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(f) During each collision, the energy of the neutron is reduced by the factor 1 –
0.89 =
0.11. If
E
i
is the initial energy, then the energy after
n
collisions is given by
E
= (0.11)
n
E
i
.
We take the natural logarithm of both sides and solve for
n
. The result is
ln( /
)
ln(0.025 eV/1.00 eV)
7.9
8.
ln 0.11
ln 0.11
i
EE
n
==
=
≈
The energy first falls below 0.025 eV on the eighth collision.
23. (a) Let
v
ni
be the initial velocity of the neutron,
v
nf
be its final velocity, and
v
f
be the
final velocity of the target nucleus. Then, since the target nucleus is initially at rest,
conservation of momentum yields
m
n
v
ni
= m
n
v
nf
+ mv
f
and conservation of energy yields
1
2
2
1
2
2
1
2
2
mv
nn
i
f
f
=+
.
We solve these two equations simultaneously for
v
f
. This can
be done, for example, by using the conservation of momentum equation to obtain an
expression for
v
nf
in terms of
v
f
and substituting the expression into the conservation of
energy equation. We solve the resulting equation for
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 Spring '08
 Any
 Physics, Momentum, Neutron

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