# Momentum in elastic collision

## Homework Statement

Uranium fission within a nuclear reactor produces high-speed neutrons. These must be slowed down for them to trigger a chain reaction within the uranium. In a heavy-water reactor the neutrons of mass m make multiple elastic collisions with deuterium nuclei of 2m, losing speed at each collision.

(a) Suppose a fast-moving neutron of speed v makes a head-on elastic collision with a stationary deuterium nucleus. Show that its speed is reduced to v/3.

## Homework Equations

p=mv - We probably have to use this
Ek=0.5mv2 - Not sure if we have to use this, but the question IS in a section about kinetic energy. It just seems logical to use p=mv because using kinetic energy makes no sense.

## The Attempt at a Solution

Ok it took me absolutely AGES to do this, but I don't think it's right:

PD1 = mv = 2m x 0 = 0 N s
PN1 = mv = 1m x v = mv

PD1 + PN1 = PD2 + PN2 if no external forces are applied

0 + mv = PD2 + PN2
mv = PD2 + PN2
PD2 = mv - PN2

PD2 = PN2 - not sure

Substitute above two equations together to show PN2:
PN2 = mv - PN2
2PN2 = mv
PN2 = mv/2
vN2 = PN2/2m
vN2 = (mv/2)/2m
vN2 = mv/4m - my rearranging may be wrong
vN2 = v/3m

But the answer is just v/3 not v/3m. Is this still right? Or is it just coincidence I got an answer so similar?

v/3m cannot be right. it has units of m.kg/s

The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations

The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations

Could you post your calculation please? You don't even have to explain it, but I've been trying to do this all night and I don't have a clue. Got an exam on thursday l0L!

bah fine. I'm a lazy individual.
v is initial velocity
v1 is final velocity of the neutron
v2 is the final velocity of the deuterium

mv = m . v1 + 2m . v2 ------------- (momentum)
mv² = m . v1² + 2m . v2² ------------- (energy)

All the masses cancel:

v = v1 + 2 . v2 ------------- (momentum)
v² = v1² + 2 . v2² ------------- (energy)

isolate the momentum equation for v2
v2 = (v - v1)/2
v2² = (v² - 2*v1*v + v1²)/4

sub that into your energy equation and isolate for v1.