Question

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u. (a) What is the speed of a neutron, expressed as a fraction of its original speed, after a head-on, elastic collision with a deuteron that is initially at rest(b) What is its kinetic energy, expressed as a fraction of its original kinetic energy? (c) How many such successive collisions will reduce the speed of a neutron to 1/59,000 of its original value?

Answer 1

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 $v_{n+1} = (\frac{1}{3}) ^{n}v_{1}$

for $v_{n+1}/v_{1} = 1/59000$

$(1/3)^{n} = 1/59000$

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

Answer 2

Answer:

A) Vn2 = (1/3)(Vn1)

B) Kn2 = (1/9)Kn1

C) 10 Collisions

Explanation:

From elastic collision, when there is no net external force acting on the system, we know that the momentum before collision is equal to the momentum after collision. Thus;

MaVa1 + MbVb1 =MaVa2 +MbVb2

We also know that their relative velocities will be;

Va1 - Vb1 = Vb2 - Va2

From conservation of momentum, momentum is conserved because there is no electromagnetic force between the neutron and deutron.

Since we are dealing with neutron and deutron, let's rewrite the elastic collision equation earlier to reflect this;

MnVn1 + MdVd1 =MnVn2 +MdVd2

Where;

Mn = mass of neutron

Md = Mass of deutron

Vn1 = Initial Speed of neutron

Vn2 = Final speed of neutron

Vd1 = Initial Speed of deutron

Vd2 = Final Speed of deutron

Also for the relative velocities ;

Vn1 - Vd1 = Vd2 - Vn2

Thus;

Let's substitute the following for the for the equation above;

Mn = 1u, Md = 2u, Vd1 = 0

Thus;

Vn1 = Vn2 +2Vd2 ------ eq 1

Now, from the relative velocities of;

Vn1 - Vd1 = Vd2 - Vn2

Since Vd1 = 0,then Vd2 = Vn1 + Vn2

And so putting that in eq 1 to get;

Vn1 = Vn2 +2(Vn1 + Vn2)

So, Vn1 = Vn2 + 2Vn1 + 2Vn2

-Vn1 = 3Vn2

Thus, Vn2 = - (1/3)(Vn1)

The minus sign implies that after collision, the neutron moves in a reverse direction.

Thus, we'll take the absolute value which is Vn2 = (1/3)(Vn1)

B) kinetic energy before collision is;

Kn1 = (1/2)(Mn1)(Vn1)²

Kinetic energy after collision = (1/2)(Mn2)(Vn2)²

From answer (a) above, Vn2 = (1/3)(Vn1)

So kinetic energy after collision =

(1/2)(Mn2)((1/3)(Vn1))²

= (1/9) (1/2)(Mn1)(Vn1)²

Thus, since Kn1 = (1/2)(Mn1)(Vn1)²,

Therefore, Kn2 cam be written as;

(1/9)Kn1

C) Following the same conditions in option a above and running another collision, we'll arrive at;

Vn3 = (1/3)Vn2

From earlier, we saw that vn2 = (1/3)(Vn1)

Thus; Vn3 = (1/3)((1/3)(Vn1))

Vn3 = (1/3²)(Vn1)

So after n number of collisions,

We can derive a formula of;

V(n+1) = (1/(3^(n)))(Vn1)

So, for us to get the number of collisions that will reduce the speed of a neutron to 1/59,000 of its original value ; we will substitute (1/59,000)Vn1 for V(n+1)

Thus;

(1/59,000)Vn1 = (1/(3^(n)))(Vn1)

Vn1 will cancel out and we have;

3^(n) = 59,000

Thus, n In3 = In 59000

So n = (In59000) /(In3) = 10.9853/1.0986 = 10 collisions