Introductory Physics I - PHYS 200 LAB 5-COLLISION IN TWO DIMENSIONS COMPLETE NEW UPDATE VERSION 2026

Introductory Physics I - PHYS 200| LAB 5: COLLISION IN TWO DIMENSIONS COMPLETE NEW UPDATE VERSION 2026 ATHABASCA UNIVERSITY

PHYSICS 200

LAB 5: COLLISION IN TWO DIMENSIONS

Introduction The main purpose of this lab is to examine the two-dimensional collision of two identical bodies using Tracker software. This lab will be performed with one body initially at rest, and the other moving in the positive x-axis. We can hypothesize that total momentum will be conserved, and can therefore describe the momentum of the two objects on both axis as such: px = p′x v1 = v′1cosθ + v′2cosϕ

py = p′y 0 = v′1sinθ − v′2sinϕ Because the objects are of identical masses, they are not considered in these equations during the lab. If the collision in this lab is perfectly elastic, then we know that total kinetic energy is conserved, and the below equations are satisfied: ½mv12 + 0 = ½mv′1 2 + ½mv′22 v12 = v′12 + v′22 θ + ϕ = 90°

Materials • • •

Two identical spherical objects Measuring tape Digital camera

Procedure Place one of the spheres motionless on a smooth, hard surface. Position the measuring tape near the motionless ball where it will not interfere with the experiment, but still provides necessary information. Place the second sphere about 20cm from the other, and bump it towards the motionless sphere.

Figure 1: Lab setup – two identical masses on a hard surface with the measuring tape perpendicular to the path of ball one.

Data

Figure 2: Time and displacement of ball 1 after being bumped. It takes 0.133 seconds for the collision to take place.

Figure 3: Time vs. displacement along x-axis of ball 1 prior to the collision occurring.

Figure 4: Data collection of ball 1 after the collision. Graph and table on right display the time and displacement of the ball along the x and y-axis.

Figure 5: Graph of time vs. displacement along x-axis of ball 1 after the collision.

Figure 6: Graph of time vs. displacement along y-axis of ball 1 after the collision.

Figure 7: Data collection of ball 2 after the collision. Graph and table on right display the time and displacement of the ball along the x and y-axis.

Figure 8: Graph of time vs. displacement along x-axis of ball 2 after the collision.

Figure 9: Graph of time vs. displacement along y-axis of ball 2 after the collision.

By reviewing the table and graph above, I have observed a linear trend in the raw data of the experiment. I will process these findings in the calculations below to conclude whether or not the collision that occurred in the above procedure was elastic. My hypothesis is that the collision was not elastic, as factors such as friction and air drag make the likelihood of a perfectly elastic collision extremely low.

Analysis By using the best fit equation y = At + B and the corresponding parameters for A and B (provided in the figures above of the final graphs) to solve for the velocity, I will find the following velocities. I will also check these results using the kinematic equation = ∆/∆t

v1 = At + B v1 = (1.057)(0.133) + (2.686 x 10-2) v1 = 1.17 m/s

v2 = 0

v′1x = At + B v′1x = (2.633 x 10-1)(0.267) + (1.613 x 10-1) v′1x = 0.232 m/s

v′1y = At + B v′1y = (-3.119 x 10-1)(.267) + (-3.787 x 10-2) v′1y = -0.121 m/s

v’1 = 0.262 m/s

v′2x = At + B v′2x = (4.173 x 10-1)(0.334) + (1.982 x 10-1) v′2x = 0.338 m/s

v′2y = At + B v′2y = (-9.861 x 10-3)(0.334) + (9.11 x 10-4) v′2y = -0.00238 m/s

v’2 = 0.338 m/s

Using the above results, I will calculate θ with the equation θ = tan-1(vy/vx)

θ1 = tan-1(-0.121/0.232) θ1 = 27.6° below the x-axis

θ2 = tan-1(-0.00238/0.338)

θ2 = 4.03° below the x-axis

These results are acceptable, as the video shows the ball 1 colliding into ball 2 and scattering at an angle below the x-axis, while ball 2 scatters very close to the x-axis upon colliding with ball 1.

Regardless of whether or not the collision was elastic, the two equations below should always be satisfied. These equations describe a situation in which total momentum of the two-body system is conserved in each direction: px = p′x v1 = v′1cosθ + v′2cosϕ v1 = (0.262)cos(27.6°) + (0.338)cos(4.03°) = 0.231 m/s

py = p′y 0 = v′1sinθ − v′2sinϕ 0 = (0.262)sin(27.6) + (0.338)sin(4.03°) = 0.145 m/s Based on the results of my lab, these two equations have not been satisfied, and therefore does not support the assumption that linear momentum has been conserved. I believe that the errors occurred when I performed the procedure on my wood floor, which was potentially not 100% level or that the grooves in the flooring altered the results. There is also potential for error when using the Tracker software, as the results rely heavily on the resolution of the video, which was not totally clear when reviewing under such small frames.

Based on these results and my understanding of this unit, I do not believe that this collision was elastic. The following equations are satisfied when a collision is perfectly elastic, so I will pursue my hypothesis that these will not be satisfied and further prove that the collision in this test was not elastic. v12 = v′12 + v′22 (1.17)2 = (0.262)2 + (0.338)2 1.37 ≠

0.183

If momentum had been conserved, the sum of the square the v’1 and v’2 would equal v1.

θ + ϕ = 90° 27.3° + 4.03° ≠

90°

When a collision is perfectly elastic, the angle between both bodies involved 90°. I believe that the reason that this was not satisfied in my procedure was that ball 1 did not contact ball 2 perfectly straight on, therefore skewing the angles of both of the balls after the collision.

v′1 = v1cosθ 0.262 = (1.17)cos(27.3°) 0.262 ≠

1.04

v′2 = v2cosθ 0.338 = (0)cos(4.03°) 0.338 ≠

0.0

Based on the above results of the above relations that only apply to perfectly elastic collisions of bodies of equal mass, we can see that they have not been satisfied.

Conclusion When examining the results of the above analysis, we can see that the collision that occurred during the procedure of this lab was not elastic. This only happens when the bodies are involved are identical in mass and there are no other forces acting upon the system. However, in this situation, there would’ve been factors such as friction acting upon the balls, and despite choosing two balls that appeared identical in mass, potentially were not. This could be the result of manufacturing discrepancies. I believe that other sources of uncertainty in my results came from performing the lab on a surface that might’ve not been totally level, and that the software used to analyze the results was not accurate enough (resolution not clear enough, manually selecting the centre of the body in each frame not accurate enough).

Questions Below is my derivation of the equations used in the lab analysis: When linear momentum is conserved, it is conserved separately in both the x and y-axis, and when the bodies involved in the collisions are identical in mass, their masses cancel in the equations below. These relationships can be written as such:

px = p′x mv1 + 0 = mv′1cosθ + mv′2cosϕ v1 = v′1cosθ + v′2cosϕ

py = p′y 0 + 0 = mv′1sinθ − mv′2sinϕ 0 = v′1sinθ − v′2sinϕ v′1sinθ = v′2sinϕ v′1 = (v′2sinϕ)/(sinθ)

When an elastic collision occurs, internal kinetic energy is conserved. As a result, the sum of kinetic energies before the collision equals the sum after the collision. Because the masses involved are equal, they cancel.

KEi = KEf ½mv12 + 0 = ½mv′12 + ½mv′22 ½mv12 + 0 = ½mv′12 + ½mv′22 v12 = v′12 + v′22

By substituting the final line of equation two into the final line of equation one, we get:

v1 = [(v′2sinϕ)/(sinθ)] cosθ + v′2cosϕ v1sinθ= v′2(sinθcosϕ + cosθsinϕ) v1sinθ= v′2sin(θ + ϕ) v1sinθ= v′2sin90° v1sinθ= v′2(1) v1sinθ= v′2

Because we have already established v′1 in the final line of equation 2, we can simplify this as such: v′1 = (v′2sinϕ)/(sinθ) v′1 = [(v1sinθ)sinϕ]/sinθ v′1 = v1sinϕ v′1 = v1sin(90° - θ) v′1 = v1cos θ

For a perfectly elastic collision between bodies of equal mass, their angles can be solved for as such: θ1 − θ2 = 90° ϕ = 90° - θ θ + ϕ = 90°