An unstable atomic nucleus of mass 1.78 Times 10^-26 kg initially at rest disint

Question

An unstable atomic nucleus of mass 1.78 Times 10^-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.17 Times 10^_27 kg, moves along the y axis with a speed of 6.00 x 106 m/s. Another particle, of mass 8.48 Times 10^-27 kg, moves along the x axis with a speed of 4.00 Times 10^6 m/s. Find the velocity of the third particle. (Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.? + j) Find the total kinetic energy increase in the process.

Explanation / Answer

a) initially nucleus was at rest so initial momentum was zero.

ans disintegration happens due to internal forces hence momentum will be conserved.

initial = final momentum = 0

finally:

v1 = 6 x 10^6 j m/s

v2 = 4 x 10^6i m/s

so m1v1 + m2v2 + m3v3 = 0

(5.17 x10^-27 x 6 x 10^6 j ) + (8.48 x 10^-27 x 4 x 10^6i) + (( 17.8 - 5.17 - 8.48) x 10^-27 x v3 ) = 0

v3 = 8.17 x 10^6 i - 6.89 x 10^6 j

b) speed of third particle.

v3 =sqrt [ 8.17^2 + 6.89^2] x 10^6 = 10.68 x 10^6 m/s

increase in KE = m1v1^2 /2 + m2 v2^2 /2 + m2 v3^2 /2

= 3.97 x 10^-13 J

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