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Page 1
Edurev123
4. Work & Energy
4.1 A body is describing an ellipse of eccentricity e under the action of a central
force directed towards a focus and when at the nearer apse the centre of the force
is transferred to the other focus. Find the eccentricity of the new orbit in terms of
the eccentricity of the original orbit.
(2009: 12 Marks)
Solution:
Approach : The motion in an ellipse in under inverse square force and the velocity
determines the eccentricity of the ellipse. So balance the velocity between the two orbits.
Let ?? and ?? '
be the foci of the original orbits and the force initially be directed towards ?? .
The nearer apse is at ?? .
Since the motion is in an ellipse the velocity at any point is
given by
?? 2
=?? (
2
?? -
1
?? )
At ?? ,?? =???? =???? -???? =?? -???? =?? (1-?? )
? ?? 2
=?? (
2
?? (1-?? )
-
1
?? )=
?? (1+?? )
?? (1-?? )
When the centre of force is changed to ?? '
the velocity remains same. Let ?? '
be the semi
major axis of the new orbit.
Then: ?? 2
=
?? (1+?? )
?? (1-?? )
=?? (
2
?? -
1
?? '
)
?? =?? ?? '
=???? +???? =?? +???? =?? (1+?? )
Page 2
Edurev123
4. Work & Energy
4.1 A body is describing an ellipse of eccentricity e under the action of a central
force directed towards a focus and when at the nearer apse the centre of the force
is transferred to the other focus. Find the eccentricity of the new orbit in terms of
the eccentricity of the original orbit.
(2009: 12 Marks)
Solution:
Approach : The motion in an ellipse in under inverse square force and the velocity
determines the eccentricity of the ellipse. So balance the velocity between the two orbits.
Let ?? and ?? '
be the foci of the original orbits and the force initially be directed towards ?? .
The nearer apse is at ?? .
Since the motion is in an ellipse the velocity at any point is
given by
?? 2
=?? (
2
?? -
1
?? )
At ?? ,?? =???? =???? -???? =?? -???? =?? (1-?? )
? ?? 2
=?? (
2
?? (1-?? )
-
1
?? )=
?? (1+?? )
?? (1-?? )
When the centre of force is changed to ?? '
the velocity remains same. Let ?? '
be the semi
major axis of the new orbit.
Then: ?? 2
=
?? (1+?? )
?? (1-?? )
=?? (
2
?? -
1
?? '
)
?? =?? ?? '
=???? +???? =?? +???? =?? (1+?? )
?
?? (1+?? )
?? (1-?? )
=?? (
2
?? (1+?? )
-
1
?? '
)
?
1+?? 1-?? =
2
1+?? -
?? ?? '
Again the direction of velocity is not changed, the velocity still being perpendicular to the
direction of motion the point ?? is an apse for the new orbit as well.
? ?? '
?? =?? '
(1-?? '
)=?? (1+?? )
?
?? ?? '
=
1-?? '
1+??
?
1+?? 1-?? =
2
1+?? ·
1-?? '
1+?? =
1+?? '
1+??
? ?? '
=
(1+?? )
2
1-?? -1=
?? 2
+3?? 1-?? =
?? (?? +3)
1-??
4.2 A particle moves with a central acceleration ?? (?? ?? -?? ?? ) , being projected from
an apse at a distance v?? with velocity ?? v?? ?? . Show that its path is the curve ?? ?? +
?? ?? =?? .
(2010 : 20 Marks)
Solution:
The equation of motion is
h
2
(?? +
?? 2
?? ?? ?? 2
)=
?? ?? 2
where ?? =
1
??
?? = central accel
=?? (
1
?? 5
-
9
?? )
? Equation becomes h
2
(?? +
?? 2
?? ?? ?? 2
) =
?? ?? 2
(
1
?? 5
-
9
?? )
? h
2
(?? +
?? 2
?? ?? ?? 2
) =?? (
1
4
7
-
9
4
3
)
Multiplying both sides by 2
????
????
and integrating, we get
Page 3
Edurev123
4. Work & Energy
4.1 A body is describing an ellipse of eccentricity e under the action of a central
force directed towards a focus and when at the nearer apse the centre of the force
is transferred to the other focus. Find the eccentricity of the new orbit in terms of
the eccentricity of the original orbit.
(2009: 12 Marks)
Solution:
Approach : The motion in an ellipse in under inverse square force and the velocity
determines the eccentricity of the ellipse. So balance the velocity between the two orbits.
Let ?? and ?? '
be the foci of the original orbits and the force initially be directed towards ?? .
The nearer apse is at ?? .
Since the motion is in an ellipse the velocity at any point is
given by
?? 2
=?? (
2
?? -
1
?? )
At ?? ,?? =???? =???? -???? =?? -???? =?? (1-?? )
? ?? 2
=?? (
2
?? (1-?? )
-
1
?? )=
?? (1+?? )
?? (1-?? )
When the centre of force is changed to ?? '
the velocity remains same. Let ?? '
be the semi
major axis of the new orbit.
Then: ?? 2
=
?? (1+?? )
?? (1-?? )
=?? (
2
?? -
1
?? '
)
?? =?? ?? '
=???? +???? =?? +???? =?? (1+?? )
?
?? (1+?? )
?? (1-?? )
=?? (
2
?? (1+?? )
-
1
?? '
)
?
1+?? 1-?? =
2
1+?? -
?? ?? '
Again the direction of velocity is not changed, the velocity still being perpendicular to the
direction of motion the point ?? is an apse for the new orbit as well.
? ?? '
?? =?? '
(1-?? '
)=?? (1+?? )
?
?? ?? '
=
1-?? '
1+??
?
1+?? 1-?? =
2
1+?? ·
1-?? '
1+?? =
1+?? '
1+??
? ?? '
=
(1+?? )
2
1-?? -1=
?? 2
+3?? 1-?? =
?? (?? +3)
1-??
4.2 A particle moves with a central acceleration ?? (?? ?? -?? ?? ) , being projected from
an apse at a distance v?? with velocity ?? v?? ?? . Show that its path is the curve ?? ?? +
?? ?? =?? .
(2010 : 20 Marks)
Solution:
The equation of motion is
h
2
(?? +
?? 2
?? ?? ?? 2
)=
?? ?? 2
where ?? =
1
??
?? = central accel
=?? (
1
?? 5
-
9
?? )
? Equation becomes h
2
(?? +
?? 2
?? ?? ?? 2
) =
?? ?? 2
(
1
?? 5
-
9
?? )
? h
2
(?? +
?? 2
?? ?? ?? 2
) =?? (
1
4
7
-
9
4
3
)
Multiplying both sides by 2
????
????
and integrating, we get
h
2
(?? 2
+(
????
????
)
2
)=?? (
-2
6?? 6
+
9×2
2?? 2
)+?? ? h
2
(?? 2
+(
????
????
)
2
)=?? (
9
?? 2
-
1
3?? 6
)+?? =?? 2
? ?? (9×3-
1
3
×27)+?? =18?? ? ?? +18?? =18?? ??? =0
? h
2
(?? 2
+(
????
????
)
2
)=?? (
?? ?? 2
-
1
3?? 6
)
???????? ,???? ???????? , (
????
????
)=0
h
2
(
1
3
)=?? (9×3-
1
3
×27)=18?? h=54??
? Equation becomes
54(?? 2
+(
????
????
)
2
)=?? (
9
4
2
-
1
3?? 6
)
? ?? 2
+(
????
????
)
2
=
1
5?? (
9
?? 2
-
1
3?? 6
)
? (
????
????
)
2
=
27?? 4
-1-162?? 8
162?? 6
?
????
????
=
v
27
162
?? 4
-
1
162
-?? 8
? =
?? 3
????
v
?? 4
6
-
1
162
-?? 8
????
integrating both sides, we get
?
36?? 3
????
v1-(36?? 4
-3)
2
=? ????
?
sin
-1
(36?? 4
-3)
4
=?? +??
Now, at ?? =
1
?? =
1
v3
,?? =0 (apse)
Page 4
Edurev123
4. Work & Energy
4.1 A body is describing an ellipse of eccentricity e under the action of a central
force directed towards a focus and when at the nearer apse the centre of the force
is transferred to the other focus. Find the eccentricity of the new orbit in terms of
the eccentricity of the original orbit.
(2009: 12 Marks)
Solution:
Approach : The motion in an ellipse in under inverse square force and the velocity
determines the eccentricity of the ellipse. So balance the velocity between the two orbits.
Let ?? and ?? '
be the foci of the original orbits and the force initially be directed towards ?? .
The nearer apse is at ?? .
Since the motion is in an ellipse the velocity at any point is
given by
?? 2
=?? (
2
?? -
1
?? )
At ?? ,?? =???? =???? -???? =?? -???? =?? (1-?? )
? ?? 2
=?? (
2
?? (1-?? )
-
1
?? )=
?? (1+?? )
?? (1-?? )
When the centre of force is changed to ?? '
the velocity remains same. Let ?? '
be the semi
major axis of the new orbit.
Then: ?? 2
=
?? (1+?? )
?? (1-?? )
=?? (
2
?? -
1
?? '
)
?? =?? ?? '
=???? +???? =?? +???? =?? (1+?? )
?
?? (1+?? )
?? (1-?? )
=?? (
2
?? (1+?? )
-
1
?? '
)
?
1+?? 1-?? =
2
1+?? -
?? ?? '
Again the direction of velocity is not changed, the velocity still being perpendicular to the
direction of motion the point ?? is an apse for the new orbit as well.
? ?? '
?? =?? '
(1-?? '
)=?? (1+?? )
?
?? ?? '
=
1-?? '
1+??
?
1+?? 1-?? =
2
1+?? ·
1-?? '
1+?? =
1+?? '
1+??
? ?? '
=
(1+?? )
2
1-?? -1=
?? 2
+3?? 1-?? =
?? (?? +3)
1-??
4.2 A particle moves with a central acceleration ?? (?? ?? -?? ?? ) , being projected from
an apse at a distance v?? with velocity ?? v?? ?? . Show that its path is the curve ?? ?? +
?? ?? =?? .
(2010 : 20 Marks)
Solution:
The equation of motion is
h
2
(?? +
?? 2
?? ?? ?? 2
)=
?? ?? 2
where ?? =
1
??
?? = central accel
=?? (
1
?? 5
-
9
?? )
? Equation becomes h
2
(?? +
?? 2
?? ?? ?? 2
) =
?? ?? 2
(
1
?? 5
-
9
?? )
? h
2
(?? +
?? 2
?? ?? ?? 2
) =?? (
1
4
7
-
9
4
3
)
Multiplying both sides by 2
????
????
and integrating, we get
h
2
(?? 2
+(
????
????
)
2
)=?? (
-2
6?? 6
+
9×2
2?? 2
)+?? ? h
2
(?? 2
+(
????
????
)
2
)=?? (
9
?? 2
-
1
3?? 6
)+?? =?? 2
? ?? (9×3-
1
3
×27)+?? =18?? ? ?? +18?? =18?? ??? =0
? h
2
(?? 2
+(
????
????
)
2
)=?? (
?? ?? 2
-
1
3?? 6
)
???????? ,???? ???????? , (
????
????
)=0
h
2
(
1
3
)=?? (9×3-
1
3
×27)=18?? h=54??
? Equation becomes
54(?? 2
+(
????
????
)
2
)=?? (
9
4
2
-
1
3?? 6
)
? ?? 2
+(
????
????
)
2
=
1
5?? (
9
?? 2
-
1
3?? 6
)
? (
????
????
)
2
=
27?? 4
-1-162?? 8
162?? 6
?
????
????
=
v
27
162
?? 4
-
1
162
-?? 8
? =
?? 3
????
v
?? 4
6
-
1
162
-?? 8
????
integrating both sides, we get
?
36?? 3
????
v1-(36?? 4
-3)
2
=? ????
?
sin
-1
(36?? 4
-3)
4
=?? +??
Now, at ?? =
1
?? =
1
v3
,?? =0 (apse)
?
sin
-1
(
36
9
-3)
4
=0+?? ?
sin
-1
(1)
4
=?? ??? =
?? 8
?
sin
-1
(36?? 4
-3)
4
=?? +
?? 8
? sin
-1
(36?? 4
-3)=4?? +
?? 2
? 36?? 4
-3=sin (
?? 2
+4?? )=cos 4?? ? (cos
2
2?? -sin
2
2?? )=36?? 4
-3
? 2cos
2
2?? =36?? 4
-2
? cos
2
2?? =18?? 4
-1
? (cos
2
?? -sin
2
?? )
2
+(cos
2
?? +sin
2
?? )
2
=18?? 4
? cos
4
?? +sin
4
?? -2cos
2
?? sin
2
?? +cos
4
?? +sin
4
?? +2cos
2
?? sin
2
?? =18?? 4
? 2cos
4
?? +2sin
4
?? =18?? 4
? cos
4
?? +sin
4
?? =9?? 4
=
9
?? 4
? ?? 4
cos
4
?? +?? 4
sin
4
?? =9
? ?? 4
+?? 4
=9
? Path is the curve ?? 4
+?? 4
=9.
4.3 A particle moves in a plane under a force, towards a fixed centre, proportional
to the distance. If the path of the particle has two apsidal distance ?? ,?? (?? >?? ) , then
find the tyuation of the path.
(2015 : 15 Marks)
Solution:
Page 5
Edurev123
4. Work & Energy
4.1 A body is describing an ellipse of eccentricity e under the action of a central
force directed towards a focus and when at the nearer apse the centre of the force
is transferred to the other focus. Find the eccentricity of the new orbit in terms of
the eccentricity of the original orbit.
(2009: 12 Marks)
Solution:
Approach : The motion in an ellipse in under inverse square force and the velocity
determines the eccentricity of the ellipse. So balance the velocity between the two orbits.
Let ?? and ?? '
be the foci of the original orbits and the force initially be directed towards ?? .
The nearer apse is at ?? .
Since the motion is in an ellipse the velocity at any point is
given by
?? 2
=?? (
2
?? -
1
?? )
At ?? ,?? =???? =???? -???? =?? -???? =?? (1-?? )
? ?? 2
=?? (
2
?? (1-?? )
-
1
?? )=
?? (1+?? )
?? (1-?? )
When the centre of force is changed to ?? '
the velocity remains same. Let ?? '
be the semi
major axis of the new orbit.
Then: ?? 2
=
?? (1+?? )
?? (1-?? )
=?? (
2
?? -
1
?? '
)
?? =?? ?? '
=???? +???? =?? +???? =?? (1+?? )
?
?? (1+?? )
?? (1-?? )
=?? (
2
?? (1+?? )
-
1
?? '
)
?
1+?? 1-?? =
2
1+?? -
?? ?? '
Again the direction of velocity is not changed, the velocity still being perpendicular to the
direction of motion the point ?? is an apse for the new orbit as well.
? ?? '
?? =?? '
(1-?? '
)=?? (1+?? )
?
?? ?? '
=
1-?? '
1+??
?
1+?? 1-?? =
2
1+?? ·
1-?? '
1+?? =
1+?? '
1+??
? ?? '
=
(1+?? )
2
1-?? -1=
?? 2
+3?? 1-?? =
?? (?? +3)
1-??
4.2 A particle moves with a central acceleration ?? (?? ?? -?? ?? ) , being projected from
an apse at a distance v?? with velocity ?? v?? ?? . Show that its path is the curve ?? ?? +
?? ?? =?? .
(2010 : 20 Marks)
Solution:
The equation of motion is
h
2
(?? +
?? 2
?? ?? ?? 2
)=
?? ?? 2
where ?? =
1
??
?? = central accel
=?? (
1
?? 5
-
9
?? )
? Equation becomes h
2
(?? +
?? 2
?? ?? ?? 2
) =
?? ?? 2
(
1
?? 5
-
9
?? )
? h
2
(?? +
?? 2
?? ?? ?? 2
) =?? (
1
4
7
-
9
4
3
)
Multiplying both sides by 2
????
????
and integrating, we get
h
2
(?? 2
+(
????
????
)
2
)=?? (
-2
6?? 6
+
9×2
2?? 2
)+?? ? h
2
(?? 2
+(
????
????
)
2
)=?? (
9
?? 2
-
1
3?? 6
)+?? =?? 2
? ?? (9×3-
1
3
×27)+?? =18?? ? ?? +18?? =18?? ??? =0
? h
2
(?? 2
+(
????
????
)
2
)=?? (
?? ?? 2
-
1
3?? 6
)
???????? ,???? ???????? , (
????
????
)=0
h
2
(
1
3
)=?? (9×3-
1
3
×27)=18?? h=54??
? Equation becomes
54(?? 2
+(
????
????
)
2
)=?? (
9
4
2
-
1
3?? 6
)
? ?? 2
+(
????
????
)
2
=
1
5?? (
9
?? 2
-
1
3?? 6
)
? (
????
????
)
2
=
27?? 4
-1-162?? 8
162?? 6
?
????
????
=
v
27
162
?? 4
-
1
162
-?? 8
? =
?? 3
????
v
?? 4
6
-
1
162
-?? 8
????
integrating both sides, we get
?
36?? 3
????
v1-(36?? 4
-3)
2
=? ????
?
sin
-1
(36?? 4
-3)
4
=?? +??
Now, at ?? =
1
?? =
1
v3
,?? =0 (apse)
?
sin
-1
(
36
9
-3)
4
=0+?? ?
sin
-1
(1)
4
=?? ??? =
?? 8
?
sin
-1
(36?? 4
-3)
4
=?? +
?? 8
? sin
-1
(36?? 4
-3)=4?? +
?? 2
? 36?? 4
-3=sin (
?? 2
+4?? )=cos 4?? ? (cos
2
2?? -sin
2
2?? )=36?? 4
-3
? 2cos
2
2?? =36?? 4
-2
? cos
2
2?? =18?? 4
-1
? (cos
2
?? -sin
2
?? )
2
+(cos
2
?? +sin
2
?? )
2
=18?? 4
? cos
4
?? +sin
4
?? -2cos
2
?? sin
2
?? +cos
4
?? +sin
4
?? +2cos
2
?? sin
2
?? =18?? 4
? 2cos
4
?? +2sin
4
?? =18?? 4
? cos
4
?? +sin
4
?? =9?? 4
=
9
?? 4
? ?? 4
cos
4
?? +?? 4
sin
4
?? =9
? ?? 4
+?? 4
=9
? Path is the curve ?? 4
+?? 4
=9.
4.3 A particle moves in a plane under a force, towards a fixed centre, proportional
to the distance. If the path of the particle has two apsidal distance ?? ,?? (?? >?? ) , then
find the tyuation of the path.
(2015 : 15 Marks)
Solution:
?? =?? 2
h
2
[?? +
?? 2
?? ?? ?? 2
]
? -
?? 1
=?? 2
h
2
(?? +
?? 2
?? ?? ?? 2
)?-
?? ?? 3
=h
2
(?? +
?? 2
?? ?? ?? 2
)
? -
?? ?? 3
·
2?? ????
=h
2
[?? ·2
????
????
+
?? 2
?? ?? ?? 2
·
2????
????
]
? ?? +
?? ?? 2
=h
2
[?? 2
+(
????
????
)
2
]
or ?? 2
=h
2
[?? 2
+(
????
????
)
2
]=
?? ?? 2
+?? At ?? =
1
?? ,
????
?? 6
=0??? +?? ?? 2
=h[
1
?? 2
]
? ?? +??ˆ
?? 2
=
h
2
?? 2
Similarly, ?? +?? ?? 2
=
h
2
?? 2
? ?? (?? 2
-?? 2
)=h
2
(
1
?? 2
-
1
?? 2
)?
?? h
2
=
-1
?? 2
?? 2
? ?? +?? ?? 2
=
-?? ?? 2
?? 2
?? 2
??? =-?? (?? 2
+?? 2
)
Thus, -?? ?? 2
?? 2
[?? 2
+(
????
????
)
2
]=-?? (?? 2
+?? 2
)+
?? ?? 2
? ?? 2
(
????
????
)
2
=
(?? 2
+?? 2
-
1
?? 2
)
?? 2
?? 2
? (
????
????
)
2
=
?? 2
+?? 2
?? 2
?? 2
-
1
?? 2
?? 2
?? 2
-?? 2
=
1
?? 2
[
?? 2
+?? 2
?? 2
?? 2
?? 2
-
1
?? 2
?? 2
-?? 4
]
=
1
?? 2
[-(
1
2
?? 2
+?? 2
?? 2
?? 2
-?? 2
)
2
+
1
4
(
?? 2
+?? 2
?? 2
?? 2
)
2
-
1
?? 2
?? 2
]
?????? ?? ?? 2
=
1
4
(
?? 2
+?? 2
?? 2
?? 2
)
2
-
1
?? 2
?? 2
=
1
4(?? 2
?? 2
)
2
[(?? 2
+?? 2
)
2
-4?? 2
?? 2
]
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FAQs on Work & Energy - Mathematics Optional Notes for UPSC
| 1. What is the relationship between work and energy? |  |
Ans. Work is defined as the transfer of energy from one object to another, or the movement of an object by a force. Energy, on the other hand, is the ability to do work. The relationship between work and energy is that work is a way to transfer energy from one object to another or transform energy from one form to another.
| 2. How is work calculated in physics? | |
Ans. In physics, work is calculated by multiplying the force applied to an object by the distance the object moves in the direction of the force. Mathematically, work (W) is equal to force (F) multiplied by displacement (d) in the direction of the force, expressed as W = Fd.
| 3. What are the different forms of energy? | |
Ans. Energy can exist in various forms, including kinetic energy (energy of motion), potential energy (stored energy), thermal energy (heat energy), chemical energy (energy stored in bonds of chemical compounds), and electromagnetic energy (energy carried by light and other electromagnetic waves).
| 4. How is mechanical energy conserved in a system? | |
Ans. Mechanical energy is conserved in a system when the sum of the kinetic energy (energy of motion) and potential energy (stored energy) remains constant, assuming no external forces such as friction or air resistance are present. This is known as the conservation of mechanical energy.
| 5. What is the work-energy theorem? | |
Ans. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, it can be expressed as the work done (W) on an object is equal to the change in its kinetic energy (∆KE), represented by the equation W = ∆KE.
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Nov 24, 2024 Last updated |
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