Answer:
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
