# Toy Popper Lab – Update #1

Posted January 5, 2011

on:We finished the lab today. I gave the kids two days to do it. Most of them figured out the initial velocity by the end of the first day. The start of the second day, I put two hints on the board. For question 2, I put up t=d/Vagv. For question 3, I told them they needed to calculate the acceleration of the popper.

I decided to be only somewhat helpful. At the start of day 2, I told them the initial velocity should be in the range of 5 m/s. I told them I would not answer questions about their numbers if the formulas were not there and units were not shown. I generally only told them they were either on the right track or wrong, nothing more. Most of them had a tough time making the leap to the distance in part 2 was how far the inverted popper moved from rest to the calculated initial velocity. Once they got that, they were well on their way to solving the problem.

I did an interesting experiment while they worked. I set up a LabQuest to sample at 1 ms intervals. I build a tiny tray from cardboard and string and attached it to the force sensor. I set the meter to trigger at a force greater than 2.5 N, zeroed the sensor, and let it rip. It showed a nice impulse function that took 23 ms and a peak force of close to 7 N.

I could use some help with my interpretation of the graph. I believe the integral of the Force v. Time curve gives me the impulse (the LabQuest gave me a value of 47 N*ms). If I divide that value by the mass of the popper (9.1 g), I get a delta v of 5.16 m/s. This is in agreement with the numbers the kids got in the experiment.

Now if I divide the delta v by the time, I should have the acceleration. The LabQuest samples every millisecond and there are 23 points, so I think the time is either 22 ms or 23 ms. The acceleration works out to be 235 m/s^2. Doing this, I only get a force of 2.1 N, but the graph shows close to 7 N. The students calculated forces in the 6-7 N range. I think the discrepancy has to do with using the integral (which should be more accurate) and getting a peak force compared to an average force. Can someone either confirm this or correct it for me please?

### 5 Responses to "Toy Popper Lab – Update #1"

I love your site! I’m trying this one out with my classes tomorrow. Like you said, the poppers were a little difficult to find. I tried the local Walgreens and dollar store and came up empty handed but they had them at the toy store (Stellabella Toys). I second Jake’s average vs peak force explanation.

I think Jake’s comment above got it right. Dividing the change in velocity by the time you are calculating an average acceleration for the time interval. Which then gives you an average force.

When impulse is calculated as force times change in time, it really is an average force that is being used.

I’m thinking about doing a “popper” lab next year and I stumbled onto your site. Looks like a lot of good stuff.

One additional comment about your analysis here…it is the NET impulse that equals the change in momentum. I think the impulse shown in the graph is the impulse due to upward force of the surface on the popper and does not take the force of gravity acting on the popper into account. But, since the weight of the popper is so much smaller than the average launching force it doesn’t make much difference in this case. From a work energy perspective I think you can neglect the force of gravity if h>>d (where d is the 1 cm or so to compress the stopper).

1 | Jake

January 6, 2011 at 6:25 am

Your calculations seem good. The discrepancy between the forces is due to the fact that you’re comparing the maximum force and the average force. When you look at your graph you see that the force is not 7 N the entire time. Actually, it’s less than 7 N everywhere except the peak.