- #1

kenzieleigh

## Homework Statement

We did a lab where we had one cart with an unknown mass stationary on a friction-less surface. Another cart with a mass of 378.9g, was pushed down the track (which moves at a constant speed due to the friction-free surface) and collides with the second cart - so it is a hit and stick collision. We used photogate timers to determine time values immediately before and after which allowed us to find initial and final velocities. We graphed these values, found the best fit line and the slope from that. Then stacked the equations y=mx + b with m1v1 + m2v2 = (m1 + m2)vf to try and find the second mass. So we ended up having a slope of 0.8, thus we had the equation: 0.8 = 378.9 / (378.9 + m2). We calculated the second cart's mass to be 94.73 g, but after weighing it our mass is almost half of the actual mass. I attached a picture of our data and a diagram of the lab.

## Homework Equations

- y = mx + b

- slope = rise/run

- hit and stick collision

## The Attempt at a Solution

We now have to explain why we got the results we did. Momentum is clearly not conserved or we would have been able to determine the second mass accurately. The question is why not? Could it be air drag? Or is there some other force acting?

#### Attachments

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