Collision and momentun lab 6 (1)

Johanna Gantier Lab Partner: Simrun Karin & Averseen Sous Instructor: Professor King Leung Lab: PHYS 11000- 1L12 March 31, 2023 Laboratory #6: Collisions and Momentum Objectives: To study the relation between force and change in momentum, and to determine whether momentum is conserved in inelastic collisions Equipment: two carts, two force probes, motion detector, electronic balance, masses. Introduction: For collision and momentum, we can use Newton’s third law but instead of writing Newton’s Second Law as F=ma, we can write it as F=∆p/∆t. This means that if a constant force is acting on an object for a time ∆t , then its momentum will change by an amount F∆t . Let's assume that the force starts acting at time t0 . If we plot the force as a function of time, it will be zero until t0 . The force between t0 and t0+∆t , and then zero again after that. The area under the force versus time plot is just F∆t , which is the change in momentum . If the force is not constant, it is still true that the area under the force versus time plot will be the change in the momentum of the object. When two objects collide in the absence of external forces, Newton’s Third Law guarantees that the total momentum is unchanged. If the two colliding objects are object A and object B , then when they hit, the force exerted by A on B will have the same magnitude as the force that B exerts on A , but its direction will be opposite. This means that the change in momentum of object A will be the opposite of the change in the momentum of object B so that the sum of their changes in momentum is zero. Therefore, the total momentum before the collision is the same as the total momentum after the collision. In terms of equations, if p A =m A v A is the initial momentum of object A p B =m B v B is the initial momentum of object B, P A’ =m A’ V A’ is the final momentum of object A , and P B’ =m B’ V B’ is the final momentum of object B, then p A +p B =P A’ +P B’ . In this lab, we want to verify Newton’s Third Law, show that the change in momentum is equal to the area under the force versus time graph, and study momentum conservation in an inelastic collision. The collision will involve two carts, one initially moving and the other sitting still, that stick together when they hit. This means that if the initially stationary cart is object B , then v B =0 and V A =V B  p A +p B =P A’ +P B’ Procedure: 1. You should have two cars with force probes mounted on them. Place them both on the track with the parts sticking out of the force probes facing each other. Open the file Collisions; you should see two sets of axes on the screen, a force versus time plot for each force probe. You are going to push the carts together (gently!) so that they collide. The force probes will hit each other and measure the forces on each of the carts. Sketch predictions for what the force versus time plots for each cart will look like. Zero all sensors by 28 clicking the button “Zero all sensors” before every measurement. Now do the experiment and sketch the results.

Figure 1. The experiment was set up with two cars that move on a track and motion sensors at the end of the track. Each of these cars has a probe detector on the end of contact. Predictions: Graph 1. Prediction of Force on Cart 1 vs. Time graph before and after collision Time (s) Force on Cart 1 Graph 2. Prediction of Force on Cart 2 vs. Time graph before and after collision Time (s) F o rce o n C art 2

Graph 4. Prediction of Force on Cart 1 vs. Time graph with 499.6g (0.4996kg) mass on cart 1, before and after collision. Time (s) Force on Cart 2 Graph 5. Prediction of Force on Cart 2 vs. Time graph with 499.6g (0.4996kg) mass on cart 1, before and after collision Part 2 Experimental results

graph 6. Experimental data of force on cart 1 and 2 with mass before and after collision. Observations: we observed that on part 1 on our first experiment graph “cart without mass” the force of cart 2 went up to a peak of 2.56N in t= 0.10s whereas on our second experiment graph “cart with a mass” the force of cart 2 peak went higher to 3.001N in t= 0.12s. Same data for cart 1 but in the negative axis. 2b.What do your results tell you about Newton’s Third Law? According to Newton’s third law of motion every action force has an equal and opposite reaction force. These forces act on different objects therefore they don’t cancel. If object A exerts a force on object B , then object B must exert a force of equal magnitude and opposite direction back on object A . These results tell me that Cart 2 (moving cart with mass) exerts a maximum force of 3.001N on cart 1 (stationary cart with no mass), However, cart 1 exerts an equal opposite force on cart 2. According to Newton's third law, when the moving car exerted a force on the stationary car, the stationary car exerted an equal and opposite reaction force back on the moving car. 3. Now open the file Impulse and Momentum. You should see two sets of axes, one for a velocity-time plot and one for a force time plot. The force probe that will be recorded is the one that goes onto channel 2. You are going to start with one cart moving and the other standing still, and then they will collide. The force probe going into channel 2 should be on the car that is initially motionless, and the motion detector would also be recording the motion of this cart. After doing the experiment what you will see is a plot of the force acting on the initially motionless cart and a plot of its velocity. Sketch a prediction for what you think these plots will look like. Zero all sensors. Now do the experiment. Sketch your results. Impulse and Momentum . Sketch a prediction for the velocity and force plots on initially motionless cart. Graph 7 & 8 Time (s) V e lo c ity (m /s )

Run #1 Mass of cart and force probe = 1209.8g, 1.2098kg Change in momentum of cart due to collision = ΔP= m • ΔV  (1.2098kg) *(-0.4m/s – 0.0m/s) = -0.484kg*m/s Area under the force versus time plot = -0.2 N*S Run #2 Mass of cart and force probe = 1209.8g, 1.2098kg Change in momentum of cart due to collision = ΔP= m • ΔV  (1.2098kg) *(-0.2m/s – 0.0m/s) = -0.242kg*m/s Area under the force versus time plot = -0.2 N*S Compare this area to the change in momentum . Area under the curve gives change in momentum. change in momentum should be =-0.2N*s. However, For the first run the area under the force (N) vs time graph (s) graph is about ½ the result we obtained in the change in momentum of the cart due collision. Moreover, for the second run the area under the force versus time plot and the change in momentum are the same (as it should be).

*These results could be inaccurate due to technical issues with the equipment and methods utilized during the experiment. 5. Inelastic collision: Remove the force probes from the carts making sure not to lose any screws. Find the mass of each cart by using the electronic balance. Open the file Inelastic Collision, and you should see axes for a velocity-time plot on the screen. Notice that on each cart there are Velcro pads on one end. You want to have one cart initially moving and one initially at rest so that the two Velcro ends are facing each other. This will cause the two carts to stick together when they hit (again, gently!). The motion detector should be on the same side of the track as the initially moving cart. When you start the experiment, the motion detector will see the initially moving cart, and, after the collision, it will see the two carts moving together. You can use the velocity data to find the momentum before and after the collision. Now do the experiment and compare the initial and final momentum. Graph 11. Velocity of inelastic collision of carts with no mass added to the carts. Run #1 Inelastic collision without added mass. Results of inelastic collision of carts without an added mass Mass of cart 1= 529.9g (1kg/1000g)  0.5299kg Mass of cart 2= 677.5g (1kg/1000g)  0.6775kg Values Mass cart 1 (MA) 0.5299kg

6. Repeat step 5 except with a mass in one of the carts. Graph 12. Velocity of inelastic collision of carts with a mass on one of the carts Results of inelastic collision of carts with a mass on one of the carts Mass of cart 1 (initially moving) = 0.5299kg Mass of cart 2 (initially motionless) =0.6775kg Mass= 498g (1kg/1000g)  0.498kg on which cart? Cart 2 (motionless) From the graph, calculate and compare the initial and final momentum. Values Mass cart 1 (MA) 0.5299kg Mass cart 2 (MB) motionless 0.6775kg Added mass 0.498kg Initial velocity cart 2 (motionless) 0m/s Initial velocity cart 1(V0) 0.165m/s Final Velocity (VF) cart 1+ cart 2 0.063m/s Initial time (T0) cart 1 0.050s Final time (TF) cart 1 1.453s Initial momentum (P0) 0.0874kg*m/s  rounded = 0.1kg*m/s Final momentum (PF) 0.107kg*m/s  rounded =0.1kg*m/s Calculations

pA=mAvA  PA’=mA’VA’ pB=mBvB  PB’=mB’VB’ P= mAvA+ mBvB= mA’VA’+ mB’VB’  P= mAvA+ mBvB= (mA’+mB’) V’ P= (0.5299kg) *(0.165m/s) + (1.1755kg + 0.498kg) *(0m/s) = (0.529+0.6775kg+0.498kg) *(0.063m/s) P  0.0874335kg*m/s= 0.107kg*m/s From the graph, calculate and compare the initial and final momentum. In this experiment, we examine the impulse in inelastic collision between two carts on a track to test the change in force, velocity, and momentum over time. Momentum is conserved in inelastic collisions, but energy is not; kinetic energy converts into thermal energy. From the results above we can see that momentum is conserved on both sides of the equation. On the left side we have P= 0.087 kg*m/s (rounded 0.1 kg*m/s) and on the right side we have P= 0.107kg*m/s (rounded 0.1kg*m/s) Conclusion: In this experiment, we examine the impulse in an elastic and inelastic collision between two carts on a track to test the change in force, velocity, and momentum over time. Momentum is conserved in both elastic and inelastic collisions, but energy is not conserved on inelastic collision; kinetic energy converts into thermal energy. The collision of two carts on a track can be described in terms of momentum conservation and, in some cases, energy conservation. If there is no net external force experienced by the system of two carts, then we expect the total momentum of the system to be conserved. This is true regardless of the force acting between the carts. Collision and momentum happen in everyday life. Collision of atoms and particles, collision of two cars, billiard balls, a bullet coming in contact with a target, two hockey players colliding with each other or two football players colliding with each other, or just a small collision between two people walking in opposite directions of the walkway can be calculated with this formula which makes it important to resolve cases that go from scientific information, a car accident or physics application. Some inaccuracies that may have occurred during this lab experiment can be due to issues with the recording equipment and sensors, errors in graphing and collection of data, and calculation errors.