Elastic collision between neutrons and deuterons

[Edwardo_Elric]

Homework Statement

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?

Homework 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

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
I honestly don't know about this

 

[Astronuc]

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

Intially, the deuteron at rest, V_d, 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

 

[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?...

 

[learningphysics]

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?...

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