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Page 1
Edurev123
Dynamics & Statics
1. Rectilinear Motion
1.1 One end of a light elastic string of natural length ?? and modulus of elasticity
?????? is attached to a fixed point ?? and the other end to a particle of mass ?? . The
particle initially held at rest at ?? is bit fall. Find the greatest extension of the string
during motion and show that the particle will reach ?? again after a time.
(?? +?? -?????? -?? ?? )v
?? ?? ??
(2009 : 20 Marks)
Solution:
Let ?? be the equilibrium position of the body and ???? =?? .
In position of equilibrium
???? =2???? ·
?? ?? ??? =
?? 2
When particle is dropped from ?? it free falls till ?? .
?? ?? =0+2?? ×????
? ?? ?? =v2????
After ?? the tension in the string starts acting which is balanced at ?? . Beyond ?? the
particles moves due to its velocity till it comes to stop at ?? .
At any point ?? with ???? =?? .
Page 2
Edurev123
Dynamics & Statics
1. Rectilinear Motion
1.1 One end of a light elastic string of natural length ?? and modulus of elasticity
?????? is attached to a fixed point ?? and the other end to a particle of mass ?? . The
particle initially held at rest at ?? is bit fall. Find the greatest extension of the string
during motion and show that the particle will reach ?? again after a time.
(?? +?? -?????? -?? ?? )v
?? ?? ??
(2009 : 20 Marks)
Solution:
Let ?? be the equilibrium position of the body and ???? =?? .
In position of equilibrium
???? =2???? ·
?? ?? ??? =
?? 2
When particle is dropped from ?? it free falls till ?? .
?? ?? =0+2?? ×????
? ?? ?? =v2????
After ?? the tension in the string starts acting which is balanced at ?? . Beyond ?? the
particles moves due to its velocity till it comes to stop at ?? .
At any point ?? with ???? =?? .
?? ?? 2
?? ?? ?? 2
=???? -(2???? )
?? +?? ?? =-2????
?? ?? ?
?? 2
?? ?? ?? 2
=-
2?? ?? ??
So, the body performs SHM with centre ?? .
Multiplying with 2
????
????
and integrating
(
?? ?? ????
)
2
=-
2?? ?? ?? 2
+?? ???? ?? , ?? ?? =v2???? ,?? =-
?? 2
2???? =-
2?? ?? ?? 2
4
+?? ??? =
5
2
????
? (
????
????
)
2
=
5
2
???? -
2?? ?? ?? 2
(??)
At ?? ,
????
????
=0
? ?? 2
=
5
4
?? 2
??? =
v5
2
??
So greatest distance through which particle falls
=???? =???? +???? +????
=?? +
?? 2
+
v5?? 2
=
(3+v5)?? 2
Greatest extension =
(1+v5)?? 2
From (i),
????
????
=
v
2?? ?? [
5
4
?? 2
-?? 2
]
1/2
where positive sign is taken as particle is moving in direction of increasing ?? .
v
?? 2?? ????
v
5
4
?? 2
-?? 2
=????
Page 3
Edurev123
Dynamics & Statics
1. Rectilinear Motion
1.1 One end of a light elastic string of natural length ?? and modulus of elasticity
?????? is attached to a fixed point ?? and the other end to a particle of mass ?? . The
particle initially held at rest at ?? is bit fall. Find the greatest extension of the string
during motion and show that the particle will reach ?? again after a time.
(?? +?? -?????? -?? ?? )v
?? ?? ??
(2009 : 20 Marks)
Solution:
Let ?? be the equilibrium position of the body and ???? =?? .
In position of equilibrium
???? =2???? ·
?? ?? ??? =
?? 2
When particle is dropped from ?? it free falls till ?? .
?? ?? =0+2?? ×????
? ?? ?? =v2????
After ?? the tension in the string starts acting which is balanced at ?? . Beyond ?? the
particles moves due to its velocity till it comes to stop at ?? .
At any point ?? with ???? =?? .
?? ?? 2
?? ?? ?? 2
=???? -(2???? )
?? +?? ?? =-2????
?? ?? ?
?? 2
?? ?? ?? 2
=-
2?? ?? ??
So, the body performs SHM with centre ?? .
Multiplying with 2
????
????
and integrating
(
?? ?? ????
)
2
=-
2?? ?? ?? 2
+?? ???? ?? , ?? ?? =v2???? ,?? =-
?? 2
2???? =-
2?? ?? ?? 2
4
+?? ??? =
5
2
????
? (
????
????
)
2
=
5
2
???? -
2?? ?? ?? 2
(??)
At ?? ,
????
????
=0
? ?? 2
=
5
4
?? 2
??? =
v5
2
??
So greatest distance through which particle falls
=???? =???? +???? +????
=?? +
?? 2
+
v5?? 2
=
(3+v5)?? 2
Greatest extension =
(1+v5)?? 2
From (i),
????
????
=
v
2?? ?? [
5
4
?? 2
-?? 2
]
1/2
where positive sign is taken as particle is moving in direction of increasing ?? .
v
?? 2?? ????
v
5
4
?? 2
-?? 2
=????
If ?? 1
is time from ?? to ??
v
?? 2?? ? ?
v5//2
-?? /2
?
????
v
5
4
?? 2
-?? 2
=??
?? 0
?????
? v
?? 2?? [sin
-1
?? v5??/2
]
-?? /2
v5//2
=?? 1
?? 1
=v
?? 2?? [
?? 2
-sin
-1
-
1
v5
]=v
?? 2?? [
?? 2
+sin
1
v5
]
=v
?? 2?? [
?? 2
+tan
-1
1
2
]
=v
?? 2?? [
?? 2
+
?? 2
-tan
-1
2]
=v
?? 2?? [?? -tan
-1
2]
Time in falling from ?? to ??
1
2
?? ?? 2
2
=?? ??? 2
=v
2?? ??
? Total time taken to come back to
?? =2v
?? 2?? [?? -tan
-1
2+2]
=v
2?? ?? [?? +2-tan
-1
2]
1.2 The velocity of a train increases from 0 to ?? at a constant acceleration ?? ?? , then
remains constant for an interval and again decreases to 0 at a constant
retardation ?? ?? . If the total cistance described is ?? ?? find the total time taken.
(2011 : 10 Marks)
Solution:
(i) When velocity increases from 0 to ?? . Using ?? =?? +???? ,?? =?? ,?? =0,?? =?? ?? ,?? =?? 1
, we
get,
?? =0+?? 1
?? 1
Page 4
Edurev123
Dynamics & Statics
1. Rectilinear Motion
1.1 One end of a light elastic string of natural length ?? and modulus of elasticity
?????? is attached to a fixed point ?? and the other end to a particle of mass ?? . The
particle initially held at rest at ?? is bit fall. Find the greatest extension of the string
during motion and show that the particle will reach ?? again after a time.
(?? +?? -?????? -?? ?? )v
?? ?? ??
(2009 : 20 Marks)
Solution:
Let ?? be the equilibrium position of the body and ???? =?? .
In position of equilibrium
???? =2???? ·
?? ?? ??? =
?? 2
When particle is dropped from ?? it free falls till ?? .
?? ?? =0+2?? ×????
? ?? ?? =v2????
After ?? the tension in the string starts acting which is balanced at ?? . Beyond ?? the
particles moves due to its velocity till it comes to stop at ?? .
At any point ?? with ???? =?? .
?? ?? 2
?? ?? ?? 2
=???? -(2???? )
?? +?? ?? =-2????
?? ?? ?
?? 2
?? ?? ?? 2
=-
2?? ?? ??
So, the body performs SHM with centre ?? .
Multiplying with 2
????
????
and integrating
(
?? ?? ????
)
2
=-
2?? ?? ?? 2
+?? ???? ?? , ?? ?? =v2???? ,?? =-
?? 2
2???? =-
2?? ?? ?? 2
4
+?? ??? =
5
2
????
? (
????
????
)
2
=
5
2
???? -
2?? ?? ?? 2
(??)
At ?? ,
????
????
=0
? ?? 2
=
5
4
?? 2
??? =
v5
2
??
So greatest distance through which particle falls
=???? =???? +???? +????
=?? +
?? 2
+
v5?? 2
=
(3+v5)?? 2
Greatest extension =
(1+v5)?? 2
From (i),
????
????
=
v
2?? ?? [
5
4
?? 2
-?? 2
]
1/2
where positive sign is taken as particle is moving in direction of increasing ?? .
v
?? 2?? ????
v
5
4
?? 2
-?? 2
=????
If ?? 1
is time from ?? to ??
v
?? 2?? ? ?
v5//2
-?? /2
?
????
v
5
4
?? 2
-?? 2
=??
?? 0
?????
? v
?? 2?? [sin
-1
?? v5??/2
]
-?? /2
v5//2
=?? 1
?? 1
=v
?? 2?? [
?? 2
-sin
-1
-
1
v5
]=v
?? 2?? [
?? 2
+sin
1
v5
]
=v
?? 2?? [
?? 2
+tan
-1
1
2
]
=v
?? 2?? [
?? 2
+
?? 2
-tan
-1
2]
=v
?? 2?? [?? -tan
-1
2]
Time in falling from ?? to ??
1
2
?? ?? 2
2
=?? ??? 2
=v
2?? ??
? Total time taken to come back to
?? =2v
?? 2?? [?? -tan
-1
2+2]
=v
2?? ?? [?? +2-tan
-1
2]
1.2 The velocity of a train increases from 0 to ?? at a constant acceleration ?? ?? , then
remains constant for an interval and again decreases to 0 at a constant
retardation ?? ?? . If the total cistance described is ?? ?? find the total time taken.
(2011 : 10 Marks)
Solution:
(i) When velocity increases from 0 to ?? . Using ?? =?? +???? ,?? =?? ,?? =0,?? =?? ?? ,?? =?? 1
, we
get,
?? =0+?? 1
?? 1
? ?? 1
=
?? ?? 1
.
Using ?? =???? +
1
2
?? ?? 2
We get,
?? 1
=0+
?? 2
·?? 1
·?? 1
2
=
1
2
·?? 1
·
?? 2
?? 1
2
=
?? 2
2?? 1
(ii) When velocity decreases from ?? to 0 . Using ?? =?? +?? ?? 3
?? =0,?? =?? ,?? =-?? 2
,?? =?? 2
,
we get, 0=?? -?? 2
?? 2
? ?? 2
=
?? ?? 2
Using
?? =???? +
1
2
?? ?? 1
2
?? 2
=?? ?? 2
-
1
2
·?? 2
?? 2
2
=
?? 2
?? 2
-
1
2
?? 2
?? 2
=
?? 2
2?? 2
(iii) Since the total distance travelled is ' ?? ', the distance travelled during the constant
velocity phase
=?? -
?? 2
2?? 1
-
?? 2
2?? 2
Time taken to travel this distance, ?? 3
=
?? -
?? 2
2?? 1
-
?? 2
2?? 2
??
=
?? ?? -
?? 2?? 1
-
?? 2?? 2
? Total time taken =?? 1
+?? 2
+?? 3
=
?? ?? 1
+
?? ?? 2
+
?? ?? -
?? 2?? 1
-
?? 2?? 2
=
?? ?? +
?? 2?? 1
+
?? 2?? 2
Page 5
Edurev123
Dynamics & Statics
1. Rectilinear Motion
1.1 One end of a light elastic string of natural length ?? and modulus of elasticity
?????? is attached to a fixed point ?? and the other end to a particle of mass ?? . The
particle initially held at rest at ?? is bit fall. Find the greatest extension of the string
during motion and show that the particle will reach ?? again after a time.
(?? +?? -?????? -?? ?? )v
?? ?? ??
(2009 : 20 Marks)
Solution:
Let ?? be the equilibrium position of the body and ???? =?? .
In position of equilibrium
???? =2???? ·
?? ?? ??? =
?? 2
When particle is dropped from ?? it free falls till ?? .
?? ?? =0+2?? ×????
? ?? ?? =v2????
After ?? the tension in the string starts acting which is balanced at ?? . Beyond ?? the
particles moves due to its velocity till it comes to stop at ?? .
At any point ?? with ???? =?? .
?? ?? 2
?? ?? ?? 2
=???? -(2???? )
?? +?? ?? =-2????
?? ?? ?
?? 2
?? ?? ?? 2
=-
2?? ?? ??
So, the body performs SHM with centre ?? .
Multiplying with 2
????
????
and integrating
(
?? ?? ????
)
2
=-
2?? ?? ?? 2
+?? ???? ?? , ?? ?? =v2???? ,?? =-
?? 2
2???? =-
2?? ?? ?? 2
4
+?? ??? =
5
2
????
? (
????
????
)
2
=
5
2
???? -
2?? ?? ?? 2
(??)
At ?? ,
????
????
=0
? ?? 2
=
5
4
?? 2
??? =
v5
2
??
So greatest distance through which particle falls
=???? =???? +???? +????
=?? +
?? 2
+
v5?? 2
=
(3+v5)?? 2
Greatest extension =
(1+v5)?? 2
From (i),
????
????
=
v
2?? ?? [
5
4
?? 2
-?? 2
]
1/2
where positive sign is taken as particle is moving in direction of increasing ?? .
v
?? 2?? ????
v
5
4
?? 2
-?? 2
=????
If ?? 1
is time from ?? to ??
v
?? 2?? ? ?
v5//2
-?? /2
?
????
v
5
4
?? 2
-?? 2
=??
?? 0
?????
? v
?? 2?? [sin
-1
?? v5??/2
]
-?? /2
v5//2
=?? 1
?? 1
=v
?? 2?? [
?? 2
-sin
-1
-
1
v5
]=v
?? 2?? [
?? 2
+sin
1
v5
]
=v
?? 2?? [
?? 2
+tan
-1
1
2
]
=v
?? 2?? [
?? 2
+
?? 2
-tan
-1
2]
=v
?? 2?? [?? -tan
-1
2]
Time in falling from ?? to ??
1
2
?? ?? 2
2
=?? ??? 2
=v
2?? ??
? Total time taken to come back to
?? =2v
?? 2?? [?? -tan
-1
2+2]
=v
2?? ?? [?? +2-tan
-1
2]
1.2 The velocity of a train increases from 0 to ?? at a constant acceleration ?? ?? , then
remains constant for an interval and again decreases to 0 at a constant
retardation ?? ?? . If the total cistance described is ?? ?? find the total time taken.
(2011 : 10 Marks)
Solution:
(i) When velocity increases from 0 to ?? . Using ?? =?? +???? ,?? =?? ,?? =0,?? =?? ?? ,?? =?? 1
, we
get,
?? =0+?? 1
?? 1
? ?? 1
=
?? ?? 1
.
Using ?? =???? +
1
2
?? ?? 2
We get,
?? 1
=0+
?? 2
·?? 1
·?? 1
2
=
1
2
·?? 1
·
?? 2
?? 1
2
=
?? 2
2?? 1
(ii) When velocity decreases from ?? to 0 . Using ?? =?? +?? ?? 3
?? =0,?? =?? ,?? =-?? 2
,?? =?? 2
,
we get, 0=?? -?? 2
?? 2
? ?? 2
=
?? ?? 2
Using
?? =???? +
1
2
?? ?? 1
2
?? 2
=?? ?? 2
-
1
2
·?? 2
?? 2
2
=
?? 2
?? 2
-
1
2
?? 2
?? 2
=
?? 2
2?? 2
(iii) Since the total distance travelled is ' ?? ', the distance travelled during the constant
velocity phase
=?? -
?? 2
2?? 1
-
?? 2
2?? 2
Time taken to travel this distance, ?? 3
=
?? -
?? 2
2?? 1
-
?? 2
2?? 2
??
=
?? ?? -
?? 2?? 1
-
?? 2?? 2
? Total time taken =?? 1
+?? 2
+?? 3
=
?? ?? 1
+
?? ?? 2
+
?? ?? -
?? 2?? 1
-
?? 2?? 2
=
?? ?? +
?? 2?? 1
+
?? 2?? 2
=
?? ?? +
?? 2
(
1
?? 1
+
1
?? 2
)
1.3 A mass of ?????? ???? moving with a velocity of ?????? ?? /?????? strikes a fixed target
and is brought to rest in
?? ??????
sec. Find the impulse of the biow on the target and
assuming the resistance to be uniform throughout the time taken by the body in
coming to rest, find the distance through which it penetrates.
(2011: 20 Marks)
Solution:
Initial velocity, u=240 m/s
Final velocity, v=0(as the mass finally comes to rest)
Time taken to come to rest, t=0.01s
Using, v=u+at, we get
0=240+a*0.01
? a=-24000m/?? 2
The negative sign indicates that the velocity of the bullet is decreasing.
Using, ?? 2
-?? 2
=2???? ,???? h??????
0-(240)
2
=2*?? *(-24000)
? ?? =
-240×240
-2×24000
=0.1 m
Hence, the distance of penetration of the mass into the fixed target is 0.1 m.
impulse of the blow on the target,
?? = Change in momentum of the mass
=???? -???? 4
=?? (?? -4)=560×(-240)=-134400 kg m/s
1.4 (i) After a ball has been falling under gravity for 5 seconds it passes through a
pane of glass and loses half its velocity. If it now reaches the ground in 1 second,
find the height of glass above the ground.
(2011 : 10 Marks)
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