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An unstable atomic nucleus of mass 1.74 10^-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.14 x10^-27 kg, moves in the y direction with a speed of 6.00 x 10^6 m/s. Another particle, of mass 8.50 x10^-27 kg, moves in the x direction with a speed of 4.00 10^6 m/s.

a. Find the velocity of the third particle.
b. Find the total kinetic energy increase in the process.

Respuesta :

Answer:

a)    v₂ = (-9.04 i ^ - 8.20 j ^) 10⁶ m / s , b) K_{f} = 440.52 10⁻¹⁵ J

Explanation:

We can solve this exercise using the conservation of the moment, remember that being a vector quantity must be conserved in each axis

X axis

Initial. Before the breakup (v = 0)

           p₀ₓ = 0

Final. After disintegration

           [tex]p_{fx}[/tex] = m₁ v₁ + m₂ v₂

           m₁ = 8.50 10⁻²⁷ kg

            v₁ = 4.00 10⁶ m / s

            p₀ₓ = p_{fx}

             0 = m₁ v₁ + m₂ v₂ₓ

            v₂ₓ = -m₁ / m₂   v₁

The total mass is M is divided into three parts m1, m2 and m3

             M = m₁ + m₂ + m₃

             m₂ = M - m₁ - m₃

             m₂ = 1.74 10⁻²⁶ - 5.14 10⁻²⁷ - 8.50 10⁻²⁷

             m₂ = 3.76 10⁻²⁷ kg

We replace

            v₂ₓ = - 8.50 / 3.76 4.00 10⁶

           v₂ₓ = - 9.04 10⁶ m / s

Y Axis  

Initial

             [tex]p_{oy}[/tex] = 0

Final

             p_{fy} = m₃ v₃ + m₂ [tex]v_{2y}[/tex]

            p_{oy} = p_{fy}

             0 = m₃ v₃ + m₂ v_{2y}

             v_{2y} = - m₃ / m₂   v₃

            v_{2y} = -5.14 / 3.76 6.00 10⁶

            v_{2y} = - 8.20 10⁶ m / s

a) the speed of the particle is

           v₂ = (-9.04 i ^ - 8.20 j ^) 10⁶ m / s

b) the change in kinetic energy is

        K₀ = 0

        [tex]K_{f}[/tex] = K₁ + K₂ + K₃

        K_{f} = ½ m₁ v₁² + ½ m₂ v₂² + ½ m₃ v₃²

K_{f} = ½ 8.5 10⁻²⁷ (4 10⁶)²+½ 3.76 10⁻²⁷(9.04² + 8.20²) 10¹²+½ 5.14 10⁻²⁷ (6 10⁶)²

        K_{f} = 68 10⁻¹⁵ + 280 10⁻¹⁵ + 92.52 10⁻¹⁵

        K_{f} = 440.52 10⁻¹⁵ J

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