Plz help me how to find angle "a" , and how do the log function is dissapeared from U/ Vo?

sorry for the late solution....

here i use the technique of momentum conservation.....at any given time let us shift our x-axis along the line of the velocity vector of the particle.....

maan le ki at this time the velocity is parallel to our x-axis....after a time 'dt' the space ship emmits a mass of say Adt where A is the rate of decrement of mass..... and let us assume the spaceship tilts by an angle d@ after the emmision of the small mass Adt......the new mass of the spaceship = (m-Adt)....the new velocity be incremented by a value of dv and become(V+dV)

now we have the new scenario as : the velocity = (V + dV)

new mass = (m-Adt)

reletive inclination= d@

here the initial momentum along the y-axis= 0 (as the spaceship was moving along the x-axis)

final momentum = (Adt)U - { (m-Adt) (V + dV) d@}................. (multiplying by d@ we get vertical component of vel)

P_initial = P_final      or

we have AdtU = mVd@ + mdVd@ + AVdtd@

but we know dat mdVd@ + AVdtd@  tends completely to 0 so we neglect it......

and thus have AUdt = mV d@

but m(t) = m - At

integrating we have  ₀S^(m-M)/A dt/(m- At)  = ₀S^$ (V/UA) d@

where $ is the final angle of deviation (Sorry for usage of S for integration as my formula edittor isnt working)

integrating and substituting the limits we have our final $ = (U/V) log{ m/M}......

Hope it helps....

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