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				   a neuton of mass m collides with a stationery deutron of mass 2m ,the percentage loss of k.e. of colliding neutron is

6 years ago


Answers : (1)


in every collision momentam is conserved

  (Pn)i +0 = (Pn)f + (Pd)f ......1                           (Pn,Pd are momentam of proton and deutron)

    by using KE=p^2/2m

sqrt{(KE)ni.Mn}  = sqrt{(KE)nf.Mn} + sqrt{(KE)df.Md}             ((KE)n,(KE)d are kinetic energies of neutron and deutron)  


  sqrt{(KE)ni}  = sqrt{(KE)nf  + sqrt2{(KE)df}..............2            


now applying conservation of KE

   (KE)ni = (KE)nf  +  (KE)df...........3

after solving 2 and 3

(KE)nf =0 so (KE)df=(KE)ni/sqrt2

            (KE)ni - x%(KE)ni=(KE)ni/sqrt2

    x=29.37% ans



6 years ago

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