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JEE Advanced 2020 Concept Recapitulation Test 4 Paper 2
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© examsnet.com
Question : 3
Total: 54
Two blocks kept on a rough plank are moving with the plank with velocity of
2
m
∕
s
. The plank is retarded uniformly for
1
s
e
c
and brought to rest:
Assuming the plank to be sufficiently long then:
The velocity of block of mass
2
k
g
after collision with
1
k
g
block is
1
m
∕
s
The velocity of block of mass
2
k
g
after collision with
1
k
g
block is
1
2
m
∕
s
The final separation between two blocks is
1
16
m
The final separation between two blocks is
1
8
m
Validate
Solution:
As the plank retards
1
k
g
block will move relative to the plank but
2
k
g
block will remain at rest distance travelled by
1
k
g
block in
1
s
e
c
.
S
r
=
1
2
×
a
r
×
t
2
=
1
2
m
Relative velocity of
1
k
g
after
1
s
e
c
V
r
=
1
m
∕
s
Then using conservation of linear momentum and Newtons experimental Law
V
1
+
2
V
2
=
1
−
−
−
−
(
1
)
V
1
−
V
2
=
−
1
2
−
−
−
−
(
2
)
⇒
V
1
=
0
&
V
2
=
1
2
m
∕
s
Final separation between blocks
=
(
1
∕
2
)
2
2
×
2
=
1
16
m
© examsnet.com
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