Question

Asked Oct 26, 2019

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An unstable nucleus of mass 1.7 ✕ 10−26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of

m1 = 1.0 ✕ 10−27 kg,

moves in the positive y-direction with speed

v1 = 5.2 ✕ 106 m/s.

Another particle, of mass

m2 = 9.0 ✕ 10−27 kg,

moves in the positive x-direction with speed

v2 = 3.0 ✕ 106 m/s.

Find the magnitude and direction of the velocity of the third particle. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

Step 1

Given values:

Mass of the nucleus, M = 1.7 ✕ 10^{-26} kg = 17.0 ✕ 10^{-27} kg

m_{1} = 1.0 ✕ 10^{-27} kg,

v_{1} = 5.2 ✕ 10^{6} m/s. (In positive Y - direction)

m_{2} = 9.0 ✕ 10^{-27} kg,

v_{2} = 3.0 ✕ 10^{6} m/s. (In positive X - direction)

Step 2

Let the velocity of the third particle be v_{3}

Mass of the third particle m_{3 }= mass of the nucleus – mass of the other two particle

= 17.0 ✕ 10^{-27} - (1.0 ✕ 10^{-27} + 9.0 ✕ 10^{-27})

m_{3} = 7✕ 10^{-27} kg

Step 3

Total initial momentum = 0 (Nucleus is at rest)

Total final momentum = P1...

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