A body is released from the top ofan inclined
planeofinclination (). It reaches the bottom
with velocity (v). Ifkeeping the length same
the angle of inclination is doubled, what will
be the velocity of the body on reaching the
ground
Answers
Answer:
nswer:
v
1
=
√
4
⋅
H
⋅
g
cos
θ
Explanation:
my own work
my own work
let the height of incline be initially be
H
and length of the incline be
l
.and let
θ
be the initial angle.
The figure show Energy diagram at the different points of the inclined plane.
there for
sin
θ
=
H
l
...
...
...
...
.
.
(
i
)
and the
cos
θ
=
√
l
2
−
H
2
l
...
...
...
...
.
(
i
i
)
but ,now after change new angle is (
θ
∘
)=
2
⋅
θ
Let
H
1
be the new height of triangle.
sin
2
θ
=
2
sin
θ
cos
θ
=
h
1
l
[since length of the inclined has not yet changed.]
using (i) and (ii)
we get the new height as ,
h
1
=
2
⋅
H
⋅
√
l
2
−
H
2
l
by conserving the Total mechanical energy,
we get,
m
g
h
1
=
1
2
m
v
2
1
[let
_
v
1
be new speed]
putting
h
1
in this ,
v
1
=
√
4
⋅
H
⋅
g
⋅
√
l
2
−
H
2
l
or (to reduce variables)
v
1
=
√
4
⋅
H
⋅
g
cos
θ
but the initial velocity is
v
=
√
2
g
H
v
1
v
=
√
2
⋅
cos
θ
or
v
1
=
v
⋅
√
2
⋅
cos
θ
Hence ,the velocity becomes
√
2
cos
θ
times the initial.
Explanation: