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JEE Advanced 2020 Concept Recapitulation Test 2 Paper 1
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© examsnet.com
Question : 5
Total: 54
In the adjoining figure block
A
is of mass
m
and block
B
is of mass
2
m
. The spring has a force constant
k
. All the surfaces are smooth and the system is released from rest with spring unstretched
The maximum extension of the spring is
4
m
g
k
The speed of block A when extension in spring is
2
m
g
k
, is
2
g
√
2
m
3
k
Net acceleration of block B when the extension in the spring is maximum, is
2
3
g
.
Tension in the thread for extension of
2
m
g
k
in spring is mg.
Validate
Solution:
(A) At maximum extension
V
A
=
V
B
=
0
2
m
g
x
=
1
2
k
x
2
x
m
=
4
m
g
k
(B)
2
m
g
2
m
g
k
=
1
2
k
(
2
m
g
k
)
2
+
1
2
(
m
+
2
m
)
v
2
V
=
2
g
√
2
m
3
k
(C) Spring force
=
4
m
g
k
×
k
=
4
m
g
net downward force
=
2
m
g
a
=
4
m
g
−
2
m
g
3
m
=
2
3
g
(D) for
x
=
2
m
g
k
,
a
=
0
∴
T
=
2
m
g
© examsnet.com
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