Experiment 2-2. Colllision

Transcription

1 Physics Laboratory Last update: Experiment 2-2. Colllision Purpose of experiment Physics shows us one way of understanding the natural phenomena. But in reality, they are so complicated and subtle that it is impossible to know all the information in the nature exactly. (So far, at least.) Often it is more meaningful for us to understand a phenomenon by intuition than gaining all the information precisely. That is, although many physical laws are explained under some assumptions, we can understand and make a prediction even when the real system doesn t accord with the assumptions. For example, there are Newton s laws of motion and momentum conservation. We look into the method of rough understanding of collision and confirm the law of momentum conservation by one dimensional collision experiment.

2 - From Fundamentals of physics, Halliday & Resnick The picture shows an orange golf ball colliding with and recoiling from ground. The decreased height when it recoils shows the collision is inelastic. Is momentum not conserved in the picture? Outline Confirm Newton s first law of motion. - Is the device truly frictionless? Is the collision on the glider elastic? - Is momentum conserved both in an elastic collision and in an inelastic one? What relationship is there between the masses of colliding bodies and the change of velocities after the collision? Is it possible to confirm the third law of motion from this? Experiment method There are apparatuses in the lab as follows. (in the parentheses are the numbers prepared)

3 CCD CAMERA (1) A fan : AC 220V, motor fan (1) A frictionless motion table : squared aluminum pipe, length 2100mm (scale length 2000mm, height 280mm (1) A glider : weight 200g (approx., including the staff), a recoiler can be attached.(2) An ancillary box (1) A weight : metallic 30g (4)

4 A rubber band recoiler (3) An inelastic collision connector set : A needle and a pair of male/female connector (1)

5 A connector, misc. (3) A scale (shared) If you need anything else, ask the teaching assistant, visit the experiment preparation room (19-114) or prepare them yourself. Study the photogate detector and make up a plan in advance. There s no specification of the experiment procedure although the following is recommended. -How to Experiment 1 Put the wind adjustment knob of the fan in the OFF position and connect the power line. Adjust the power of fan enough to float gliders. If the power is connected when the knob is ON then the fuse may be damaged. Please take heed. 2 Adjust the height of frictionless motion table not to move the gliders because of inclination. (if it vibrates, decrease fan power) Make sure that you should balance the gliders to set the motion table exactly parallel to ground. 3 Prepare two gliders and set rubber ban recoiler on both side of one glider. For elastic collision, set recoiler on another. Turn the fan on and make it exhale enough wind to float the gliders.

6 4 Capture with program(i-ca) - How to capture : turn on I-CA. press camera button on the top of left - Adjust brightness, zoom, exposure and focus and select the location to save a file. - Press capture and proceed the experiment. - Press stop button and save the video file. 5 Analyze the video file with I-CA - How to analyze : press the red arrow button on the top of left in I-CA - Bring the video file. Play the video and check whether it is right file or not. - Choose the frame to start analyze and press next button. - Choose the frame to end analyze and press next button. - Select the analysis option. Normal analyze consumes more time but more precise. - Select the number of subject. Press test button and select the color sticker corresponding to subject. Press next and start analyze. (if it fail to analyze, make sure that you choose appropriate setting options like brightness, zoom, exposure and focus for I-CA to identify the sticker - Check whether it has constant Y. (if not, the motion table should be adjusted) - Press graph and T-X graph button. Select the subject you want to analyze and press nest button. See whether it has linear graph and save the file. - You can also save the data in text form. Get the slope of linear graph with Excel. 6 Experiment again for inelastic collision with the glider. To get collision coefficient, proceed the

7 process written above again. Also do the experiment with weighted gliders. 7 Reanalyze the data with I-CA in the case of inelastic collision and elastic collision, Is energy and momentum conserved? If not, why? Is energy and momentum conserved in the case of collision with weighted glider? If not, why? How much the collision coefficient with one glider? Now we have the collision coefficient in C. Classify whether its collision is elastic or inelastic. Does the data have error? How much does it have? How can you diminish the error? Do you think that we confirm Newton s Law of motion with this experiment? Why? Background theory The total momentum of the point mass system is an important mechanical quantity that is conserved if external forces don t act on the system whether there s interaction among the particles consisting the system or whether the interaction is conservative or not. If there s an external forces acting on it, then the total momentum of the system changes, and the time rate of its change equals the total sum of the external forces acting on the system. Accordingly Newton s second law of motion can be expressed in another way using the total momentum of the system. Accordingly, collision among the particles in the point mass system and the motion of the bodies with changing mass can be easily described.

8 The momentum p of a point mass with mass m and velocity v is defined as p mv (1) If the mass m is constant then the time rate of change of p is dp dt = m dv dt (2) and the second law of motion of the point mass is expressed as F = dp dt (3) That is, the time rate of change in momentum of a body is the same as the total sum of forces acting on it and its direction is the same as that of the combined forces. The law of motion expressed as (3) is a general rule that can even be applied to the case of the mass change. We are considering the case where the masses are constant. The momentum P of the point mass system consisting of many particles is defined as the vector sum of the momentum of each point mass. P = p 1 + p 2 + = p i = m 1 v 1 + m 2 v 2 + = m i v i (4) Taking the time derivative of P and applying the equation (3) to each point mass, dp dt = dp 1 dt + dp 2 dt + = dp i dt = F 1 + F 2 + = F i In this expression, the forces among the particles constituting the system cancel by the third law of motion and only the external force F e remains, then the relation

9 F e = dp dt (5) holds. This is generalization of the equation (3) to the point mass system. On the other hand, if we rewrite the equation (4) using the definition of center of mass, P = d dt m ir i = M dr dt = MV (6) holds. That is, the total momentum of the point mass system equals the total mass M of the system times the velocity V of the center of mass. If there s no external force acting on the given point mass system, then in the equation that is, dp dt = 0 P = p 1 + p 2 + = constant vector (7) holds. Therefore, if the sum of the external forces acting upon the given system is zero then the total momentum of the given point mass system is conserved. This means that if the momentum of some part of the system increases then the corresponding reduction of momentum should come about in another part. The total momentum can only be changed by external forces on the system. This law of momentum conservation is the generalized form of the first law of Newton and one of the most significant laws of physics proven in a lot of phenomena. Now consider an isolated system of two particles with respective masses m 1 and m 2 without any external force. Since there is no external force, the total momentum P = p 1 + p 2 = m 1 v 1 + m 2 v 2 (8) is constant independent of time. However, if there is an interaction between the two particles, the momentums of each particle can change as time varies. The increase in the momentum of a particle means the decrease in the momentum of the other. Therefore the relationships Δp1 + Δp2 = 0 or Δp1 = - Δp2 are satisfied. If the change in the momentum of these particles occurs in an infinitesimal time dt then the time rate of the change will be

10 dp 1 dt = dp 2 dt (9) As the time rate of the change in the momentum of a particle equals the force acting upon the particle, the equation (9) is F 1 = F 2 (10) where F 1 is the force of the body of mass m2 upon the body of mass m1 and F2 is the force of the body of mass m1 upon the body of mass m2, which means the equation (1) is the very third law of equation of motion of Newton. That is, the law of momentum conservation that applies to any system implies the third law of Newton. Meanwhile, the third law of Newton may not hold between the particles spatially separated. Then, does the law of momentum conservation not hold in this case? It is known that the law of momentum conservation is one of the fundamental laws that hold without any exception so far. Therefore if the force between the two particles above does not satisfy the third law of Newton then this system cannot be regarded as consisting of just two particles. The system is interpreted as there s a field that mediates the interaction between them and if we take into account of this mediating field then the momentum of the system is still conserved. Consider the collision of two bodies moving one dimensionally. Suppose the external force can be ignored but only the impact force acts on them so that the state of motion of each body changes in the process. Let the speed of the bodies of mass m1 and m2 be u1 and u2 before the collision and v1 and v2 after the collision respectively then since the momentum is conserved, the relationship m 1 u 1+ m 2 u 2 = m 1 v 1 + m 2 v 2 (11) holds. Meanwhile the kinetic energy is generally not conserved in the process of collision. However, if the kinetic energy is conserved before and after the collision and the relationship 1 2 m 1u m 1u 1 2 = 1 2 m 1v m 1v 1 2 (12) holds then we say this collision is an elastic collision. If the collision is an elastic one then from the equation (11) and (12), the relation

11 v 2 v 1 = (u 2 u 1 ) (13) is obtained. As you can see, in a one dimensional elastic collision, the magnitude of the relative speed of two bodies is equal and opposite before and after the collision. If the kinetic energy decreases after the collision then we say it is an inelastic collision. In this case the relation v 2 - v 1 < u 2 - u 1 holds. If we define the coefficient of restitution e as e v 1 v 2 u 1 u 2 (14) then the coefficient of restitution e represents the elasticity of the body. That is, if e=1 it becomes the equation (13) and means that the collision is elastic, if e=0 then after the collision v2 - v1 = 0 so the two bodies become one which is a perfectly inelastic collision and if 0<e<1 then v2 - v1 < u1 - u2 so it is a general inelastic collision.