A ball of mass m1 makes a head on elastic collision with a ball of mass m2 which is initially at rest. The transfer of kinetic energy to the second ball is maximum when
Answers
Answer:
Let initial velocity of particle
m
1
be
u
1
, final velocity of particle
m
1
be
v
1
and that of particle
m
2
be
v
2
.
Since collision is an elastic head on, from Law of Conservation of Momentum we have
m
1
u
1
=
m
1
v
1
+
m
2
v
2
⇒
m
2
v
2
=
m
1
u
1
−
m
1
v
1
.........(1)
From the Law of Conservation of kinetic energy we have
1
2
m
1
u
2
1
=
1
2
m
1
v
2
1
+
1
2
m
2
v
2
2
.......(2)
Rewriting (1) as
m
2
v
2
=
m
1
(
u
1
−
v
1
)
.......(3)
Rewriting (2) as
m
2
v
2
2
=
m
1
(
u
2
1
−
v
2
1
)
.......(4)
Dividing (4) with (3)
v
2
=
u
1
+
v
1
.......(2)
Eliminating
v
2
from (1) and (5) we get
m
1
u
1
−
m
1
v
1
m
2
=
u
1
+
v
1
⇒
(
m
1
u
1
−
m
1
v
1
)
=
m
2
u
1
+
m
2
v
1
⇒
m
1
u
1
−
m
2
u
1
=
m
1
v
1
+
m
2
v
1
⇒
m
1
−
m
2
m
1
+
m
2
u
1
=
v
1
.........(6)
Fraction of KE of
m
1
carried by
m
2
K
E
i
−
K
E
f
K
E
i
=
1
−
K
E
f
K
E
i
=
1
−
1
2
m
1
v
2
1
1
2
m
1
u
2
1
=
1
−
(
v
1
u
1
)
2
Using (6) in above we get
Fraction of KE of
m
1
carried by
m
2
is
1
−
[
m
1
−
m
2
m
1
+
m
2
]
2
=
4
m
1
m
2
(
m
1
+
m
2
)
2
Explanation: