Please complete the following lab.

Procedure A. mass of red PasCar 254 q 1 VI V 2 (m) (m) (m) (s) (m/s) (s ) (m/s) 261 . 135 126 0 . 14 0. 65 P : 74 - 298 . 16 . 137 57 L. la 178 097 0. 45 0 .75 20 109 2 , 236 . 12 4 112 5 . 12 0. 75 0 . 65 a VS % difference (m) (m/$2 ) (m/s) (J) (J) .1126 2.4 137 J. 75 1 15 109 2 1. 77 . (12 2. 12 Ah = 41 - 72 hex = amvf V2 - V , (see related question - top of next page) a = Kf = AU VF = V2 aD 6 -9Do you see any trend in the % differences between the change in kinetic energy and the magnitude of the change in the potential energy as the track's angle of inclination changes? If so, what is that trend? Are there any other interesting features in your data? cond object at rest Procedure B. 242.3 9 Viot ritzmg ( vio - vif ) mass of blue PasCar. 2423 hg m, = Blue Ma = Red " 2 = 1. 89 Blue m 1 m2 M Vo pred. v % (kg) (kg) (kg) (m/s) (m/s) (m/s) difference 2423 75 9 9963 0. 79 0.18 .19 3.9 " 2423 5 04 0, 73 0. 23 237 3 243 0. 77 0. 37 492 2 . 254 0 . 46 0.29 7422 . 254 0 . 5 D. 37 1 4923 101 6 . 3 m M Ko Klost (kg) (kg) (J) Klost /K. (J) (J) V1: = mvf (see related question - top of next page) 6 - 10Do you see any trend between the fraction of the kinetic energy lost during the collision and the masses of the objects involved in the collisions? If so, what is that trend? Procedure C m1 Vo pred. v difference (kg) (kg) (m/s) (m/s) (m/s) (m/s 0. 72 - . 83 - 137 104 72 -. 21 -325 . 79 -. 01 . 59 ./18 the 0.54 . 27 0 .49 O hilar m1 m2 Kif K trans Ktrans / K 10 (kg) (kg) (J ) (J) (J) Band wasting the is wrapped around 6 - 11Experiment 6. Conservation of Energy and Momentum Purpose To study the conservation of mechanical energy by observing the motion of an object on an inclined plane, and to study conservation of momentum by observing objects undergoing collisions Apparatus Pasco LXI Data Logger, motion sensor, PasCars, 1.2-m dynamics track, meterstick, stand with base, electronic balance Theory Conservation of Energy The kinetic energy of an object is given by: K = Imv2 Equation (6 - 1) where m is the mass of the object and v is its velocity. An object in a uniform gravitational field also has a potential energy given by: U = mgh Equation (6 - 2) where h is the height relative to some arbitrarily chosen zero reference height. gusto .most'll essar to The sum of the kinetic and potential energies is referred to as mechanical energy and if there are no non- conservative forces acting on an object its mechanical energy is constant. Thus, in the absence of non- conservative forces any change in potential energy is accompanied by an equal and opposite change in the kinetic energy. We express this fact in the form of an equation as: AK = -AU. Equation (6 - 3) Conservation of Momentum (Collisions) In all collisions, the total system momentum, which is the sum of the products of mass and velocity for the components of the system, is conserved. For the special case in which there are two objects, one of which is initially at rest, the sum of the momenta of the two objects after the collision has the same value as the momentum of the single moving object prior to the collision. 6 - 1In the case of a completely inelastic collision, the objects join together and move as one after the collision. For a completely inelastic collision in which the "target" is initially at rest, conservation of momentum may be expressed by: my. = MVs Equation (6 - 4) where m is the mass of the initially moving object, v. is its velocity, M is the mass of the combined object after the collision, and v, is the velocity of the combined object. Equation (6-4) can be solved for any one of the quantities that appear in it if all of the others are known. In this experiment we'll use it to obtain predicted final velocities, which will be compared to observed final velocities. In completely inelastic collisions, and in many collisions that are not completely inelastic, kinetic energy is lost. [It is also possible to have a collision in which kinetic energy is gained; consider the collision between a "pop ball" and the pavement.] Collisions in which the kinetic energy of the system after the collision is the same as the system kinetic energy was before the collision are called elastic collisions. In all elastic collisions, the individual objects remain separate. Thus, the conservation-of-momentum equation contains more terms. Again considering the case for which one object is initially at rest, we will now have: wa el Boldon Equation (6 -5) The additional requirement that is imposed on elastic collisions, conservation of kinetic energy, may be expressed as: Equation (6 - 6) In our experiment, we will be monitoring the velocity of the initially-moving object. In order to be able to compare our experimental results with the predictions of theory, we need to obtain the final velocity of mass 'l' from these two equations. First, solve Equation (6-5) for vy This produces: v2, = m my ( v1 0 - VIS ) . m 2 Substitute this result into Equation (6-6) and multiply all terms by '2'. From this we obtain: mv? = mv?, + my|my [v. - vis])". The equation can be rewritten: m (vi - v],)= =[v. - vy, ]. The difference of squares factors into the product of the sum and the difference. differeen Then, the difference term can be cancelled to produce: 1 + 1 = mi (v. - Vis ) 6 -2