An unstable atomic nucleus of mass 1.80 10-26 kg initially at rest disintegrates into three particles. one of the particles, of mass 5.10 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) find the velocity of the third particle. î + ĵ m/s (b) find the total kinetic energy increase in the process. j

a)   v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s  and b)  ΔK = 3.96 10⁻¹³ J

Explanation:

We can work this process of disintegration as a conservation problem of the moment.

We create a system formed by the initial nucleus and the three final particles, in this case all the forces involved are internal and the amount of movement is conserved. We will write the equations each axis

X axis

      p₀ = 0

      pf = m1 v1x + m vfx

      m₁ = 8.50 10-27 kg

      vf₁ = 4.00 106 m / s

     p₀ = pf

     0 = m₁ v₁ₓ + m₃ vfₓ

     vfₓ = -m₁ / m₃ v1ₓ

Y Axis

     P₀ = 0

     Pf = m₂ v₂ + m₃ vfy

     m₂ = 5.10 10⁻²⁷ kg

     v₂ = 6.00 10⁶ m / s

     p₀ = pf

     0 = m₂ v₂ + m₃ vfy

     vfy = -m₂ / m3₃v₂

We have the initial particle mass and it decomposes into three parts after disintegration

     m = m₁ + m₂ + m₃

     m₃ = m-m₁-m₂

     m₃ = 1.80 10⁻²⁶ - 5.10 10⁻²⁷ - 8.50 10⁻²⁷ = (1.80 -0.510 -0.850) 10⁻²⁶

     m3 = 0.44 10⁻²⁶ kg = 4.4 10⁻²⁷ kg

Let's replace and calculate

    vfₓ = -m₁ / m₃ v₁ₓ

    vfₓ = - 8.50 10⁻²⁷/4.4 10⁻²⁷   4.00 10⁶

    vfₓ = -7.73 10⁶ m / s

    vfy = -m₂ / m₃ v₂

    vfy = - 5.10 10⁻²⁷ /4.4 10⁻²⁷   6.00 10⁶

    vfy = -6.95 10⁶ m/s

We set the speed vector

     v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s

b) Let's calculate the kinetic energy

Initial

   K₀ = 0

Final

   Kf = K₁ + K₂ + K₃

   Kf = ½ m₁ v₁² + ½ m₂ v₂² + ½ m₃ v₃²

   v₃² = (7.73 10⁶)²+ (6.95 10⁶)²  = 108.5 10⁻¹²

  Kf = ½ 8.50 10⁻²⁷(4 10⁶) 2 + ½ 5.1 10⁻²⁷ (6 10⁶) 2 + ½ 4.4 10⁻²⁷ 108.05 10⁻¹²

  Kf = 68 10⁻¹⁵ + 91.8 10⁻¹⁵ + 237.7 10⁻¹⁵

  Kf = 396.8 10⁻¹⁵ J

   Kf = 3.96 10⁻¹³ J

  ΔK = Kf -K₀

  ΔK = 3.96 10⁻¹³ J