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chakrala mounica Grade: 11
        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)

vikas askiitian expert
510 Points
										

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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