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JEE Advanced 2023 Paper 1 Solutions
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© examsnet.com
Question : 1
Total: 51
A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height
3
h
from the ground, as shown in the figure. A spherical ball of mass
m
is released on the slide from rest at a height
h
from the top of the terrace. The ball leaves the slide with a velocity
→
u
0
=
u
0
^
x
and falls on the ground at a distance
d
from the building making an angle
θ
with the horizontal. It bounces off with a velocity
→
v
and reaches a maximum height
h
1
. The acceleration due to gravity is
g
and the coefficient of restitution of the ground is
1
∕
√
3
. Which of the following statement(s) is(are) correct?
→
u
0
=
√
2
g
h
^
x
→
v
=
√
2
g
h
(
^
x
−
^
z
)
θ
=
60
∘
d
∕
h
1
=
2
√
3
Validate
Solution:
u
0
=
√
2
g
h
v
z
=
√
2
g
(
3
h
)
tan
θ
=
v
z
u
=
√
3
θ
=
60
∘
d
=
u
0
T
=
u
0
√
2
(
3
h
g
)
=
√
(
2
g
h
)
√
(
2
)
(
3
h
g
)
Velocity after collision, only velocity along z-direction change
v
1
=
e
v
z
=
√
2
g
h
→
v
=
v
1
^
k
+
u
0
^
i
=
√
2
g
h
[
^
i
+
^
k
]
h
1
=
v
1
2
2
g
=
h
Finally,
u
0
=
√
2
g
h
,
θ
=
60
∘
,
d
h
=
2
√
3
© examsnet.com
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