Question

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

Answer 1

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

Step-by-step 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