The spontaneous decay of a heavy atomic nucleus of mass M = 16.0 10 27 kg that i

Question

The spontaneous decay of a heavy atomic nucleus of mass M = 16.0 1027 kg that is initially at rest produces three particles. The first particle, with mass m1 = 6.64 1027 kg, is ejected in the positive x direction with a speed of 1.90 107 m/s. The second particle, with mass m2 = 3.10 1027 kg, is ejected in the negative y direction with a speed of 2.00 107 m/s.

(a) What is the velocity of the third particle? (Enter the magnitude.)? __________ m/s

(b) What is the increase in the kinetic energy of the system after the decay? _________J

Explanation / Answer

m1 = 6.4 x 10^-27 kg

v1 = 1.90 x 10^7 i

m2 = 3.10 x 10^-27 kg

v2 = -2 x 10^7 j

m3= (16 - 6.64 - 3.10) x 10^-27 = 6.26 x 10^-27 kg

Applying momentum conservatiom,

0 = m1 v1 + m2 v2 + m3 v3

putting values,

v3 = -1.94 x 10^7 i + 0.99 x 10^7 j m/s

(A) magniude , |v3| = sqrt(1.94^2 + 0.99^2) = 2.18 x 10^7 J

(B) Ki = 0

Kf = m1 v1^2 /2 + m2 v2^2 /2 + m3 v2^2 / 2

Kf = 3.26 x 10^-12 J

increase = Kf - Ki = 3.26 x 10^-12 J

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