A neutron with mass ?m? makes a head-on, elastic collision with a nucleus of mass M?, which is initially at rest. (a) Show that if the neutron’s initial kinetic energy is K?0, the kinetic energy that it loses during the collision is 4?mMK?0/(?M? + ?m?)2. (b) For what value of ?M? does the incident neutron lose the most energy? (c) When ?M has the value calculated in part (b), what is the speed of the neutron after the collision?

Solution 93P Step 1: (a). = 0 1 2 Let us consider m = 1and m = M 2 2 K = neutrons eergy lost = energy gain of nucleus = 1/2 MV 2 Since u = 0 u = v v 2 1 2 1 2 K = 1/2 M(u + V 1 1 2 2 K = 1/2 MV 1 (1 + (m M/m + M)) Since V = (m m /m + m )u ,so 1 1 2 1 2 1 2 2 K = 1/2 Mu (2m 1m + M) K = 1/2 Mu 2 4mM/(m + M) 2 1 * 4mMk /(m 0 M) 2 Where k = 1/2 mu 2 0 1 Step 2: (b).value of M does the incident neutron lose the most energy 2 K = 4mMk /(m 0 M) d(K)/dM = 4mMk (m + 0) 4mMk 2 0 *2(M + m)/(M + m) 4 4mMk (m + M) = mMk 0 0 m + M = 2M M = neutron lost more energy. 1