An unstable atomic nucleus of mass 1.58 times 10^-26 kg initially at rest disint

Question

An unstable atomic nucleus of mass 1.58 times 10^-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.08 times 10^-27 ky, moves in the y direction with a speed of 6.00 times 10^6 m/s. Another particle, of mass 8.50 times 10^-27 kg, moves in the x direction with a speed of 4.00 times 10^6 m/s. (a) Find the velocity of the third particle. -15740000 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. i + -14110000 Your response Is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake In your calculator Carry out all Intermediate results to at least four-digit accuracy to minimize roundoff error. J) (b) Find the total kinetic energy Increase in the process. J

Explanation / Answer

initial momentum of the system Pi = 0

after disintegration

final momentum Pf = m1*v1 + m2*v2 + m3*v3

m1 = 5.08*10^-27 kg

m2 = 8.5*10^-27 kg

m3 = M - (m1+m2) = (1.58*10^-26) - (5.08*10^-27+8.5*10^-27 )

m3 = 2.22*10^-27 kg

v1 = 6*10^6 i m/s

v2 = 4*10^6 j m/s

0 = (5.08*10^-27*6*10^6)i + (8.5*10^-27*4*10^6)j + (2.22*10^-27*v3)

v3 = -13.7*10^6 i - 15.3*10^6 j m/s    <<<<---------answer

=================

kinetic energy

Ktot = (1/2)*m1*v1^2 + (1/2)*m2*v2^2 + (1/2)*m3*v3^2

v3 = sqrt((13.7*10^6)^2+(15.3*10^6)^2) = 20.4 *10^6 m/s

Ktot = ((1/2)*5.08*10^-27*(6*10^6)^2) + ((1/2)*8.5*10^-27*(4*10^6)^2) + ((1/2)*2.22*10^-27*(20.4 *10^6)^2)

Ktot = 6.21*10^-13 J <<<<----------answer

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