2.3 Solutions
85
v A =
3u
4
; v B =
u
4
(b) From (1), Q = m(u 2 − v 2
A − 5v 2
B )
= m
u
2
−
9u 2
16
−
5u 2
16
=
mu 2
8
∴
Q
k A
=
mu 2/8
mu 2 =
1
8
Since Q is positive, energy is lost in the collision process.
2.38 (a) In the elastic collision (head-on) of a particle of mass m 1 and kinetic
energy K 1 with a particle of mass m 2 initially at rest, the fraction of
kinetic energy imparted to m 2 is
K 2
K 0
=
4m 1 m 2
(m 1 + m 2 ) 2 =
4 × 1 × 12
(1 + 12) 2 =
48
169
(b)
1
2 m 2 v 2
2
1
2 m 1 v 2
1
=
12
1
v 2
2
v 2
0
=
48
169
∴ v 2 =
2
13
v 0
K 1 = K 0 − K 2 = K 0 −
48
169
K 0 =
121
169
K 0
∴
1
2
m 1 v
2
1 =
121
169
×
1
2
m 1 v
2
0
∴ v 1 = −
11
13
v 0
Negative sign is introduced because neutron being lighter then the carbon
nucleus will bounce back.
2.39 Let the heavy body of mass M with momentum P 0 collide elastically with a
very light body of mass m be initially at rest. After the collision both the bodies
will be moving in the direction of incidence, the heavier one with velocity v H
and the lighter one with velocity v L .
Momentum conservation gives
p 0 = p L + p H
(1)
Energy conservation gives
p 2
0
2M
=
p 2
L
2m
+
p 2
H
2M
(2)