Volume 04 Issue 02-2024
34
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
A
BSTRACT
This article is devoted to one such theory the theory of collision of bodies. The impact phenomenon is well
described by a simple mathematical model. This article also discusses the issue of joint use of two
conservation lows
–
energy and momentum
–
to solve various physical problems.
K
EYWORDS
Energy, impulse, elastic, inelastic, theory of collision of bodies, absolutely elastic impact, total impulse, total
kinetic energy.
I
NTRODUCTION
Some physical theories are built on the model
adopted in mathematics
–
form a small number of
physical statements that play the same role as
axioms in mathematical theories, various
consequences are logically strictly deduced. This
article is devoted to one such theory
–
the theory
of collision of bodies. The impact phenomenon is
well described by a simple mathematical model.
Journal
Website:
http://sciencebring.co
m/index.php/ijasr
Copyright:
Original
content from this work
may be used under the
terms of the creative
commons
attributes
4.0 licence.
Research Article
METHODS FOR SOLVING PROBLEMS BASED ON THE LAWS
OF CONSERVATION OF ENERGY AND MOMENTUM
Submission Date:
February 05, 2024,
Accepted Date:
February 10, 2024,
Published Date:
February 15, 2024
Crossref doi:
https://doi.org/10.37547/ijasr-04-02-06
Saipnazarov Shaylovbek Aktamovich
Candidate Of Pedagogical Sciences, Uzbekistan
Nurmanov Ma’ruf Shaymardonovich
Tashkent University Of Economics, Uzbekistan
Fayziev Javlon Abduvoxidovich
Tashkent State University Of Economics, Uzbekistan
Volume 04 Issue 02-2024
35
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
This article discusses the issue of joint use of two
conservation laws
–
energy and momentum
–
to
solve various physical problems. During exams in
physics, a situation often arises when students,
knowing well each conservation law separately,
experience “psychological” difficulties when it is
necessary to combine these laws together within
the framework of one problem. Moreover, more
often than not, the simpler, in our opinion, law of
conservation of momentum escapes attention.
Having written down the corresponding equation
for energy, the student no longer remembers the
impulse
–
and gets into trouble. Let’s consider
several specific examples of how the joint efforts
of energy and impulse lead to the desired result.
Example 1.
Two balls, made of the same
material and having masses
1
m
and
2
m
, move
towards each other with speeds
1
V
and
2
V
. Now
much will the temperature of the balls increase
after a frontal absolutely inelastic impact if the
specific heat capacity of the material of the balls is
C
? The initial temperatures of the balls were the
same.
Solution.
The change in the temperature of the balls is determined by the increase in the their internal
energy:
(
)
t
m
m
c
E
+
=
2
1
(1)
Many student mistakenly believe that as a result of an impact, all the initial kinetic energy of the
system
2
2
2
2
2
2
1
1
V
m
V
m
+
is converted into integral energy. At the same time, they forget that the balls cannot
stop after impact, since this would contradict the law of conservation of momentum
–
the initial momentum
of the system
2
2
1
1
V
m
V
m
+
, generally speaking, is not equal to zero. This means that when calculating energy,
it is necessary to take into account the kinetic energy of the balls in the final state.
Let us denote the speed of the balls stuck together after an absolutely inelastic impact by
v
and write
the laws of conservation of energy and momentum for the direction of motion of the first ball:
(
)
E
V
m
m
V
m
V
m
+
+
=
+
2
2
2
2
2
1
2
2
2
2
1
1
(2)
(
)
V
m
m
V
m
V
m
2
1
2
2
1
1
+
=
−
(3)
Solving the three equations obtained together, we find the desired temperature increase:
(
)
(
)
2
2
1
2
2
1
2
1
2
m
m
c
V
V
m
m
t
+
+
=
(4)
Volume 04 Issue 02-2024
36
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
Example 2.
Two cars, the masses of which are
1
M
and
2
M
, are moving towards each other with
speed
1
V
and
2
V
. During a collision, four identical
buffer springs move apart. Find the maximum
deformation of each spring if its spring constant
is
k
.
Solution.
In this problem, unlike the previous
one, you can use the law of conservation of
mechanical energy (assuming that friction is low
and the springs are ideal), equating the initial
energy of the cars to the energy of the system at
the moment when the deformation of the springs
“
x
” is maximum. In this case, the desired value “
x
”
will be included in the potential energy of elastic
deformation of the springs:
2
4
2
kx
E
p
=
(5)
However, in addition to this energy, it is also
necessary to take into account the kinetic energy
of the cars.
The fact that at maximum approach the cars do
not stop (which, unfortunately, many students
forget) follows, as in the previous problem from
the law of conservation of momentum. The only
peculiarity of this moment is that at maximum
deformation of the springs from the speed of the
cars are the same:
V
V
V
=
=
2
1
. Therefore, the laws
of conservation of energy and momentum are as
follows:
(
)
2
4
2
2
2
2
2
2
1
2
2
2
2
1
1
kx
V
M
M
V
M
V
M
+
+
=
+
(6)
(
)
V
M
M
V
M
V
M
2
1
2
2
1
1
+
=
−
(7)
From here we get
(
) (
)
2
1
2
1
2
1
2
1
V
V
M
M
K
M
M
X
+
+
=
(8)
Example 3.
A block of mass
M
, hanging on parallel threads of length
l
, is hit by a horizontally flying bullet
of mass
m
and gets stuck in it (Fig.1). As a result of the impact, each thread is deflected by angle
. Find
the initial speed of the bullet
V
. the threads are considered ideal (weightless an inextensible).
Solution.
As can be seen from the figure, the angle of deflection of the threads
is related to the height
h
to which the block rises:
(
)
cos
1
−
=
l
h
and the height
h
can be related to the potential energy of the block and bullet in the final state:
(
)
gh
m
M
E
p
+
=
(9)
Volume 04 Issue 02-2024
37
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
a)
b)
Fig. 1.
The question arises: is the law of conservation of
mechanical energy satisfied in this situation? In
other words, is the energy of the system in the
final state equal to its initial energy, i.e. kinetic
energy of the bullet
2
2
mv
? The answer, of course,
is negative. After all, we know that during an
inelastic impact, part of the mechanical energy
transforms into internal energy.
Now to be? Let‘s consider another,
intermediate state of the system
–
immediately
after the end of the impact, when the bullet is
already stuck in the bar, but the threads are still
vertical. The energy of the system this state is
simply the kinetic energy of the block with the
bullet:
(
)
2
2
V
M
m
E
k
+
=
, where
V
- is their total
speed. After the inelastic impact has already
ended, no more energy will be lost, and we can
write
p
k
E
E
=
, or
(
)
(
)
gh
m
M
V
M
m
+
=
+
2
2
The speed
V
can be related to the initial
speed of the bullet using the law of conservation
of momentum
(
)
V
M
m
mV
+
=
From the last two equations, taking into
account the expression for
h
, we have
2
sin
1
2
gl
m
M
V
+
=
Note that in this problem the laws of
conservation of momentum and energy do not
work simultaneously, but as if in turn. It turns out
that this is not so easy to understand, and many
students solve problems of this type using only
the law of conservation of energy, obtaining, of
course, incorrect results.
Example 4.
On a block of length
l
and
mass
M
, located on a smooth horizontal surface,
lies a small div of mass
m
(fig.2). The coefficient
of friction between the div and the block
. At
what speed
V
must the system move so that after
an elastic impact of the block on the wall the div
falls from the block?
Solution.
The impact of the block against the wall
will cause its speed to abruptly change to the
opposite. The speed of the div will not have time
l
m
V
M
l
h
M+m
Volume 04 Issue 02-2024
38
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
to change during the impact, and it will begin to
slide along the impact, and it will begin to slide
along the block. Let us find the distance
x
the div
will move relative to the block before the end of
sliding. It is clear that the condition
l
x
will be
the condition for the div to fall from the block.
The work done by the distance
x
Fig. 2.
mgx
A
−
=
,
which, in turn, is equal to the change in the kinetic energy of the system:
(
)
+
−
+
=
2
2
2
2
2
2
MV
mV
V
M
m
A
here
V
- is the speed of the block with the div at the moment when the div stops relative to the block.
This speed can be found from the law of conservation of momentum.
(
)
V
m
M
mV
MV
+
=
−
Solving all three equations together, we get
(
)
.
2
2
m
M
g
MV
x
+
=
The condition
l
x
allows you to find the required speed:
+
M
m
gl
V
1
2
1
Example 4.
At the left edge of the tree with length
m
L
2
,
0
=
and mass
kg
M
1
=
lies a cube with
mass
kg
m
3
,
0
=
(Fig.3). the cube is given a push to the horizontal speed
s
m
V
/
1
0
=
to the right. Assuming
that the cart is stationary at the initial moment, determine at what distance from the left edge of the cart
the cube will be after its sliding relative to the cart stops. The coefficient of friction between the cube and
the walls is assumed to be absolutely elastic. The cart rides on the table without friction.
m
l
V
M
Volume 04 Issue 02-2024
39
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
Solution.
The easiest way to solve this problem is form energy considerations. According to the
law of conservation of energy, the decrease in the kinetic energy of the system is equal to the amount of
heat released, which, in turn, is equal to the work of the sliding friction force on the braking distance
l
:
(
)
.
2
2
2
0
2
mgl
l
F
Q
mV
u
m
M
E
fr
x
−
=
=
=
−
+
=
The speed of the system
U
after the cessation of slipping is easy to find from the law of conservation
of momentum
(
)
U
m
M
mV
+
=
0
Fig. 3.
After simple transformations we get
38
,
0
2
2
0
+
=
M
m
g
V
l
This means that the cube will stop at a distance
(
)
m
L
l
L
x
02
,
0
=
−
−
=
from the left edge of the cart.
C
ONCLUSION
Galileo also carried out a series of experiments to
clarify the laws of the collision of bodies. These
experiments, however, did not lead him to
definite conclusions. A contemporary of Galileo,
the Prague professor Marzi, published in his work
some of the results of his research into the
phenomenon of impact. In particular, he knew
that a div, having elastically hit an identical
div at rest, loses its speed, imparting it to this
div. The first detailed study of the laws of
impact was undertaken in 1668 at the suggestion
of the Royal Society of London. Three outstanding
mechanics and mathematicians Wallis, Rehn and
Huygens presented their works in which they
outlined the laws of motion of colliding bodies.
John Wallis limited himself, without specifying
this, to considering a completely inelastic impact.
He proceeded from the hypothesis of
conservation of the total momentum of colliding
bodies. Christopher wren outlined the rules for
calculating elastic impact. Wren, like Wallies, did
not provide any theoretical considerations, but to
test his rules he performed a number of simple
Volume 04 Issue 02-2024
40
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
02
Pages:
34-40
SJIF
I
MPACT
FACTOR
(2021:
5.478
)
(2022:
5.636
)
(2023:
6.741
)
OCLC
–
1368736135
and convincing experiments. Newton referred to
these experiments in his famous “Mathematical
Principles of Natural Philosophy” (1687).
Christian Huygens’s competition memoir was the
most complete study of impact theory. It outlined
the derivation of the relations of the impact
theory
, based on Galileo’s principle of relativity.
The Royal Society of London published only the
memoirs of Wallis and Renault. In Huygens
memoir on the motion of bodies under the
influence of an impact, the law of conservation of
momentum is derived from Gali
leo’s principle of
relativity. From a mathematical point of view,
Huygens’s reasoning may not be
considered
completely rigorous.
R
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1.
Трофимова Т.И. Курс физики: Учебное
пособие для вузов. 7
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е изд. стер М.: Высш.
шк., 2002,
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542 ст.
2.
Савельев И.В. Курс общей физики. Учебное
пособие для вузов. СПб.: СпецЛит, 2002.
336 с.
3.
Волькенштейн В.С. Сборник задач по
общему курсу физики. Изд. Доп. и перераб.
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Чертов А.Г., Воробьев А.А. Задачник по
физике: Учеб. пособие для вузов. –
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