# Elastic collision between neutrons and deuterons

1. Sep 18, 2007

### Edwardo_Elric

1. The problem statement, all variables and given/known data
Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between neutrons and deuterons of mass 2.0u.
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 which 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 1/6600 of its original value?

2. Relevant equations
If the 2nd particle is at rest
$$V_{A} = \frac{m_{A} - m_{B}}{m_{A}+m_{B}}$$* V2
$$V_{B} = \frac{2m_{A}}{m_{A} + m_{B}}$$* V2

3. The attempt at a solution
let n = neutrons , dn = deuterons
a.)
VnMn = MnVn2 + MdnVdn2

using equation above:
$$V_{n2} = \frac{2.0u - 2.0u}{2.0u+2.0u} V_{n}$$
Vn2 = 0

b.) K2 = ?
since Vn2 =0
VnMn = Vdn2
K2 = 1/2 (2.0u)(Vdn2)2
K2 = u(Vdn2)^2 <<<< K equals velocity of deuterons squared times u

c.) (2.0)(1/6600Vn) =

3300 collisions

Last edited: Sep 18, 2007
2. Sep 18, 2007

### Staff: Mentor

For an elastic collision one applies conservation of momentum and energy.

Intially, the deuteron at rest, Vd, is zero, so it has not momentum or KE.

The neutron will not lose all its energy or momentum, but will recoil 180°.

So write the momentum and energy equations and one should end up with

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c1

3. Sep 18, 2007

### Edwardo_Elric

but i thought masses are the same:
which makes
v{initial speed of neutron} = (m1 - m2) / (m1 + m2) * V{final speed of deuteron}

equal to zero? and the deuteron goes to move which has a speed equal to the initial speed of neutron?...

4. Sep 18, 2007

### learningphysics

deuteron has a mass of 2.0u. Neutro has a mass of 1.0u