T.C. MANİSA CELAL BAYAR UNIVERSITY PHYSICS I LABORATORY MANUALS

Transcription

1 T.C. MANİSA CELAL BAYAR UNIVERSITY PHYSICS I LABORATORY MANUALS 2016

2 EXPERIMENT 1: ONE-DIMENSIONAL MOTION OBJECTIVE: Study of one-dimensional rectilinear and uniformly accelerated motion UNIFORMLY ACCELERATED MOTION THEORY It is called the motion with constant acceleration if the change in velocity during equal time intervals is always same for an object which moves along a straight line. Average acceleration is equal to instantaneous acceleration of the object when the acceleration is constant. In this type of motion, the velocity increases or decreases at the same amount from beginning to end of motion. If we take t i =0 and t f =t, then the following equation can be written or a x = v xs v xi t v xs = v xi + a x t ( constant a x ) (1.1) The displacement of the motion can be written as the function of time as follows Δx = x s x i x s x i = v ix t a xt 2 (constant a x ) (1.2) Finally, we replace the value of t in Equation 1.1 into Equation 1.2, time independent velocity equation which depends on displacement is obtained as follows v 2 v 0 2 = 2a(x x 0 ) (1.3) The position-time, velocity-time and acceleration-time graph of the object which moves with constant acceleration have been given in Figure 1.1. Figure 1.1: Variation graph for position, velocity and acceleration with respect to time for the motion with constant acceleration

3 Another example to the motion with constant acceleration is the motion of an object on an inclined plane. Figure 1.2: The external forces exerted on a block lying on a frictionless incline plane F x = mgsinθx (1.4) N = mgcosθy (1.5) N = N (1.6) The forces acting on an object which moves on an inclined plane without friction are as shown in Figure 1.2. When an object which is as rest on an inclined plane is falling down, it speeds up in the direction of force and with the proportional to the magnitude of the acting force according to Newton s Second Law that is it makes a motion with acceleration. A general relation between the force acting on an object and acceleration is given by F = ma (1.7) The acceleration of an object which moves on an inclined plane along x-direction is calculated by a x = gsinθ x (1.8) Given the integral of acceleration with respect to time, velocity and position v x = gt sinθ x (1.9) x = 1 2 gt2 sinθ x (1.10) On the above equations, it is assumed that the initial velocity of an object is zero.

4 THE EXPERIMENTAL PROCEDURE: Do not touch the metal parts of the table during the measurements with spark. Never switch on the spark if one or two pucks are out of table. Do not keep one of the pucks at your hand when spark is on. Otherwise you may get an harmless but disturbing electric shock. Incline the air table as shown in Figure 1.3 with the help of a wooden block. Figure 1.3: Schematic drawing of an inclined air table sinα = h 2 h 1 L Measure h 1 and h 2 which are the heights of the inclined plane s both ends from the bottom and L is the length of the inclined plane. Write the results to the table below. Period (ms) h 1 (cm) h 2 (cm) L (cm) α( 0 ) Measured values Place the carbon and measurement paper on to the surface of the air table. Place one of the pucks on top edge of the air table in order to use it for measuring and place the other puck to the top corner so that it doesn t disturb the other s motion and then fix it at that place by a plastic fixer.

5 o Before starting the compressor, make fixed the air puck on top edge of the air table. Run the air pump. Switch on the spark timer and choose a frequency value either 40 or 60 or 80 Hz. Press the red spark button in synchronization with the motion. Keep the button pressed until the end of the motion Switch off the air pump and spark timer to see your measurement results. Pull out the paper on the surface of the air table and observe the burn points. At least one of your measurements should be confirmed by the laboratory supervisor. On the carbon paper there should be lots of burn points on the same line. While choosing the starting point of your measurement you don t need to take the first burn point on the paper as the starting point of your measurement. Since the puck has a small acceleration just before it set in motion? Select a few more points from the reference point. Measure the distance between points and then record the frequency value. o The elapsed time between the two consequent points is calculated by the formula t = 1 Spark Frequency

6 CALCULATIONS Record the x and t values to the table below x i (cm) t i (s) Use gsinα instead of g in all projectile motion equations since the projectile motion is carried out with an inclined air table. Calculate the theoretical acceleration value belongs to the motion with the formula; a T = g sin θ Plot x t 2 graph according to the table above. If x i = 0 and v ix = 0, the equation (1.2) becomes x s = 1 2 a xt 2 where, a x = 2x t 2 this value gives the slope of x t 2 graph and it is experimental acceleration. Calculate the error for the result you obtained by using; % Error = (Theoretical Experimental) 100 Theoretical

7 1.2. RECTILINEAR MOTION THEORY The motion of an object which moves with a constant velocity along a line is called rectilinear motion. If the velocity of an object is constant, then its acceleration is zero. When the acceleration on above equations is taken to be as a=0, the formula for uniform, linear motion is obtained x = x 0 + v 0 t (1.12) If we want to know the distance traveled by an object, we can write this formula in terms of distance s as follows s = x x 0 = vt (1.13) Figure 1.4: Variation of the position, velocity and acceleration of a disc in a rectilinear motion THE EXPERIMENTAL PROCEDURE Level the air table off. Place the carbon and measurement paper on to the surface of the air table. Place one of the pucks on top edge of the air table in order to use it for measuring and place the other puck to the top corner so that it doesn t disturb the other s motion and then fix it at that place by a plastic fixer. Run the air pump. Switch on the spark timer and choose a frequency value either 40 or 60 or 80 ms. Press the red spark button in synchronization with the projectile motion that you have started with the launcher. Keep the button pressed until the end of the motion. Switch off the air pump and spark timer to see your measurement results. Pull out the paper on the bottom of the air puck and observe the burn points. At least one of your measurements should be confirmed by the laboratory supervisior.

8 On the carbon paper there should be lots of burn points on the same line. While choosing the starting point of your measurement you don t need to take the first burn point on the paper as the starting point of your measurement. Since the puck has a small acceleration just before set in motion. Select a few more points from the reference point. Measure the distance between points and then record the frequency value. The elapsed time between the two consequent points is calculated by the formula t = 1 Spark Frequency CALCULATIONS Record the x and t values to the table below. x i (cm) t i (s) Plot x t graph according to the table above. If x i = 0, the equation (1.12) becomes where, x = v 0 t v 0 = x t this value gives the slope of x t graph and it is experimental velocity.

9 Calculate velocity value belong to the motion using following equation x = v 0 t Calculate the error for the result you obtained by using; % Error = (Theoretical Experimental) 100 Theoretical QUESTIONS 1. Is it possible to have a case in which the acceleration of an object is zero if its velocity is not zero? Explain. 2. Is the acceleration of an object different from zero if its velocity is zero? Explain. 3. An automobile moving with a speed of 75km/h is getting slow down by braking and its speed getting fall down to 36 km/h in 5 seconds. a) What is the constant acceleration of an automobile? b) How much distance does an automobile travel during this time? c) What is time passed to stop an automobile exactly from the beginning of the motion?

10 EXPERIMENT 2: PROJECTILE MOTION OBJECTIVE: Study of a projectile motion on an inclined plane. THEORY PROJECTILE MOTION Projectile motion of an object is simple to analyze if we make two assumptions: (1) the free-fall acceleration is constant over the range of motion and is directed downward, and (2) the effect of air resistance is negligible. With these assumptions, we find that the path of a projectile, which we call its trajectory, is always a parabola as shown in Active Figure 2.1. We use these assumptions throughout this experiment. Figure 2.1: The parabolic path of a projectile that leaves the origin with a velocity v i. The velocity vector v changes with time in both magnitude and direction. This change is the result of acceleration a = g in the negative y direction. The expression for the position vector of the projectile as a function of time, with its acceleration being that due to gravity, a = g. r s = r i + v i t g t2 (2.1) where the initial x and y components of the velocity of the projectile are v xi = v i cosθ i v yi = v i sinθ i (2.2)

11 For a projectile launched from the origin, so that r i = 0. The final position of a particle can be considered to be the superposition of its initial position r i ; the term v i t, which is its displacement if no acceleration were present; and the term 1 g 2 t2 that arises from its acceleration due to gravity. In other words, if there were no gravitational acceleration, the particle would continue to move along a straight path in the direction of v i. Therefore, the vertical distance 1 g 2 t2 through which the particle falls off the straight-line path is the same distance that an object dropped from rest would fall during the same time interval. Two-dimensional motion with constant acceleration can be analyzed as a combination of two independent motions in the x and y directions, with accelerations a x and a y. Projectile motion can also be handled in this way, with zero acceleration in the x direction and a constant acceleration in the y direction, a y = g. Therefore, when analyzing projectile motion, model it to be the superposition of two motions: (1) motion of a particle under constant velocity in the horizontal direction and (2) motion of a particle under constant acceleration (free fall) in the vertical direction. The horizontal and vertical components of a projectile s motion are completely independent of each other and can be handled separately, with time t as the common variable for both components. HORIZONTAL RANGE AND MAXIMUM HEIGHT OF A PROJECTILE Let us assume a projectile is launched from the origin at t i = 0 with a positive v yi component as shown in Figure 2.2 and returns to the same horizontal level. Figure 2.2: A projectile launched over a flat surface from the origin at t i = 0 with an initial velocity v i. The maximum height of the projectile is h, and the horizontal range is R. At, the peak of the trajectory, the particle has coordinates (R/2, h).

12 Two points in this motion are especially interesting to analyze: the peak point A, which has Cartesian coordinates (R/2, h), and the point B, which has coordinates (R, 0). The distance R is called the horizontal range of the projectile, and the distance h is its maximum height. Let us find h and R mathematically in terms of v i, θ i, and g. We can determine h by noting that at the peak v ya = 0. Therefore: v yf = v yi + a y t 0 = v i sinθ i gt A t A = v isinθ i g (2.3) Substituting this expression for t A into the y component of Equation 2.1 and replacing y = y A with h, we obtain an expression for h in terms of the magnitude and direction of the initial velocity vector: h = (v i sinθ i ) v isinθ i g g (v isinθ i ) g h = v i 2 sin 2 θ i 2g (2.4) The range R is the horizontal position of the projectile at a time that is twice the time at which it reaches its peak, that is, at time t B = 2t A. Using the x component of Equation 2.1, noting that v xi = v xb = v i cosθ i, and setting x B = R at t = 2t A, we find that R = v xi t B = (v i cosθ i )2t A R = (v i cosθ i ) 2v isinθ i g = 2v i 2 sinθ i cosθ i g Using the identity sin2θ = 2sinθcosθ, we can write R in the more compact form R = v i 2 sin2θ i g (2.5)

13 INCLINED PLANE The external forces exerted on a block lying on a frictionless incline plane is shown in Figure 2.3. Figure 2.3: The external forces exerted on a block lying on a frictionless incline plane. The net force acting on the block in y direction, according to the Newton s law is: F y = ma y = mgsinα y Hence the net acceleration of the block in y direction is: a y = gsinα (2.6) When the projectile motion is studies on an inclined plane, the gravitational acceleration g in the Equations 2.1 to 2.5 must be replaced by gsinα. That is: t A = v isinθ i gsinα (2.7) h = v i 2 sin 2 θ i 2gsinα (2.8) R = v i 2 sin2θ i gsinα (2.9) THE EXPERIMENTAL PROCEDURE Do not touch the metal parts of the table during the measurements with spark. Never switch on the spark if one or two pucks are out of table. Do not keep one of the pucks at your hand when spark is on. Otherwise you may get an harmless but disturbing electric shock.

14 Incline the air table as shown in Figure 2.4 with the help of a wooden block. Figure 2.4: Schematic drawing of an inclined air table Measure the lengths h 1, h 2 and L of the inclined table. Write the results to the table below. Place the carbon and measurement paper to the air table. Place the air puck launcher to the BOTTOM RIGHT corner of the air table with an angle you desire. Place the rubber band to the second position on the launcher. Place one of the air puck to the launcher and fix the other puck to the top corner with the help of a plastic fixer in such a way that it doesn t block the motion of the other. (Figure 2.5). Figure 2.5: Projectile method. Run the air pump and try projectile motion few times until you get enough experience. Switch on the spark timer and choose a frequency value (40,60 or 80 ms). Press the red spark button in synchronization with the projectile motion that you have started with the launcher. Keep the button pressed until the end of the motion. Switch off the air pump and spark timer to see your measurement results. The laboratory supervisor should confirm your chosen measurement when you are sure about your measurement. Draw the x and y coordinate projections as shown in Figure 2.6. Take the second measurement point after starting point since the starting point can be problematic and place

15 the corner of your coordinate axes to this point. Measure h max and R values and record it to the table below. Figure 2.6: Measurement of the projectile motion data. Determine t and t A flight times by counting the points on projectile motion. The time elapsed between the two consequent points is ; T period (ms). Measure the lengths r sθ, y sθ, r eθ, y eθ in order to determine the starting (θ s ) and the end (θ e ) angles as shown in figure 2.7. Use more points as much as possible in such a way that the linearity of the line is not violated. Extend the line in order to increase the precision of the measurement r sθ, r eθ. Figure 2.7: Angle measurement of the initial and final velocities.

16 CALCULATIONS Determine the angles θ s and θ e with the help of equations given below. sinθ = y θ (2.10) r θ Determine the inclination angle of the air table by using the equation below. sinα = h 2 h 1 L (2.11) Use gsinα instead of g in all projectile motion equations since the projectile motion is carried out with an inclined air table. Figure 2.8: A sample projectile motion data view. Take the average of the equal distances on x axes between the points. Calculate the x component of the initial velocity v xi by using the equation below: v xi = x t (2.11) where t is the multiplication of the number of points N on paper and T period (ms).this gives us the flight time experimentally. Calculate the initial velocity experimentally by using the equation v si = v xi sinθ s = v xir sθ y sθ. Calculate also the end velocity v ei using the same equation. Can θ s and θ e angles be different from each other? Make comments on the corresponding reasons and the results. Calculate the experimental value of the time to reach maximum height by using number of points N: t A = NxT(period) (2.12) Calculate the theoretical value of the time to reach maximum height by using the equation (2.7). Calculate the time t = 2t A to reach the range R experimentally.

17 Measure the experimental value of the range R exp by using a ruler. Calculate the same value theoretically by using the Equation 2.9. Measure the experimental value of the maximum height by using a ruler. Calculate the same value theoretically by using the Equation 2.8. Write your results to the corresponding table below and make the error calculation for each result you obtained by using; % Error = (Theoretical Experimental) Theoretical 100 (2.14) MEASUREMENTS TABLE Measured Values T Period (ms) h 1 (cm) h 2 (cm) L (cm) r sθ (cm) y sθ (cm) r eθ (cm) y eθ (cm) Time to reach maximum height (t A ) (s) Flight time t(s) Range (R) (cm) Maximum height (h max ) (cm) RESULTS TABLE Experimental Theoretical % Error θ s ( 0 ) θ e ( 0 ) α ( 0 ) v si (cm/s) v ei (cm/s) Time to reach maximum height (t A ) (s) Flight time t(s) Range (R) (cm) Maximum height (h max ) (cm)

18 QUESTIONS 1. Which parameter remains constant for an object making a projectile motion? a. Magnitude of its velocity, b. Acceleration, c. Horizontal projection of its velocity, d. Vertical projection of its velocity. 2. A ball hit by a soccer player has an initial velocity of 30 m/s and it makes 37 o angle with horizontal axes. a. Write the equation of motion of the ball. b. What is the position and velocity components of the ball at t=1s? c. How long it takes to reach the maximum height? d. What is the maximum height that ball reaches? e. How far the ball goes before hitting the ground? 3. A ball is thrown away from the roof of a 33 m high building, with an angle of 53 o with horizontal and with an initial speed of 5 m/s. At ground, a children starts running horizontally at the same time with a constant acceleration of a. a. Write the equation of motion of the ball by choosing a coordinate system. b. What is the flight time of the ball? c. How far away from the building the ball goes? d. What should be the acceleration a of the children to be able to catch the ball before hitting the ground?

19 EXPERIMENT 3: DYNAMIC PURPOSE: In this lab, we will explore the relationship between position, velocity and acceleration. In this experiment, friction will be neglected. Constant (uniform) acceleration will be investigated NEWTON S LAW INTRODUCTION Speed is just how fast something is moving. It is measured in how far something moves in some time. Speed is given by the equation: Speed=x/t (3.1) The symbol indicates a change is some quantity such as position ( x ) in this case. The unit of speed is always distance divided by time. We might measure speed in miles/hour, meters/second etc. In this lab we will use meters/second (m/s). Speed is how far something moves in some time. Obviously this depends on the time period we use. Things rarely move at constant speed. We measure the speed of something over a very short time, we can talk about the instantaneous speed. Instantaneous speed is like taking a snap shot of the speed. The average speed is what we have in equation 2.1 for some time period that may not be very small. Just as the position changing gives us velocity, when velocity changes we have an acceleration. Acceleration is: Acceleration= a = change in velocity / time interval = v/t (3.2) Acceleration has units of distance divided by time. For this lab, we will use meter/second 2 (m/s 2 ) When we specify a speed and the direction, we are giving a velocity. Velocity is a vector. Like all vectors it has magnitude (how much) and direction (which way). Acceleration being the change in velocity is also a vector. In this lab, we will restrict the motion to one dimension so the vector nature of velocity and acceleration are no so important. However, you should remember that velocity and acceleration are vectors. If the acceleration is constant, there are relatively simple relationships between position, velocity and acceleration in one dimension. If an object at time zero (t = 0) is at the origin

20 (x = 0), the the velocity and position are given by: v(t) = v 0 + a.t (3.3) x(t) = v 0. t a. t2 (3.4.) where a is the constant acceleration, t is the time and v 0 is the initial velocity of the object at time zero. For this experiment, we will have motion in one dimension with zero initial velocity. Next we should define force. Force is simply a push or a pull. It can be from gravity i.e. gravity pulls on a mass, a magnetic pull between a north and south magnetic pole or simply you pushing a book across a table. Newton s 2 nd law relates mass, acceleration and force. In words: The acceleration of n object is directly proportional to the net force acting on the object, is in the direction of the net force and is inversely proportional to the mass of the object. F= m.a (3.5) The units of force in the metric system are the newton. There are several things we should notice about this statement. The equation for Newton s 2 nd law tells us force is a vector like velocity or acceleration. When you push something, you push in a direction e.g, north or down, as well as some amount e.g. 10 N. We can use Newton s 2 nd law to figure out what a newton (N) is in terms of mass, time and distance. Since force is mass times acceleration, the units of a newton can be broken down to kg m/s 2. Finally, notice the statement of Newton s 2 nd law says net force. This means all the forces added together acting on an object. Since they are vectors, we have to add all the forces vectors. We will investigate Newton s second law and constant acceleration with the air table in the configuration shown in figure 1.2. The hanging mass, m, will pull the puck across the table as it falls. The mass of the puck is mp. The string connecting the hanging mass and the puck has a tension, T. We assume there is no friction between the table and the puck. The force on the puck is simply the tension (pull of the hanging mass). Assuming no

21 friction, Newton s 2 nd law gives: Force = T = m p.a (3.6.) Figure 3.1. Equipments of experiment. a m p T T a mg Figure 3.2. The air table arrangement. The forces on the hanging mass are T (upward) and the force of gravity, mg (downward). g is the acceleration of gravity (9.81 m/s 2 ). The acceleration of the hanging mass is downward.

22 Newton s second law gives: Force = T m.g = m (-a) (3.7) If we substitute equation 2.6 into equation 2.7 and solve for the acceleration we have: a = m.g m p +m (3.8) This equation shows that the acceleration is constant and proportional to the acceleration of gravity through the ratio PROCEDURE Special Caution: To avoid electrical shock, do not touch the pucks or air table while the spark timer is on. Use an insulator such as a rolled up sheet of paper to push or hold the puck. The air table apparatus is shown in figure 2. The white box at the back of the table is the spark unit. Two pucks are shown connected to the rubber hoses which supply the air to float the puck. Newton s Second Law; In this section of the procedure, the puck will be pulled across the table by a hanging mass connected to the puck by a string over a pulley. Attach the pulley to the side of the table if it is not already attached. Connect the string to the collar and the other end to the hanging mass. Place the string over the pulley wheel. Using the unmarked side the white paper, configure the free puck on several folds of the corner of the paper so it will not move. Hold the puck connected to the hanging mass at the far end of the table with the insulated tube. Press the sparker pad and release the puck. The puck should have zero initial speed so the puck should not be released until the sparker has started. Stop the spark timer when the mass reaches the end of the table. Turn off (for safety) the sparker power supply, remove both pucks, lift up the paper sheet and turn it over to examine the sparker dots marking the puck trajectory.

23 Qualitatively describe how the dots are spaced on the paper. Measure the distance between the first sparker dots another dot approximately 0.20 meters from the first dot. Count the number of time intervals (0.05 seconds between dots) to determine the total time between the first dot and the dot where you measure the distance. Using equation 2.4, calculate the acceleration assuming v0 = 0. Using equation 2.8, calculate the acceleration of gravity. Calculate the percentage error for your value with the standard value of 9.81 m/s 2 QUESTIONS In the constant acceleration part of the experiment (Newton s 2 nd law), if the initial velocity is not zero would the final result be serious effected? In the constant acceleration part of the experiment (Newton s 2 nd law) there may be some friction between the puck and air table. Assuming the friction is constant, would you still expect the acceleration to be constant? Would it change the value of g you calculated?

24 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 P = mv (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; F ext = ma = m dv dt (4.2) When m is constant, this equation is written as follows: F ext = d(mv ) dt = dp dt (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 F ext = 0 then dp dt = 0 (4.4) P = constant (4.5) Here the momentum not change with time, the object always have the same momentum. N- particle system consist of m 1, m 2,.m N masses can be generalized based on the above results. When we are dealing with a system of objects (m 1, m 2,.m N ), the total momentum P tot of the system is the vector sum of the individual momentums: P tot = P 1 + P P N (4.6)

25 Where are, P 1 = m 1 v 1, P 2 = m 2 v 2,.. etc. (4.7) The sum in the equation (4.6) is a vector sum process. In this situation, if a generalized equation (4.3); F ext = dp tot dt = d (P dt 1 + P P N ) (4.8) where F ext, 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; dp tot dt = d (P dt tot = P 1 + P 2 + P N ) = 0 (4.9) P tot = P 1 + P P N = constant (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.

26 Velocities of two pucks are respectively v A, v B and v A, v B before and after the collision. Momentum is conversed because the system is isolated and at any time; P tot = constant (4.11) P A+ P B = P A + P B (4.12) Where the momentums are P A = m A v A, P B = m B v B. Because the masses of the pucks are same, the above equation is converted as follows. v A + v B = v A + v B (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 v. 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 R of a system with N particles is defined as follows, (seen Figure 4.2)

27 R = m 1r 1 +m 2 r m N r N m 1 +m 2 + +m N (4.14) Where r 1, r 2,, r N are position vectors and m 1, m 2,. m N 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. V CM = dr dt (4.15) when we take the derivative of both sides of equation (4.14) for particles with constant mass. R = m 1r 1 +m 2 r m N r N m 1 +m 2 + +m N (4.16) V CM = m 1v 1 +m 2 v 2 + +m N v N m 1 +m 2 + +m N (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; R = mr A+mr B m+m R = r A+r B 2 (4.18) (4.19)

28 are obtained. Since the masses of the pucks are equal here, equation 4.20 is obtained by removing the masses. Thus velocity of CM, V CM = v A+v B 2 (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. V CM = V CM (4.21) V CM= v A + v B 2 = V CM = v A +v B 2 (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. K = 1 2 mv2 (4.23) Therefore total kinetic energy of the two-pucks system prior to the elastic collision; K = 1 2 mv A mv B 2 (4.24) and kinetic energy after the collision; K = 1 mv 2 A mv 2 2 B (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; K = 1 2 (2m)v2 = mv 2 (4.26)

29 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 = 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 have completed 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 location of the center of mass combining those points. Find the speed using the obtained recording for CM, before and after the collision.

30 Find the kinetic energy of two-pucks before and after the collision and compare them. 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?

31 EXPERIMENT 5: INELASTIC COLLISIONS OBJECTIVE: To investigate that in inelastic collisions the energy and momentum equations THEORY INELASTIC COLLISIONS Inelastic collisions; where the two colliding bodies adhere to each other upon contact and move off as a unit after the collision. Linear momentum is conserved, while not conserved mechanical energy in inelastic collision. After the collision, system moves from its axis of rotation, linear velocity and the speeds of the center of mass of both two bodies (m 1, m 2 ) are equal for each one. The velocities of first (m 1 ), and second (m 2 ) are represent v 1, v 2 ve u 1, u 2 for before and after collision. If the V is represents and the speeds of the center of mass; u 1 = u 2 = V (5.1) can be written as (5.1). Since momentum is conserved in inelastic collisions; m 1 v 1 + m 2 v 2 = m 1 u 1 + m 2 u 2 = (m 1 + m 2 )V (5.2) the equations (5.2) can be written. The speed of the center of mass(v) ; can derivative as following. V = m 1v 1 + m 2 v 2 (5.3) m 1 + m 2 If the masses m 2 and m 1 are equal to each other the speed of the center of mass(v) can be obtained (5.4); V = v 1 + v 2 2 (5.4) Inelastic collision, the total momentum of the system although there is a change in the system s kinetic energy is always a loss. So, the equations of the kinetic energies for before and after collision are written, respectively. ( 1 2 m 1v m 2v 2 2 ) > ( 1 2 (m 1 + m 2 )V 2 ) (5.5) If the m 1 = m 2 the equation (5.5) converted as following;

32 (v v 2 2 ) > 2V 2 (5.6) The energy difference is transformed into heat or other forms of energy in inelastic collisions. If K 1 and K 2 are defined the total kinetic energy before the collisions, reduced rate of the system can be expressed the following equation (5.7); THE EXPERIMENTAL PROCEDURE: e = K 1 K 2 K 1 (5.7) Do not touch the metal parts of the table during the measurements with spark. Never switch on the spark if one or two pucks are out of table. Do not keep one of the pucks at your hand when spark is on. Otherwise you may get an harmless but disturbing electric shock. Wrap the special adhesive tape around the equal mass of two disks as outward facing adhesive surfaces After turning the horizontal air table, put the pucks which is close to the corners of the table for you. Switch on the spark timer and choose a T period value either 40 or 60 or 80 ms. Press the air pedal, collision somewhere in the middle of the table makes you. You will see that pucks act together after the collision. After you set up the positions and velocity of the pucks, pressing air pedal, you repeat this motion. You must press the pedal after the motion of pucks. If you press the pedal before the motion, to find location or position (relative position in time t) of pucks can be difficult. Switch off the air pump and spark timer to see your measurement results. Place the carbon and measurement paper on to the surface of the air table. Pull out the paper on the surface of the air table and observe the burn points. At least one of your measurements should be confirmed by the laboratory supervisor.

33 According to results, draw the vector and measure the magnitude of this vector using by ruler, v 1, v 2 and u 1, u 2 before and after collision. If the velocities of after the collision, u 1 ve u 2, are not equal, this system revolves around itself. CALCULATIONS Show the vector R = v 1 + v 2 and R = u 1 + u 2, drawing by ruler. Are equal to the resultant vector (R ) before the collision and resultant vector (R ) after the collision. How can you show the conservation of linear momentum? Draw the speed of the center of mass(v). V are equal velocities u 1 ve u 2 after the collision. How is there a equation between V and v 1, v 2. Measure the magnitude of velocity vector V. You use your results of velocity the equation (5.6) to show deprotection of kinetic energy. Find the reduced rate of kinetic energy using by equation (5.7). QUESTIONS Figure 5.1: Inelastic collisions 1. Give examples for the Inelastic collisions, encountered in our everyday life. 2. B If the velocity of a particle doubles, how the momentum changes? What happens to the kinetic energy?

34 EXPERIMENT 6: SIMPLE HARMONIC MOTION OBJECTIVE: To investigate Hooke's Law and simple harmonic motion. The potential, kinetic energy and total energy conservation law to be tested in the system, simple harmonic motion. THEORY SIMPLE HARMONIC MOTION A fixed point in doing both sides of the oscillating object is called vibrational motion. In this experiment you will explore simple harmonic motion, special status of vibrational motion. Think a mass (m) attached, a spring clips attached to one end. If the system is left at rest at the equilibrium position then there is no net force acting on the mass. However, if the mass is displaced from the equilibrium position, a restoring elastic force which obeys Hooke's law is exerted by the spring. Mathematically, the restoring force F is given by F = kx (6.1) where, F is the restoring elastic force exerted by the spring (in SI units: N), k is the spring constant (N m 1 ), and x is the displacement from the equilibrium position (in m). Displacement is proportional to the force and direction is always toward the equilibrium position that is the opposite direction with displacement. Once the mass (m) is displaced from its equilibrium position (x), it experiences a net restoring force. As a result, it accelerates and starts going back to the equilibrium position. So net restoring force and acceleration of mass can be expressing the following equations (6.2), (6.3); F = kx = ma (6.2) a = k m x (6.3) The acceleration of mass is proportional with displacement from the equilibrium position and opposite direction of displacement. So, For one-dimensional simple harmonic motion, the equation of motion, which is a second-order linear ordinary differential equation with constant coefficients, could be obtained by means of Newton's second law and Hooke's law.

35 d 2 x dt 2 = k m x (6.4) where m is the inertial mass of the oscillating body, x is its displacement from the equilibrium (or mean) position, and k is the spring constant and, k m = w2, ve w, the angular frequency. Solving the differential equation above, a solution which is a sinusoidal function is obtained. x(t) = Acos(wt + φ) (6.5) In the solution, c1 and c2 are two constants determined by the initial conditions, and the origin is set to be the equilibrium position. Each of these constants carries a physical meaning of the motion: A is the amplitude (maximum displacement from the equilibrium position), ω = 2πf is the angular frequency, and φ is the phase. Using the techniques of differential calculus, the velocity and acceleration as a function of time can be found (6.7): dx dt = A d dt cos(wt + φ) = Awsin(wt + φ) (6.6) d 2 x dt 2 = Aw d dt sin(wt + φ) = w2 Acos(wt + φ) (6.7) If the equation (6.7) write on the number (6.4) equation, can see the equal (6.8). w 2 Acos(wt + φ) = w 2 Acos(wt + φ) (6.8) Period is the time taken for one full oscillation. Frequency is the number of occurrences of a repeating event per unit time. The period, usually denoted by T, T = 2π w is the duration of one cycle, and is the reciprocal of the frequency f, we can expressed the following (6.9) and (6.10); T = 2π w = 2π m k (6.9)

36 f = 1 T = 1 2π k m (6.10) As can be seen, period and frequency depends only on the mass and the spring force constant. x, is its displacement from the equilibrium (or mean) position, the moving body attached the spring. A is the amplitude, maximum displacement from the equilibrium position. In simple harmonic motion displacement (x), velocity (v) and acceleration (a) are varies sinusoidally with time, but it is not in the same phase. The acceleration (a) of particle is proportional with their displacement (x) and opposite direction. The frequency of movement is independent of the amplitude and period. In addiction; think the moving object, had simple harmonic motion, attached a disc between two springs as a system spring-disc-spring. The energy of this system is conserved, namely steady, due to the non- effect forces out of this system. E, total energy of system is equal to sum of kinetic K, and potential energy U of spring. K + U = E = Sabit (6.11) where m, the mass of disc, v the velocity of disc, k the constant of the rate or spring constant and x displacement from the equilibrium, So the kinetic energy of disc is following (6.12); K = 1 2 mv2 (6.12) and the potential energy of disc is following; U = 1 2 kx2 (6.13) The total mechanical energy E=K+U, is constant of movement in the harmonic motion, not effected conservative forces. In the other words the movement is not changed with time. In this case x displacement is given following; x = Acos(wt + φ) (6.14) According to equation of number (6.13) equation, the values of potential energy are described at any given moment;

37 U = 1 2 kx2 U = 1 2 ka2 cos 2 (wt + φ) (6.15) The maximum value of potential energy is 1 2 ka2. As shown figure-6.1a and figure-6.1b, the potential energy is change from zero to maximum value during the movement. The kinetic energy at any given moment;, 1 2 mv2. the velocity of disc is given the following equations; v = dx dt = d dt Acos(wt + φ) = wasin(wt + φ)vew2 = k m Figure 6.1 a) the kinetic and potential energy of oscillator moving simple harmonic motion, b) the variation of the kinetic and potential energy according to displacement. however the kinetic energy of disc are given (6.16); K = 1 2 ka2 sin 2 (wt + φ) (6.16) As seen, the maximum values of the kinetic energy are found as 1 2 ka2 or 1 2 m(wa)2. The kinetic energy is change from zero to maximum value during the movement. The total energy for an oscillator is the sum of its kinetic energy and potential energy. The total energy is obtained by using the equations (6.15) ve (6.16); E = K + U = 1 2 ka2 sin 2 (wt + φ) ka2 cos 2 (wt + φ)

38 E = 1 2 ka2 (6.17) As expected, the total mechanical energy is stable and its value is 1 2 ka2. The kinetic energy is zero and potential energy is 1 2 ka2 at the time of the maximum values of displacement. The potential energy is zero and kinetic energy is 1 2 ka2 in the equilibrium position. The total energy occur the sum of its kinetic energy and potential energy at the other positon. THE EXPERIMENTAL PROCEDURE Do not touch the metal parts of the table during the measurements with spark. Never switch on the spark if one or two pucks are out of table. Do not keep one of the pucks at your hand when spark is on. Otherwise you may get an harmless but disturbing electric shock. a) To determine the spring constant of a spring After turning the horizontal air table, attach one of the spring (to be found constant force) back of the air table and attached the other disc on the free hand of spring. Place the carbon and measurement paper to the air table. Press the air pedal intervally and reduce to disc equilibrium position and mark the position of disc by running the arc chronometer. After then give different slope the air table (see the Figure-6.2) and Press the air pedal intervally and reduce to disc equilibrium position and mark the position of disc by running the arc chronometer every time. Figure 6.2: Inclined plane, spring-disk system

39 CALCULATIONS Graphed displacement (x), versus sin(θ) using the result of obtained from the under the sheet. Describe obtained curves. Do you say from the shape curve (graph) the constant of spring agree with the Hook Law within the effect of forces? If the graph, displacement versus to sin(θ) is linear, write the mathematical expression. According to, how is relationship between the slope of the line and k constant spring? Calculate the k constant spring. b) The spring-disk-spring system and conservation of energy After turning the horizontal air table, set up the spring-disk-spring system like in Figure-6.3. The system must be far away the front of table nearly 10 cm and parallel. The experiment sheet must fit as keep at both ends passing the carbon sheet extracting tube. After then, put the disc side of equilibrium point, press the air pedals and put the sheet slowly and constant speed. Thus you will obtain sinusoidal curving as seen Figure-6.4. This curve is required for proper data depends on the smoothness of paper speed. Mark the result of projection data on the t and x coordinate. The path taken along the x-axis of the disk, identify whether equal along the equal time intervals. Show the velocity of direction of x V x on the graph. Determine the values of V x, changed the time and find the zero values of V x on the graph. Figure 6.3: spring-disk-spring system

40 CALCULATIONS Figure 6.4: Displacement with respect to time engaged in simple harmonic motion oscillator. Find the period T and angular velocity w = 2π T from result of data sheet. Measure the x values (up to you) of curve part of 1/4. Use the x values to find potential energy by using U = ( 1 2 ) mw2 x 2 equation. Likey find the kinetic energy of disc as following equation; K = 1 2 mw2 A mw2 x 2 (6.18) As seen Figure 6.2, find the total energy of system E=K+U for to selected x values. Plot the graph of total energy versus to x. what s the slope of obtained curve? Is curve, showed total energy, parallel? If it s parallel what can we say the conservation of total energy? QUESTIONS 1. Discuss the Hook Law for flexible movements. 2. Calculate the velocity of oscillator (disc) by using figure-4 where point of t = T 4, t = T 2, t = 3T 4 vet = T 3. There is a spring to have vibrated horizontal plane force constant: 20 N/m and mass: 0.5 kg so; a) If the amplitude of motion A= 3 cm, calculate the total energy of system and maximum values of velocity of mass. b) If the displacement x= 2 cm, calculate the kinetic and potential energy of system and compare the total energy K+U where section a.