(1) (a) Applying conservation of relativistic momentum we have
 1m1 (0.4c)   2 m2 (0.6c) where
1
1 
1  (0.4)
2
1
, 2 
1  (0.6)2
, so
1  (0.4)2 0.6
m2  1.72m2 (1)
1  (0.6)2 0.4
m1 
Conservation of total energy gives
 1m1c2   2 m2 c2  Mc2 or
1
1  (0.4)
2
m1 
1
1  (0.6)2
m2  M
so
1.09m1  1.25m2  M
Using Eq. (1) we get
3.12m2  M
so
m2  0.32M
and m1  1.72(0.32M )  0.55M
(b) Fraction of mass converted to K.E. = ( 1 – 0.55 – 0.32) = 0.13
(c) K.E. of m1  ( 1  1)m1c2  (1.09  1)(0.55)Mc2  0.05Mc2
K.E. of m2  ( 2  1)m2 c2  (1.25  1)(0.32)Mc2  0.08Mc2
Total K.E. = 0.13 Mc2 as expected.
(2) We have for an electron moving in a circular orbit perpendicular to a
magnetic field,
3.2.10 22
p  qBr  1.6.10 .10 .2kg.m.s 
MeV / c  0.6MeV / c
5.344.10 22
19
3
Total energy of electron is
1
p 2 c 2  m 2 c 4  (0.6)2  (0.511)2  0.788MeV
Total energy of positron at rest is 0.511 MeV
Total energy = (0.788 + 0.511) MeV = 1.3 MeV
Total momentum = 0.6 MeV/c (to right, say) (before annihilation)
Let us assume 2 emitted -rays have momenta to the right of p1 and –p2
respectively.
By momentum conservation, p1 – p2 = 0.6 MeV/c
By energy conservation, p1c + p2c = 1.3 MeV or p1 + p2 = 1.3 MeV/c
Solving the above 2 linear equations, we get p1c = 0.95 MeV
and p2c = 0.35 MeV as the energies of the -rays.