-v_(0))`
Since `f_("ext")=0`
`(M-dm) (v+dv)+dm (v-v_(0))=Mv implies (M-dm) dv=dmv_(0)`
`implies dv=(v_(0)dm)/((M-dm))implies (dv)/(dt)=(dm)/(dt) (v_(0)/(M-dm))=(r.v_(0))/(M-rdt)`
`v=v_(0) "ln" (M_(0))/(M_(0)-rt)`. But `M_(0)-rt =M_(0) e^(-lambdat)`
`v=v_(0)"in" M_(0)/(M_(0)e^(-lambdat))=v_(0) lambda t`