I'm looking to understand this lab and what it is asking in general Right now I'm trying to understand how to derive an equation for the force of the tension for a cart of mass m by also being pulled by a mass of m

Part 1 - Impulse and Momentum Purpose To measure the change in momentum of an object during an elastic collision and 0011113\"?! \"- t0 the Integral 0f the force exerted on the object over the time of the collision. Equipment I Impulse and Momentum setup [1) II Work-Energy setup [1] In Computer with CAPSTONE software {1:} Theory This will he completed as a class demo and discussion. From your previous activities you are familiar with the observation of carts on a track colliding With, and sometimes bouncing 03' Of, 83-611 OthEF YOLI my remember that, for the part. of the motion before and after the collision. the carts had a constant velocity. The dramatic changes in velocity took place only during the periods when the carts came very close to each other. We can thus be certain that when the carts are not VEI'J' close together, they exert. no measurable force on each other. But when they collide with each other, there is a force between them. The magnitude of this force depends very strongly on time. and acts for a very short duration. The exact form of this force can be qujte complicated, but even withOut knowing the details we can still learn a lot about the eects of the force. Recall that Newton's 2nd law states that the time rate of change of momentum of an object is equal to the force acting on the object. So if F is the force experienced by the cart; That is, the total change in momentum of an object will equal the time hitcgral cf the force acting on that object, over the time during which the interaction takes place. 9: he: Pin: CI For a constant force this expression becomes at = For This change in momentum is called an impulse. .i' = A}? The word \"impulse" is mostly used in the context of a momentum transfer occurring in a short time. such as when a baseball hat hits the ball1 or when one billiard ball strikes another. The impulse ma)r be thought of as an injection of momentum. To evaluate the integral above. we would need to knew more about how the force. F, varies with time. Fortunately1 in our experiment the computer evaluates this integral for us. We will compare these in two dicrent ways to calculate the change of momentum of an object in a collision, 8E: Procedure We use a motion sensor to plot the motion of a cart before and after its collision with a magnetic bumper which is itself mounted on a force sensor. The force sensor measures the force that the cart exerts on the bumper during the collision, Since momentum must be conserved in an isolated system, this must be the opposite of the force exerted by the bumper on the cart. The TA will perform the experiment and show realtime collection of data on the classroom sateen. 1. Find the mass of the cart using the digital scale and record it in Data Table 1. 2. Your TA will now perform the experiment. 3. Record the data displayed on screen in the appropriate columns in Data Table 1. 4. Calculate pg, pf, and Ap for the cart. I've removed the vector symbols because we are working in one dimension only, but these are still vector components, and so their signs matter! Make sure you get them right. Data Table 1 With this sign convention, how does the measured change in momentum compare to the impuISe? What could possibl cause the Chan-e in momentum to be different from the measured impulse? Part 2 - Work and Change in Energy Purpose To measure the work done on a. body and to compare it to the body's change in energy. Sine: the position of the object with respect to the direction of the Efitatlonl fame '3 unchanging. an the ObjeCt Is not deformable, lonetic energy is all we have to worry about. [We are going to neglei friction, etc.) Theory Work done on a. system by an external force is given by w = f Fi - d}. When an object experiences a constant net force film over a displacement cl: parallel to the force, the work done on it is Wrist = Feet"-1 The work done an a system by the net force acting on it equals the change in the systeru's total energy (3-5 long as no heat enters or leaves the system]. Here, our system is just a cart, which by itself only has kinetic energy. {We could also consider the system formed by the Earth and the cart, which has gravitational potential energy in addition to kinetic, but since the cart will remain on the same horizontal level this gravitational energy will not change anyway) In 0111' exP'E'=1\"iI:IJent, the net force acting on the cart will be in the same direction as the dis- placement, so the work done by that force' 15 positive and, as argued above, it should match the increase in kinetic energy of the cart. Mathematically, 1 W: fF-d mof - Ems2 where W is the work, and of the nal velocity. Procedure This will also be done as a class demo by the TA. In this activity, an approm'mately constant force is applied to a cart by a. string placed over a pulley with a known mass suspended at its end. The force sensor measures this force and the motion sensor measures the motion of the cart as it is pulled. CAPSTONE displays graphs of the force applied and the velocity of the object versus position. We will use the program to integrate the area under the force versus position curve. We will identify the quantity represented by this area, and see how this relates to properties of the system and our understanding of energy. 1. Measure the mass of the cart and force sensor; record the sum of the two values in Data Table 2 2. Your TA will now perform the experiment and select the appropriate data. 3. Record the data dismayed on screen in the appropriate columns in Data Table 2 4. Calculate Km\Derive an equation for the force of tension for a cart of mass m being pulled by a hanging mass of mass mp, similar to this setup. What are the possible sources of error in this experiment? Specifically, in which way would each of these errors affect the data