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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

chakrala mounica , 15 Years ago

Grade 11

1 Answers

vikas askiitian expert

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

Last Activity: 15 Years ago

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Mechanics> collisons...

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

chakrala mounica , 15 Years ago

vikas askiitian expert

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