A neutron with mass 1.67*10^-27 kg moving with speed 45 km/s makes a head-on col

Question

A neutron with mass 1.67*10^-27 kg moving with speed 45 km/s makes a head-on collision with a boron nucleus, originally at rest, with mass 1.66*10^-26 kg.

A) If the collision is completely inelastic, so the particles stick together, what is the final kinetic energy of the system, expressed as a fraction of the original kinetic energy? Express it as a decimal.

B) If the collision were perfectly elastic, what fraction of the original kinetic energy is transferred to the boron nucleus? Express it as a decimal.

Explanation / Answer

A) final speed of the particles after the collision is V= (1.67*10^-27 *45000)/[(0.167+1.66)*10^-26] = 4113.3 = 4.11 km/s

then final kinetic energy of the system is KEf = 0.5*(1.67+16.6)*10^-27*4113.3*4113.3 = 1.54*10^-19 J

as a fraction of original kE is (0.5*1.67*10^-27*45000*45000)/(1.54*10^-19) = 1.69*10^-18 /(1.54*10^-19 ) = 10.9

KEf = original KE/10.9
--------------------------------------------------------------

Speed of neutron after collision is (1.67-16.6)*10^-27*45000/[(1.67+16.6)*10^-27] = -36.77 km/s

KE transfered to boron is = (0.5*1.67*10^-27*36.77*36.77*10^6) -(1.69*10^-18) = (1.129-1.69*10^-18) = 0.561*10^-18 J is transfered to boron

required answer is (1.69-0.561)/1.69 = 0.668 of the original kE is trasfered to boron

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