Conservation of Momentum Lab (Carts Collide and Stick Together)

[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?

 

[kuruman]

Can you explain exactly what you plotted? How are the x and y values in your plot related to what you measured?

 

[kenzieleigh]

Can you explain exactly what you plotted? How are the x and y values in your plot related to what you measured?

We got the time value from before and after the collision with the timers. Found the length of the antennae on the first cart and used v=d/t to find the initial and final velocity. Then plotted the velocities with initial on the x-axis and final on the y.

 

[kuruman]

Then plotted the velocities with initial on the x-axis and final on the y.

And why should that give you a straight line when plotted? What is the theoretical equation that you used?

 

[kenzieleigh]

And why should that give you a straight line when plotted? What is the theoretical equation that you used?

Because the slope is equal to m1/(m1+m2) which were kept constant in the lab. Therefore there should be a straight line.

 

[kuruman]

OK. Now how did you extract the unknown mass from the slope?

 

[kenzieleigh]

OK. Now how did you extract the unknown mass from the slope?

We used that equation. We got the slope through rise/run and since we knew the mass of the first cart we could use the equation above (slope=m1/(m1+m2) ) to find the m2, but it wasn't the correct mass. It was about 90g when it actually has a mass of 180g (as calculated by a scale)

 

[kuruman]

OK. You probably found the slope by using some kind of canned linear regression and set the intercept equal to zero because that makes sense theoretically. What if you redid the regression and let it find an intercept. What kind of slope and value for the mass do you get then?

 

[haruspex]

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?

Your problem is the opposite, too little speed was lost.
The plot looks suspiciously like two separate plots. The first three datapoints have a slope closer to 0.7, the rest being more like 0.9.
That might not sound much, but do the error analysis: 0.7 gives the unknown mass as 3/7 of the known mass, so quite close to the right value, while 0.9 gives 1/9. This suggests to me some experimental error crept into the set-up.
(Should such an exercise arise again, I recommend finishing with a run without the second cart, to verify the photogates still agree.)

Looking at the diagram, is it possible the carts were making contact before the first cart had cleared the first photogate? Maybe the antenna got bent back a little.