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(f) And
2
1
2
BB
mv
K
=
yields
v
B
=
2/
i
Gm R
.
(g) The answer to part (f) is incorrect, due to having ignored the accelerated motion of
“our” frame (that of body
A
). Our computations were therefore carried out in a
noninertial frame of reference, for which the energy equations of Chapter 8 are not
directly applicable.
101. (a) Their initial potential energy is
−
Gm
2
/
R
i
and they started from rest, so energy
conservation leads to
22
2
total
total
.
0.5
iii
Gm
Gm
Gm
KK
RRR
−=−
¡
=
(b) They have equal mass, and this is being viewed in the centerofmass frame, so their
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Mass, Potential Energy

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