Calculating Inertia Ratios of Carts on a Low-Friction Track

AI Thread Summary
The discussion revolves around calculating the inertia ratio of two carts, A and B, on a low-friction track after a collision. Participants initially debated the correct ratio, with one suggesting it was 1/2, while others pointed out that cart B has a larger mass due to a smaller change in speed, indicating greater inertia. The confusion stemmed from the expectation to derive an exact ratio from a diagram, which some found problematic. Ultimately, it was clarified that the correct ratio is approximately 2.1, highlighting the inadequacy of the problem statement. The conversation emphasizes the challenges of estimating values from visual data in physics problems.
emily081715
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Homework Statement



The vx(t) curves for two carts, A and B, that collide on a low-friction track are shown below:
Mazur1e.ch4.p10.jpg

What is the ratio of their inertias?

Homework Equations


the ratio i am looking for is mB/mA

The Attempt at a Solution


I believed that the ration was 1/2 however that is not correct and i do not see how there can be another ratio
 
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How did you get the value 1/2?
Which mass is larger?
 
mfb said:
How did you get the value 1/2?
Which mass is larger?
is the cart of mass B not larger because it had a smaller change is speed, meaning it has a larger inertia and greater mass. i got the 1/2 ratio because the mass of cart A is double the mass of cart B
 
emily081715 said:
is the cart of mass B not larger because it had a smaller change is speed, meaning it has a larger inertia and greater mass.
Correct, cart B has a larger mass.
emily081715 said:
because the mass of cart A is double the mass of cart B
This is in contradiction to what you said immediately before that.
 
I meant cart B is double the mass of cart A. But the ratio isn't 2/1
 
emily081715 said:
But the ratio isn't 2/1
Why not?
You can try to estimate the changes a bit better from the diagram, that will give a value very close to 2/1, but usually you are not supposed to measure individual pixels on problems like this.
 
mfb said:
Why not?
You can try to estimate the changes a bit better from the diagram, that will give a value very close to 2/1, but usually you are not supposed to measure individual pixels on problems like this.
i already tried the ratio and it was wrong which is why I'm confused for the question
 
emily081715 said:
i already tried the ratio and it was wrong which is why I'm confused for the question
turns out the answer was 2.1
 
Okay, bad problem statement. Giving something that is so close to 2 and then expecting students to count pixels is a very bad style.
 
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