EXPERIMENT 4:MOMENTUM AND COLLISION
PURPOSE OF THEEXPERIMENT: Forelastic collisionsin anisolated system, examine theconservation oflinear momentumand kinetic energy.
ELASTIC COLLISION
GENERAL INFORMATION
Linear momentumof an object "P", is the product of mass and velocity
(4.1)
Herewe will talk aboutthemomentumbrieflyfromthe linear momentum.However, onlythe netexternal forcewhen applied,we know thatthe speed ofthe objectchanged and thismeans that themomentumchange.This factcan be seen fromNewton's second law. According toNewton's second lawfora constant massof abody;
(4.2)
When m is constant, this equationis writtenas follows:
(4.3)
Fromthe above equation, if an external forceactson an object,the object'smomentumdoes not protected.That is, momentum does not change with time. If then
(4.4)
(4.5)
Herethe momentumnot change with time, the object alwayshavethe samemomentum.
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 comprisedof the particlesrefers tothe netexternal force. Thisexternal forcesmay befrictionand gravity. Hence in the system formed byparticlesit does nothaveanytotalexternal force andthe totalmomentumof the systemwill 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 totalmomentumofthe pucks seems to bepreserved. 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 conceptthat will be encounteredin this experimentisthe center of mass(CM). The CM of symmetrical objects such as cubes (figure4.2a) and spheres (figure4.2b ) is the same with the theirs geometric center. The CM of the shape in the figure4. 2c is predictably the midpoint of the rod.
Figure 4.2: CM of somesymmetrichomogeneousobjects.
CMformass distributionin different shapesshould be redefined. CM of position vector of a system with N particles is defined as follows, (seen Figure4.2)
(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 wetake the derivativeofboth 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 theabove equationis applied tothe twopucks 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)
Thereforetotal kinetic energyofthe two-puckssystem prior to theelastic collision;
(4.24)
and kinetic energyafter the collision;
(4.25)
However,thetwo puckssticks each other in aninelastic 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 energyis preservedinalmostelastic 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 ve . Example; Find the vector addition of and . 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 locationofthe center of mass combiningthose points.
Find the speedusing the obtained recording for CM, before and afterthe collision.
Find the kinetic energy of two-pucks before and after the collision and compare them.
REVIEWS
QUESTIONS
- How to change the momentum and kinetic energy,if the velocity of a particle doubles?
- Are their momentums also equal, if the kinetic energy of two objects are equal? Explain why.
- As a result of the full elastic collision between two particles, doesthe kinetic energy of each particle change?
- Is it possible for a body,the center of mass is beingoutside of its actual volume? If your answer is "Yes", give anexample?