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51. (a) Before looking at our solution to part (a) (which uses momentum conservation), it might be ad
visable to look at our solution (and accompanying remarks) for part (b) (where a very diFerent
approach is used). Since momentum is a vector, its conservation involves two equations (along the
original direction of alpha particle motion, the
x
direction, as well as along the ±nal proton direction
of motion, the
y
direction). The problem states that all speeds are much less than the speed of light,
which allows us to use the classical formulas for kinetic energy and momentum (
K
=
1
2
mv
2
and
~p
=
m~v
, respectively). Along the
x
and
y
axes, momentum conservation gives (for the components
of
~v
oxy
):
m
α
v
α
=
m
oxy
v
oxy
,x
=
⇒
v
oxy
,x
=
m
α
m
oxy
v
α
≈
4
17
v
α
0=
m
oxy
v
oxy
,y
+
m
p
v
p
=
⇒
v
oxy
,y
=
−
m
p
m
oxy
v
p
≈−
1
17
v
p
.
To complete these determinations, we need values (inferred from the kinetic energies given in the
problem) for the initial speed of the alpha particle (
v
α
) and the ±nal speed of the proton (
v
p
). One
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 Fall '08
 SPRUNGER
 Physics, Momentum

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