(a) Find the velocity of the third particle. m/s (b) Find the total kinetic ener

Question

(a) Find the velocity of the third particle.

m/s

(b) Find the total kinetic energy increase in the process.
J

An unstable atomic nucleus of mass 1.54 j) m/s (b) Find the total kinetic energy increase in the process. J i +___ 106 m/s. (a) Find the velocity of the third particle. (___ 10-27 kg, moves in the x direction with a speed of 4.00 multiply.gif 106 m/s. Another particle, of mass 8.48 10-27 kg, moves in the y direction with a speed of 6.00 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.20

Explanation / Answer

Its given that,

nuclii Mass = M = 1.54 x 10-26 kg and V = 0 (as it is at rest) so its momentum = 0

After disintegration into three particles

m1 = 5.20 x 10-27 kg and v1 = 6.0 x 106 m/s

m2 = 8.48 x 10-27 kg and v2 = 4.0 x 106 m/s

Mass od third particle will be m3 = (M - (m1 + m2) = 1.54 x 10-26 - (  5.20 x 10-27 + 8.48 x 10-27 )

m3 = 1.54 x 10-26 - 13.68 x 10-27 = (1.54-1.368) x 10 -26 = 0.18 x 10-26 kg

In the process the total moemtum is conserved. So,

m1v1 +m2v2 +m3v3 = 0, this gives us

v3 =- (m1v1 +m2v2) / m3

=- 1/ 0.18 x 10-26 ?kg x (5.20 x 10-27 kg x 6.0 x 106 + 8.48 x 10-27 x 4.0 x 106 )

= -1 / 0.18 x 10-26 ?kg x 10-26(0.520 x 6.0 x 106 j + 0.848 x 4.0 x 106 i)

= 17.33 x 106 j - 18.84 x 106 i

Hence v 3 = 18.84 x 106 i - 17.33 x 106 j

V3 = sqrt ( 18.18)2 + (17.3)2 = sqrt (330.51 + 299.29) = 25.09 x 106 m/s

(b) KE of the process can be obtained by

KE(total) = 1/2m1v12 + 1/2 m2v22 +1/2 m3v32

KE (total) = 1/2 (m1v1 +m2v2 +m3v3) = 1/2 [ (5.20 x 10-27 x (6.0 x 106)2 + 8.48 x 10-27 x (4.0 x 106 )2 + 0.18 x 10-26x (25.09 x 106 )2

KE(total) =10-15( 187.2 + 135.68 + 113.31)/2 = 218.1 x 10-15 Joules = 2.18 x 10-13 Joules

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