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.