11.
(a) Conservation of energy gives Q = K
2
+ K
3
= E
1
− E
2
− E
3
where E refers here to the rest energies
(mc
2
) instead of the total energies of the particles. Writing this as K
2
+ E
2
− E
1
=
−(K
3
+ E
3
)
and squaring both sides yields
K
2
2
+ 2K
2
E
2
− 2K
2
E
1
+ (E
1
− E
2
)
2
= K
2
3
+ 2K
3
E
3
+ E
2
3
.
Next, conservation of linear momentum (in a reference frame where particle 1 was at rest) gives
|p
2
| = |p
3
| (which implies (p
2
c)
2
= (p
3
c)
2
). Therefore, Eq. 38-51 leads to
K
2
2
+ 2K
2
E
2
= K
2
3
+ 2K
3
E
3
which we subtract from the above expression to obtain
−2K
2
E
1
+ (E
1
− E
2
)
2
= E
2
3
.
This is now straightforward to solve for K
2
and yields the result stated in the problem.
(b) Setting E
3
= 0 in
K
2
=
1
2E
1
(E
1
− E
2
)
2
− E
2
3
and using the rest energy values given in Table 45-1 readily gives the same result for K
µ
as computed
in Sample Problem 45-1.