Chetan Mandayam Nayakar
Last Activity: 14 Years ago
Let p be momentum of bullet; p2 =2(M/n)E, therefore p=√(2(M/n)E). from law of conservation of momentum, this is equal to momentum og gun(P). therefore, P= √(2Me),(e is energy of gun), p = P implies= √(2(M/n)E) = =√(2Me). therefore e=E/n
therefore total energy = energies of gun+bullet = E+(E/n) =(1+ (1/n))E
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