2.3 Solutions
83
R =
AB
2
+ BC
2
=
p 2 + (4 p/3) 2 =
5 p
3
tan θ =
BC
AB
=
4 p/3
p
=
4
3
∴ θ = 53
◦
Thus the neutrino is emitted with momentum 5 p/3 at an angle (180 − 53 ◦ ) or
127 ◦ with respect to the β particle.
2.35 Denoting the angles with (*) for the CM system transformation of angles for
CMS to LS is given by
tan θ =
sin θ ∗
cos θ ∗ +
m
M
(1)
But θ
∗
= π − φ
∗
= π − 2φ
∴ sin θ
∗
= sin(π − 2φ) = sin 2φ
cos θ
∗
= cos(π − 2φ) = − cos 2φ
(i) becomes
tan θ =
sin 2φ
m
M
− cos 2φ
Furthermore
sin θ
cos θ
=
sin 2φ
m
M
− cos 2φ
Cross-multiplying and rearranging
m
M
sin θ = sin θ cos 2φ + cos θ sin 2φ = sin(θ + 2φ)
∴
m
M
=
sin(2φ + θ)
sin θ
2.36 In the lab system let M be projected at an angle φ with velocity v. In the
CMS the velocity v ∗ for the struck nucleus will be numerically equal to v c ,
the centre of mass velocity. Therefore, M is projected at an angle 2φ with
velocity v ∗ =
mv
M + m
. The CM system velocity v c =
mv
M + m
. The velocities
v ∗ and v c must be combined vectorially to yield v, Fig. 2.32. Since v c = v ∗
the velocity triangle ABC is an isosceles triangle. If BD is perpendicular on
AC, then