Experiment No : M2 Name of Experiment: : TWO DIMENSIONAL COLLISIONS The Purpose of The Experiment

Transcription

1 Experiment No : M2 Name of Experiment: : TWO DIMENSIONAL COLLISIONS The Purpose of The Experiment : To examine momentum and energy conservation equations for elastic collision in two dimensions. To demonstrate experimentally that momentum is vectoral and energy is scaler quantity. Theoretical Information: Conservation of Mechanical Energy Energy is a scalar quantity. It can be divided into kinetic energy and potantial. In the following equations, potential energy and kinetic energy are shown with U and K, respectively. The condition before the collision is denoted by, and after collision is denoted by subscripts. According to work-energy theorem, the change in kinetic energy of an object is equal to work that has done by the net force acting on the object. K K W 3.1 Let s assume that only conservative forces are generating the work. The work done by conservative forces is independent of the path and equal to negative of the potential energy change. When we combine these together, K K W U U 3.2 equation will be obtained. When this last equation is rearranged, K U K U 3.3 equation will be obtained. Here, the left side of the equation represents the total energy after the collision whereas the right side of the equation represents the total energy before the collision. So, as E= K + U; E E 3.4 will be obtained. If only conservative forces are generating the work, the mechanic energy of the system is also conserved, thus it is independent of time. The equation 3.4 represents the mechanic energy conservation in 1,2 or 3 dimensions. This equation shows the relationship between the velocity and position of the object. During the movement of the object, both kinetic and potential energy will change but their sum will be constant. The law of conservation of the energy states that in an insulated system total energy of the system is conserved. It is possible to write this statement in form of, de Conservation of Momentum 0 E CONSTANT Momentum, P, of an object with mass m and velocity v multiplication of mass and velocity vector. is defined by the

2 P mv 3.5 Momentum is a vector quantity. Newton s Second Law can be written as, dp F 3.6 Let s think of a system with zero net force acting on it. Then, the following equation will be obtained. d P That means the change of momentum with time is zero, or momentum is independent of time. So, the momentum of an condition and condition is the same. In other words, the momentum of the system is conserved. P P 3.8 Let s assume that not the force, but only one component of it, for example F y, is zero. Then, let s write the Newton s Law in form of components: dp x F ; 0 x dp y ; dpz F z As it can be seen, the solution of the first and the third equation is not obvious, and dependent on the open form of the forces. However the solution of the second equation is simple. P y = constant. Thus momentum of the force component is conserved. Other than that, in a closed system of particles, i.e. no external force acting on the system and particles only interact with each other, total momentum of the system is conserved. (See Proble). Here the total momentum of the system and the vector sum of the momentums should be understood. N particle system of masses,,... m N can be generalized according to the statement above. Total momentum of such system composed of particles in a given time can be written as: Here,, etc. can be written. The summation in the equation 3.9 is a vector summation. In this case, when equation 3.6 generalized; 3.9 will be obtained.

3 Here ext, defines the net external force acting on the system of particles. In other words the external force is different from interaction force of the particles between them. These external forces can be friction or gravitation. Thus if there is no external force acting on the system of particles, the total momentum of the system will be conserved. So; The equation above is also a vector quantity. In a system of particles with no net force (or in an isolated system), total momentum of the system in any given time will be the same. Quantities that does not change under certain conditions such as energy and momentum are defined as constants of the motion. These constants simplify the solution of motion equations. Momentum and energy conservation expressions of elastic collision of two particles in a plane are defined as follows; Momentum horizontal ; 3.10 Momentum - vertical ; 3.11 Kinetic energy; 3.12 Center of mass (CM) is the other quantity that will be compared and investigated in this experiment. It can be predicted that center of mass of a uniform cube, cylinder, sphere and other symmetrical objects is its geometric center. Center of mass of the two objects of the same mass will be located at the midpoint of the line that connects their geometric centers. However, if one of the objects is heavier than the other, then the center of mas will shift towards the heavier one. The mass should be redefined for various mass distributions. The position vector, of system of N partical with position vectors and masses,,...,m N is defined as in equation Here, vectors are position vectors of each particle in the coordinate system whereas represents the position vector of the center of mass. If the position of particles change with time, position of the center of mass also changes an the vector changing ratio of center of mas can be considered as the velocity of (CM) If the derivative of both sides of the equation 3.13 will be taken for constant mass particles; 3.14

4 3.15 Equation (3.16) will be obtained. The dots on of the equation 3.15 are annotation of derivative with respect to time which is the velocities of the masses. When these derived equations are adapted to our two-disc system in the experiment; and the masses in the equation 3.17 simplified because the disc masses are equal ( = =m); equation 3.18 will be obtained. In that case; if the derivative of position vectors with respect to time will be taken, the velocity of the CM will be: Equation 3.19 has important outcomes. It means that while momentum is conserved, the velocity of CM is constant (constant velocity, no change in magnitude and direction.). Thus the center of mass of an isolated system with conserved momentum always moves with constant velocity in linear motion. Thus of our two-disc system should be for before and after the collision. Experimental Method: Figure 3.1: An air table will be used to minimize the friction of the experimental setup. There will be a carbon paper on the table. On top of that there will be a White paper to mark the trajectory of particles with the help of spark generator.

5 1. On the air table there will be an air pump an two discs attached to hoses that are connected to spark generator. Fixate one of the discs to the center of the air table as much as possible. Practically, it will be a bit hard. Hit the other disc with a cue from any angle to generate an elastic collision of discs with each other. Pay attention to push to the pedal of the spark generator at the same time motion started and continue till the discs hit the sides. This way the trajectory of particle motion will be obtained at the backside of the paper. If you push for too long the points on your paper will be mixed. Another point is to avoid head-to head collision of the disc. This way there will be distinct changes in the angle and the angle measurement mistakes will be reduced. 2. Remove the data sheet and carefully observe the generated points. In close proximity to the collision, mark 4-5 consecutive points generated before and after the collision as shown in figure 3.2. Figure In order to observe that the linear momentum is conserved the velocities thus the momentums of the discs before and after the collision should be known. To do this, first define the center of coordinate system according to the trajectories before the collision. Using this coordinate system, define the positional changes of the discs and their projections to the x and y axes for before and after conditions. Measure the distances with a ruler and record them to the tables. Define the points generated at the same time before and after the collision. Connect these points and define the position CM along these lines. While doing this, you will obtain the data of the position of CM during collision using this data find the velocity and momentum before and after the collision. Before the collision; Before the collision ΔS (m) Δx (m) Δy (m) ΔS KM (m)

6 After the collision; After the collision ΔS (m) Δx (m) Δy (m) ΔS KM (m) 1. You will need mass and velocity values to calculate momentum and kinetic energy. Measure the disc masses using digital balance. =... kg =... kg 2. Record the frequency values of spark generator, calculate period frequency =... ( ) period =... ( ) Calculations 1. In order to obtain the velocity values of the discs you will need to measure position changes and the times. You can find the times using the number of points in the measured trajectory and the frequency value of the spark generator. Calculate the velocity values from position change and time and record the data to the tables below. Before the collision; v 1 =... ( ) v 2 =... ( ) v x1 =... ( ) v x2 =... ( ) v y1 =... ( ) v y2 =... ( ) v CM =... ( )

7 Before the collision t (s) v (m/s) v x (m/s) v y (m/s) v KM (m/s) After the collision; v 1 =... ( ) v 2 =... ( ) v x1 =... ( ) v x2 =... ( ) v y1 =... ( ) v y2 =... ( ) v KM =... ( ) After the collision; t (s) v (m/s) v x (m/s) v y (m/s) v KM (m/s) 2. Using the measured mass and calculated velocity values calculate the momentum and energy. Record the calculated data to the tables below. Before the collision; P 1 =... P 2 =... P x1 = P x2 = P y1 =... P y2 =... P CM =...

8 E 1 =... E 2 =... Before the collision P (kg/s) P x P y P KM E (kg /s 2 ) After the collision; P 1 =... P 2 =... P x1 = P x2 = P y1 =... P y2 =... P CM =... E 1 =... E 2 =... After the collision; P (kg/s) P x P y P KM E (kg /s 2 ) 3. Conservation of horizontal and vertical momentum and energy values of the elastic collision of the two mases in the same plane were defined in equations Experimentally satisfy these momentum and energy conservaton formulas Momentum horizontal;

9 Momentum vertical; Kinetic Energy; Why there is no potential energy component in these energy conservation equations? Please Explain. If there is a mistake in your values, discuss the source of the mistake in the experiment. Proble. In a closed system of particles, total momentum of the system will be conserved. Please show it.