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A neutron traveling with a velocity u and kinetic energy E.collides perfectly elastically head-on with the nucleusof an atom of mass number A at rest. The fraction of total energy retained by neutron is
detailed solution
Correct option is A
Let the final velocity of neutron be v1 According to theory of collisionv1=m1−m2m1+m2u1+2m2m1+m2u2 Here u1=u and u2=0. Further m1=1 and m2=A ∴ v1=1−A1+Au Final K.E. of neutron =12mv12 =12m1−A1+A2u2 Initial K.E. of neutron =12mu2 ∴ Fraction of energy retained =1−A1+A2Talk to our academic expert!
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Three particles each of mass m are located at vertices of an equilateral triangle ABC. They start moving with equal speeds each along the meridian of the triangle and collide at its center G. If after collisions A comes to rest and B returns its path along GB, then C
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