Download EXPERIMENT 4: MOMENTUM AND COLLISION PURPOSE OF THE

EXPERIMENT 4: MOMENTUM AND COLLISION
PURPOSE OF THE EXPERIMENT: For elastic collisions in an isolated system, examine
the conservation of linear momentum and kinetic energy.
ELASTIC COLLISION
GENERAL INFORMATION
Linear momentum of an object "P", is the product of mass and velocity
(4.1)
Here we will talk about the momentum briefly from the linear momentum. However, only the
net external force when applied, we know that the speed of the object changed and this means
that the momentum change. This fact can be seen from Newton's second law. According to
Newton's second law for a constant mass of a body;
(4.2)
When m is constant, this equation is written as follows:
(4.3)
From the above equation, if an external force acts on an object, the object's momentum does
not protected. That is, momentum does not change with time. If
then
(4.4)
(4.5)
Here the momentum not change with time, the object always have the same momentum.
N- particle system consist of m1, m2,…….mN masses can be generalized based on the above
results. When we are dealing with a system of objects (m1, m2,…….mN), the total momentum
of the system is the vector sum of the individual momentums:
(4.6)
Where are,
,
,…..
(4.7)
The sum in the equation (4.6) is a vector sum process. In this situation, if a generalized
equation (4.3);
(4.8)
where Fext, the system comprised of the particles refers to the net external force. This external
forces may be friction and gravity. Hence in the system formed by particles it does not have
any total external force and the total momentum of the system will be protected. So;
(4.9)
(4.10)
Above collection is a vector sum process.
If the system of interest is isolated, that is not acted upon by an external force, the total
momentum of the system remains constant.
In this study, in the air table a horizontal position will be investigated momentum
conversation with two-pucks system. In the horizontal position, any external force does not
occur on the pucks on the air table which minimizes friction. Therefore, the total momentum
of the pucks seems to be preserved. Pucks are provided collisions, the total momentum before
and after the collision are measured and compared. The spots obtained in the experiment are
given in the following figure on data paper.
Figure 4.1 : Data points of the two magnetic pucks that performed elastic collisions on the air table in the
horizontal positions.
Velocities of two pucks are respectively
,
and
,
before and after the collision.
Momentum is conversed because the system is isolated and at any time;
(4.11)
(4.12)
Where the momentums are
,
. Because the masses of the pucks are
same, the above equation is converted as follows.
(4.13)
It is shown method that the vector sum in the equation 4.13 (above) are geometrically found
in the experimental procedure section. Since the system is isolated, the momentum is
conserved at an inelastic collision. In such a collision, two pucks move sticking together, as
an object having a mass 2m and the velocity
. The points in the data sheet should resemble
the Figure 4.1.
Another concept that will be encountered in this experiment is the center of mass (CM). The
CM of symmetrical objects such as cubes (figure 4.2a) and spheres (figure 4.2b ) is the same
with the theirs geometric center. The CM of the shape in the figure 4. 2c is predictably the
midpoint of the rod.
Figure 4.2: CM of some symmetric homogeneous objects.
CM for mass distribution in different shapes should be redefined. CM of position vector
of
a system with N particles is defined as follows, (seen Figure 4.2)
(4.14)
Where
are position vectors and
are masses.
Figure4. 3: CM of R for mass distribution.
If the particles change their position with time, the position of the CM changes and the
vector exchange rate of CM considered as the center of mass velocity.
(4.15)
when we take the derivative of both sides of equation (4.14) for particles with constant mass.
(4.16)
(4.17)
are obtained. The points in the equation (4.16) means derivative so that these are only
speeds. When the above equation is applied to the two pucks systems;
(4.18)
(4.19)
are obtained. Since the masses of the pucks are equal here, equation 4.20 is obtained by
removing the masses. Thus velocity of CM,
(4.20)
There are some important consequences of the above equation. In two pucks sytem, firstly,
while maintaining momentum total on the right side of the equation are constant (compare
with equation 4.13). This situation means that velocity of CM is constant under these
conditions. In other words, the CM moves at a constant velocity. (Constant velocity means
that the magnitude and direction of the speed does not changes ). Thus CM of the system
always moves at a linear constant speed for a system isolated that the total momentum is
conserved. This situation also shows that the velocity is equal to half of total velocities of both
masses. Therefore, velocity equations are as follows for our two-pucks system, before and
after the collision.
(4.21)
(4.22)
In this experiment, It will be investigated kinetic energy conservation of the pucks for the
collision secondly. Let us remember the definition of kinetic energy K of an object that have
mass m and linear velocity v.
(4.23)
Therefore total kinetic energy of the two-pucks system prior to the elastic collision;
(4.24)
and kinetic energy after the collision;
(4.25)
However, the two pucks sticks each other in an inelastic collisions. After this collision, these
two pucks moves as an object that have mass 2m and velocity v. Thus ıts kinetic energy;
(4.26)
Since the kinetic energy is a scalar quantity, the total in equation (4.25) and (4.26) is a scalar
collection process. On the other hand, the kinetic energy is preserved in almost elastic
collision that is K =
EXPERIMENTAL PROCEDURE
Please run the pump switch (P).
Gently launch the two pucks diagonally towards each other so they closely approach
each other and repel without touching. Repeat this process several times until
sufficient degree appropriate get collision.
Now, set the period from the the spark generator (example 60 ms)
Later throw the puck to the other side by an air table when running the P key and so
run the of the spark generator as soon as the pucks remain constant
Hold both switches open until two pucks movement bave comp lated
CALCULATIONS
Remove the data sheet and then please carefully review the resulting points. Spots
should be like in Figure 4.1. Points for each pucks 0, 1, 2, …. and so on the
numbering.
Two or three of the range of each, measure the length divided by the time on the road.
Later each puck collide rate found before and after the collision. Pucks come in the
way of naming the A and B before the collision and the A´ and B ´ after the collision.
Find the vector sum
and
ve
. Example; Find the vector addition of
. For example to lengthen A and B ways for finding
.
After draw vector of this velocities size which is relation with the length where start
the crosssection
and
direction. For instance velocity of 10 m/s can be drawn for
1cm length vector. Then, find the sum of this velocities using parallelogram adding.
Make the same method for
.
Define the points that made at same time after and before collision. Specify the
location of the center of mass combining those points.
Find the speed using the obtained recording for CM, before and after the collision.
Find the kinetic energy of two-pucks before and after the collision and compare them.
REVIEWS
QUESTIONS
1. How to change the momentum and kinetic energy, if the velocity of a particle doubles?
2. Are their momentums also equal, if the kinetic energy of two objects are equal?
Explain why.
3. As a result of the full elastic collision between two particles, does the kinetic energy
of each particle change?
4. Is it possible for a body, the center of mass is being outside of its actual volume? If
your answer is "Yes", give an example?