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Question 1
Two blocks of masses m and m are interconnected by an ideal spring of
stiffiness k. if m is pushed with a velocity
mmmm
Kinetic energy of the system w.r.t CM is
2
Mm9
(M + m)
Mm9
Kinetic energy of the system w.r.t CM is
26M + m)
Kinetic energy of the system w.r.t CM
Mm9
Mml
Answers
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0
Answer:
The linear momentum of system from CM frame is zero and no external force is acting on the system. We can apply work-energy theorem from CM frame.
W
ext
+W
int
=ΔK
cm
=(K
f
−K
i
)
cm
Initial kinetic energy of system w.r.t. CM
(K
i
)
cm
=
2
1
μ(ν
rel
)
initial
2
=
2
1
(
m
1
+m
2
m
1
m
2
)(ν
0
−0)
2
⟹(K
i
)
cm
=
2
1
(
m
1
+m
2
m
1
m
2
)ν
0
2
Final kinetic energy of system w.r.t. CM
(K
f
)
cm
=
2
1
μ(ν
rel
)
final
2
=0
As both the blocks move with common velocity hence, (ν
rel
)
final
=0 which gives (K
f
)
cm
=0
Hence from (1)
0+(−
2
1
kx
2
)=0−
2
1
(
m
1
+m
2
m
1
m
2
)ν
0
2
⟹[
(m
1
+m
2
)k
m
1
m
2
]ν
0
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