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JEE Advanced Model Paper 12 with solutions for online practice
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© examsnet.com
Question : 15
Total: 60
smooth hemisphere of mass m and radius r is at rest. A smooth solid sphere of mass 2m and radius r moving with velocity V0 between two horizontal smooth surfaces separated by a distance slightly greater than 2R as shown in figure. Solid sphere collides with the hemisphere. If coefficient of restitution is
1
2
, then
The speed of hemisphere after collision is
V
0
The speed of solid sphere after collision is
V
0
2
The loss in kinetic energy of the system is
m
V
0
2
4
The final kinetic energy of hemisphere is 1/4 th the initial kinetic energy of the sphere
Validate
Solution:
2M
υ
0
= 2 M
υ
1
+ M
υ
2
and $$υ_2 cos θ -
υ
1
cos θ =
e
υ
0
cso θ on solving we get
υ
1
=
(
2
−
e
)
υ
0
3
=
υ
0
2
And
υ
2
= (1 + e)
2
υ
0
3
=
υ
0
.
K
i
=
1
2
×
2
m
υ
0
2
and
K
f
=
1
5
× 2m ×
υ
0
2
4
+
1
2
m
υ
0
2
=
3
4
m
υ
0
2
∴ loss of kinetic energy of the system =
1
4
m
υ
0
2
© examsnet.com
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