Answer:
a. -1/3v₁ b. 1/9K₁ b. 10 collisions
Explanation:
Let m₁,m₂, v₁,v₂ be the masses and initial velocity of the neutron and deuterons respectively. Let v₃ and v₄ be the final velocities of the neutron and deuterons respectively.
From the law of conservation of momentum,
m₁v₁ - m₂v₂ = m₁v₃ - m₂v₄ (since they are moving in opposite directions)
Also, since the collisions are elastic,
1/2m₁v₁² + 1/2m₂v₂² = 1/2 m₁v₃² + 1/2m₂v₄²
Since the deuterons are initially at rest, v₂ = 0 and m₁ = m₂ = 2.0 u
So, m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
m₁v₁ + 0 = m₁v₃ + m₂v₄
m₁v₁ = m₁v₃ + m₂v₄. (1)
Also, 1/2m₁v₁² + 1/2m₂v₂² = 1/2 m₁v₃² + 1/2m₂v₄²
1/2m₁v₁² + 0 = 1/2 m₁v₃² + 1/2m₂v₄²
m₁v₁² = m₁v₃² + m₂v₄² (2)
From (1) v₄ = m₁(v₁ - v₃)/m₂ (3)
Substituting (3) into (2), we have
m₁v₁² = m₁v₃² + m₂[m₁(v₁ - v₃)/m₂]²
m₁v₁² = m₁v₃² + [m₁(v₁ - v₃)²/m₂] since m₁ = 1.0 u and m₂ = 2.0 u
v₁² = v₃² + [(v₁ - v₃)²/2]
v₁² = v₃² + (v₁ - v₃)²/2
2v₁² = 2v₃² + v₃² - 2v₁v₃ + v₁²
collecting like terms
v₁² = 2v₃² + v₃² - 2v₁v₃
3v₃² - 2v₁v₃ - v₁² = 0
divide through by v₁² and let v₃/v₁ = α, we have
3α² - 2α - 1 = 0.
Using the formula method, α = [-(-2)+/- √((-2)² -4×-1×3)]/2×3 = [2 +/- √(4 + 12)]/6 = [2 +/-√16]/6 = [2+/-4]/6 = (2-4)/6 or (2 + 4)/6
= -2/6 or 6/6 = -1/3 or 1. Since v₁ ≠ v₃, we take v₃/v₁ = -1/3.
So v₃ = -1/3v₁
b. The kinetic energy of the neutron is after collision is K₂ = 1/2 m₁v₃². Its initial kinetic energy before collision is K₁ = 1/2m₁v₁². So its kinetic energy as a fraction of its initial kinetic energy is K₂/K₁ = 1/2 m₁v₃²/1/2m₁v₁² = (v₃/v₁)² = (1/3)² = 1/9.
So K₂ = 1/9K₁
c. If the neutrons speed reduces to 1/59,000 of its original value, then
v/v₁ = 1/59,000 = 1.69 × 10⁻⁵.
Since the speed of the neutron is v₂ = 1/3v₁. v₃ = 1/3v₂ = 1/3(1/3v₁) = 1/3²v₁. So the nth velocity is
for
taking log to base 10 of both sides we have,
nlog(1/3)= log(1/59000)
n = log(1/59000)/log(1/3)= 9.999≅ 10 collisions