fisica1-youn-e-freedman-exercicios-resolvidos

8.87: Let the proton be moving in the + x -direction with speed v
p
after the decay. The
initial momentum of the neutron is zero, so to conserve momentum the electron must be
moving in the − x -direction after the collision; let its speed be v .
P
x
is constant gives 0 =
v = ( m m ) v – p
e
p
e
− m v + m v .
e e p p
The total kinetic energy after decay is
momentum equation to replace v
e
gives
K =
tot
K
tot
=
1
2
1
2
v 2
p p
v 2
e e
e
m + 1
m 2
. Using the
2
p v p
m ( 1+ m ).
p
m e
K
Thus
K
p
tot
1
=
1+
m
p
m
e
=
1
1836
= 5.44×
10
−4
= 0.0544
%
1.0176
and 1.0176
0.0176
−13
-14
1
−13
−13
(6.54 × 10 J) = 1.13×
10 J and b) (6.54 × 10 J) = 6.43×
10
1.0176
1.0176
−13
0.0176
1
8.88: The ratios that appear in Eq. ( 8.42) are , so the kinetic energies are
a) J.
Note that the energies do not add to 6.54× 10 J exactly, due to roundoff.
8.89: The “missing momentum” is
22
5.60× 10
− −
kg ⋅ m s − ( 3.50×
10
25 kg)(1.14×
10
3
m/s ) = 1.61×
10
−22
kg ⋅ m
s
.
Since the electron has momentum to the right, the neutrino’s momentum must be to the
left.