PostLab 6_ Conservation of Momentum

Name : Miguel Rivera Villareal Lab/Section: Physics 180L 1009 Teaching Assistant: Nuzhat Nuari Islam Due Date: November 4th, 2022 Conservation of Momentum Objective: To understand the law of conservation of momentum via colliding objects in two dimensions. Apparatus: Digital scale, ramp, two steel balls, one hollow steel ball, C-clamp, plumb bob, sheet of easel paper, masking tape, meter stick, protractor, carbon paper Theory: The general formula that defines conservation of momentum would be that momentum equates to mass times velocity, allowing for additional objects that collide to be simply added. It is important to note that since velocity is a vector, the directions are significant before and after the collision. If an object at rest is struck with an object of same mass the resulting objects will depart from one another with a 90 degree angle. When rolling a ball down a ramp off a table the vertical component of velocity is active until it hits the ground, so the distance from when the ball leaves the table and hits the ground can be measured using the velocity and time. If masses are the same they can be ignored in the momentum equation, leaving just the velocities to be added or the distances of the collided objects can be used instead if they’re known. Procedure: Part one of the lab is searching for the aforementioned distance. Before doing so, set up the lab environment by securing the ramp system, taping paper to the ground near the ramp, aligning the plumb bob to the center of the paper, marking this point, and marking where a ball rolling down hits the paper. Repeat the rolling ball action ten times to get an estimated likely position for where the ball will drop and then mark the center of the 10 droppings. Connect the new mark with a line to the plumb and record the length of this line. Encase the 10 drop points in a circle.

To determine the distances of the colliding objects, one must first set the other steel ball on the collision plate’s first depression. Replace the locations where the ball lands with carbon paper, for another 10. Repeat the line drawing procedure that was previously mentioned for this new set of collisions while also documenting the length of lines, angles between the new lines and the first line. Label the impact patterns as S ₁ ₐ S ₁ в. Part B consists of repeating with the target in the second depression and label this new set with S ₂ ₐ and S ₂ в with part C doing it again but making the target instead a hollow or plastic ball in any depression, labeling it as S ₃ ₐ and S ₃ в. For the final section, the report, get the ratio of masses between the hollow/plastic ball and the steel ball. Then, for each collision that had two steel balls, add the Sₐ Sв vectors via graphical or analytical method and compare it to the original S length., though if the latter method is chosen add a diagram. Repeat this for the hollow/plastic ball collision but multiply the B vector by the ratio. The last step is getting the percent errors for all the x-components and angles in relation to 90 degrees. Equations: [|(S Ax + S Bx ) - S | / S] * 100% = percent difference of lengths S = S A1 + S B1 S = S A2 + S B2 S = S A3 + (M B /M A ) * S B3 S Ax, S A1, S A2, S A3 : incident ball’s x-axis length in centimeters S Bx, S B1, S B2, S B3 : collision ball’s x-axis length in centimeters M B : hollow ball’s mass in grams M A : steel ball’s mass in grams [ |(Θ A + Θ B ) - 90 ° | / 90 ° ] * 100 % = percent difference of angle Θ A : incident ball’s angle Θ B : collision ball’s angle

Notes/Computations:

I do not believe kinetic energy is conserved in the collisions because the system did have friction acting. This friction would have occurred between the incident ball and the ramp and the incident ball and the colliding ball. Question 6: Why must the displacement vector of the lighter ball be scaled by the ratio of masses of the two balls? The displacement vector of the lighterball must be scaled by the mass ratio because in the previous collision tests the masses remained the same so they were able to have their masses canceled out, leaving just the distance to be added. Since the masses do vary in this test the masses must be taken into account and that’s done via the mass ratio so only one value has to be multiplied. Question 7: Is it possible for either of the heavier balls to travel farther in the x direction than the control displacement S ? What about the lighter ball? Why? It is not possible for the heavier balls or balls of the same mass to travel farther than the control displacement because the potential energy of the incident ball is just further transferred into the heavier/same mass ball to get it out of rest. This is showcased in experimental trials one and two that had the same mass balls collide but the SX value never going past the initial S value. A lighter ball can go farther because the aforementioned energy and momentum is given into the ball’s velocity after the mass is taken care of and that is shown in the third trial which did indeed have a higher of 57.7 centimeters compared to S’s initial 48.5 centimeters.