VATION OF LINEAR MOMENTUM AND INELASTIC COLLISIONS Objective: The purpose of this experiment is to study the Law of Cons

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VATION OF LINEAR MOMENTUM AND INELASTIC COLLISIONS Objective: The purpose of this experiment is to study the Law of Conservation of Linear Momentum as well as the non-conservation of kinetic energy in imelastic concions Air track and riders, spark tape position recording system Apparatus: Theory: an object of mass moving with velocity has lineat momentum p defined as PM Since the velocity is a vector quantity, linear momentum must also be a vector quantity and, therefore, it has both magnitude and direction When a system of one or more objects is isolated, that is the net external force acting on it is zero, it can be shown that the system's total linear momentum (the sum of the momenta of its individual parts) remains constant, or is conserved". This is what is referred to as the Law of Conservation of Linear Momentum. In the context of collisions, linear momentum conservation means that the (vector) sum of the momenta of the individual parts of the system before the collision remains the same after the collision That is, for a system consisting of n objects: Of course, the momentum of parts of the system may change as long as the momentum of other parts also changes so that the vector sum of the moments of all parts remains constant Contrary to linear momentum, kinetic energy is not always conserved in collisions. Only in certain types of collisions, known as "elastic collisions", total kinetic energy is conserved. All other collisions are known as "inelastic". Here, the total kinetic energy of the system before the collision is not equal to the total kinetic energy after the collision, that is, (ŠKE, (KE) , ΣΚΕ + ΣΚΕ This experiment deals with inelastic collisions, which includes explosions. before affer
Procedure: 1. Sticking Collisions Place two riders on the track with mating Velcro tape ends facing each other. With the larger mass initially stationary, gently push the smaller mass imparting an initial velocity v to it. Only one spark timer is needed to record position before and after the collision. m M m M Before After Analyze the spark tapes to obtain the velocities and of the riders before and after the collision for the following arrangements: Arrangement Riders Type of Collision A B. Two small ones One small, one large Sticking Sticking Recall that the time interval At between consecutive sparks is 0.1 s. To minimize fluctuations in the spacing of spark marks on the tape measure the distance d corresponding to a number N (e.g. 10) of spark-to-spark spaces. The velocity you wish to determine will be given by v=d/NA. Measure the masses m and M of the riders with the exact attachments used. If linear momentum is to be conserved in these cases, then: mv = (m+M)V 2. Explosive Collision Reverse the riders so that the magnets face each other. Tie the riders together to provide a moderate compression; however, keep the riders from tilting and rubbing the track. Sever the thread by carefully cutting it or burning it. Obtain the velocities v and V with following arrangement: m M m M Before After Arrangement Riders Type of Collision C. Explosive One small, one large
Explosive Collision (continued Two spark timers, connected to two separate guide wires, are needed for recording to position of each rider in this part of the experiment. Be sure to keep track of the direction of the riders. Also, don't use too fast a collision. It may cause the rider to tip and briefly rub the track If linear momentum is to be conserved in these cases, then: 0 = mv+MV, recalling that the direction of velocities and moments for motion in one dimension is taken into account through algebraic signs Analysis: 1. For each of the three collisions above calculate the following a. Total momentum before collision or explosion b. Total momentum after collision or explosion If momentum is conserved, then the total momentum after should equal the total momentum "before" Do they? Compare the two by computing the % difference between them for each of the two sticking collision cases (it is meaningless to do so for the explosive case). In calculating the difference, use the average of the "before" and "after values in the denominator. 2. For each of the two sticking collisions calculate the fraction of the initial kinetic energy that was lost. This fraction is defined by: KE -KE КЕ, It can be shown, using KE my', KE... = (m+M)V and my = {(m+MOV, M that this definition can be rewritten as: (2) M+m The subscript in fın is used to emphasize that this is theoretical value of the fraction. Here, again, m is the mass of the small (moving) rider and M is the mass of the initially stationary) target Compare the experimental value of f (use Egn. 1 and your measured masses and velocities) to the theoretical value given by Egn. 2 by computing the % difference between the two for each of the two sticking collisions. Question: Where did the final kinetic energy in the explosion come from? Can you give another example of something giving off energy?
Trial (arrangement) Mass (kg) IA . 186 Velocity Inital final red .21 .389 186 blue .271 o red. 271 .629 .233 I blue-441 0 .238 red .271 F1.65 blue .441 1101 11.B В 2.C 0