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JEE Main 20 July 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 27
Total: 90
A rod of mass
M
and length
L
is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses
(
m
M
)
is
1
x
.
The value of '
x
' will be ______.
[20 Jul 2021 Shift 1]
Your Answer:
Validate
Solution:
Just after collision
From momentum conservation,
P
i
0
=
P
f
m
u
=
M
v
.
.
.
.
.
.
.
(i)
From angular momentum conservation about
O
,
m
u
.
L
2
=
M
L
2
12
ω
⇒
ω
=
6
m
u
M
L
........(ii)
From e
=
R.V.S
R.V.A
1
=
V
+
ω
L
2
u
v
+
ω
L
2
=
u
v
+
3
m
u
M
=
u
m
u
M
+
3
m
u
M
=
u
4
m
u
M
=
u
m
M
=
1
4
X
=
4
© examsnet.com
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