Elastic collision with moderators

[mafagafo]

From University Physics,

Homework Statement

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons* of mass 2.0 u (see Example 8.11 in Section 8.4).
(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?

Homework Equations

εcni = εcnf + εcdf
pni = pnf + pdf

The Attempt at a Solution

0.5u * vni ^ 2 = 0.5u * vnf ^ 2 + 1u * vdf ^ 2

1u * vni = 1u * vnf + 2u * vdf
2u * vdf = 1u * vni - 1u * vnf
vdf = 0.5 * vni - 0.5 * vnf
vdf = 0.5 * (vni - vnf)

0.5u * vni ^ 2 = 0.5u * vnf ^ 2 + 1u * (0.5 * (vni - vnf)) ^ 2
0.5u * vni ^ 2 = 0.5u * vnf ^ 2 + 1u * (0.5 * vni - 0.5 * vnf)) ^ 2
0.5u * vni ^ 2 = 0.5u * vnf ^ 2 + 1u * (0.25 * vni ^ 2 - 0.5 * vni * vnf + 0.25 * vnf ^ 2
vni ^ 2 = vnf ^ 2 + 2 * (0.25 * vni ^ 2 - 0.5 * vni * vnf + 0.25 * vnf ^ 2
vni ^ 2 = vnf ^ 2 + 0.5 * vni ^ 2 - vni * vnf + 0.5 * vnf ^ 2
0.5 * vni ^ 2 = 1.5 * vnf ^ 2 - vni * vnf
What can I do with "- vni * vnf"


I think I can handle b) and c), but not without the answer of a), which is, according to the book, 1/3 vni.

 

[mfb]

You can introduce the fraction of those two speeds, and express the equation just based on this fraction and the initial (or final) speed - the latter one cancels and you get an equation with a single variable, which you can solve.

Alternatively, just treat it as quadratic equation (in vni or vnf, your choice) and solve it in the usual way.

 

[mafagafo]

Alternatively, just treat it as quadratic equation (in vni or vnf, your choice) and solve it in the usual way.

0.5i ^ 2 = 1.5f ^ 2 - fi
0.5i ^ 2 - 1.5f ^ 2 + fi = 0
i ^ 2 - 3f ^ 2 + 2fi = 0
3f ^ 2 - i ^ 2 - 2fi = 0
3f ^ 2 - 2fi - i ^ 2 = 0
f = \frac{- b - \sqrt{b ^ 2 - 4ac}}{2a}
f = \frac{2i - \sqrt{4i ^ 2 - 12i ^2}}{6}
f = \frac{2i - \sqrt{16i ^ 2}}{6}
f = \frac{2i - 4i}{6}
f = -\frac{2i}{6}
f = -\frac{i}{3}

This is the answer of the book, but with opposite sign.
Should I take it as being right?
If we are interested in slowing down the neutron, it does not matter if it keeps going the same way.
Maybe it is even better if it goes the other way around, because that perhaps increases the probability of another collision in a time interval t.
Couldn't a ping-pong ball thrown in a bowling ball be an analogy for this? The neutron has a smaller mass than the deuterium, so it should reverse direction.

 

[mfb]

The sign in your calculations is right - you calculated the velocity (which has a sign, the neutron changes its direction), the question asks for the speed (the magnitude of the velocity) where the sign disappears.

 

[mafagafo]

Sorry for asking more questions but I still have one.
Should I use s for speed? That's used for seconds already.
So s = |\vec{v}| or would \vec{s} be equalt to |\vec{v}|?

Another book I used for physics would frequently say that the speed of something was -85.2km/h, because the problem used the x-axis orientation.
But here even though all particles are in the same axis (it's a head-on collision) no sign is used. Is this due to the fact that there is no rule for the orientation of this axis?

 

[mfb]

So s = |\vec{v}| or would \vec{s} be equalt to |\vec{v}|?

Where is the difference between the options?
You don't need a separate "speed calculation". Just writing 1/3 as ratio is fine.

Another book I used for physics would frequently say that the speed of something was -85.2km/h, because the problem used the x-axis orientation.
But here even though all particles are in the same axis (it's a head-on collision) no sign is used. Is this due to the fact that there is no rule for the orientation of this axis?

In 1-dimensional problems, "speed" is sometimes used with the same meaning as "velocity", but it would be better to call it "velocity".

 

[mafagafo]

Where is the difference between the options?

Formally, the absolute value of the velocity gives me a vector or a scalar?

 

[mfb]

A scalar.