Elastic Collision in One Dimention
I was told that my percentages of discrepancy are wrong for this problem. I felt like that was the easiest and most straightforward part of the problem, so I am wondering if I did anything else wrong. I was told that all of my answers were correct (except the predictions, no feedback) besides the percentages which were marked wrong. Everything was graded by a computer, so it's very picky about formatting meaning that I could just be formatting my answers wrong. I've tried a couple of different ways, so I thought I would come here for a second opinion. Please help!
In this lab you will study an elastic collision between a moving glider(m1) and a second one which is initially at rest(m2). (They are on an air track, so ignore friction.)
Given Values(Every other number is calculated.)
m1: 266.8g
m2: 474.9g
L1: 14.1cm
L2: 25.3cm
Δt1i: .122955s
Δt1f: .475583s
Δt2f: .311598s
Derive expressions for v1f and v2f in terms of v1, m1, and m2, based on the scenario above. Use conservation of momentum and conservation of kinetic energy (since we are postulating that the collision is elastic).
V1f=V1i(m1m2/m1+m2) & V2f=V1i(2m1/m1+m2) (Simplified from conservation of momentum equation.)
Calculate v1i = L1 / Δt1i. Then use the expressions you derived in the Analysis section to predict values for the final velocities v1f and v2f .
(v1i(measured): 114.68cm/s )
Predicted Measurements
v1f (predicted): −32.59939086cm/s
v2f (predicted): 82.08060914cm/s
(I got these numbers using the formulas for velocity above.)
Comparison with experiment: Calculate your experimental values of v1f and v2f using Δt1f and Δt2f. Also calculate the percent discrepancies between the predicted and measured values.(to two significant figures, as a positive number.)
v1f (measured): 29.65cm/s [/U]discrepancy 1.096944377% or 1.1%
v2f (measured): 81.19cm/s discrepancy 9.947355346% or 9.9%
Expect discrepancies in the range of 10%.
(The formula I used to calculate the percentage of discrepancy was {percent error = ( T − E /T) × 100%} where T=accepted value and E=experimental value.)
Thank you for taking the time to help me out!
