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8.80:
a) From the derivation in Sec. 8.4 of the text we have
B
A
A
B
0
B
A
B
A
A
2
and
M
M
M
V
V
M
M
M
M
V
+
=
+

=
The ratio of the kinetic energies of the two particles after the collision is
B
A
2
B
A
B
A
B
A
2
B
A
2
A
B
A
B
A
2
B
A
B
A
2
B
B
2
1
2
A
A
2
1
4
)
(
or
4
)
(
2
M
M
M
M
KE
KE
M
M
M
M
M
M
M
M
M
V
V
M
M
V
M
V
M

=

=

=
=
b)
i) For
;
0
,
A
B
A
=
=
KE
M
M
i.e., the two objects simply exchange kinetic energies.
ii)
,
5
For
B
A
M
M
=
5
4
)
)(
5
(
4
)
4
(
B
B
2
B
B
A
=
=
M
M
M
KE
KE
i.e.,
A
M
gets 4/9 or 44% of the total.
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Unformatted text preview: c) We want B A 2 B B A 2 A B A 2 B A B A 4 2 4 ) ( 1 M M M M M M M M M M KE KE +== = which reduces to , 6 2 B B A 2 A = +M M M M from which, using the quadratic formula, we get the two possibilities B A 83 . 5 M M = and B. A 172 . M M =...
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