#73, physics AAMC self-assessment

[riseNshine]

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Hi everyone, for this question, it asks about a head-on inelastic collision and then how much heat and deformation energy is produced.

It was my understanding that in an inelastic collision, kinetic energy isn't conserved because SOME of it is lost to heat/deformation/light/sound, etc so the KE final is less than the KE initial. But, in their calculations, they assume that ALL of the kinetic energy is lost to heat/deformation energy.

How is this assumption valid?

Thanks.

 

[riseNshine]

Also, for #81, can someone explain why there's an exponential rise in voltage and then an exponential drop? I thought the equation is Q/C=V.

 

[LoLCareerGoals]

They did say it was inelastic didn't they? Did they say it was partially inelastic? I mean what else could you possibly assume if no further information was given?
Also can you like post the question fully for 81? Not everyone has all the books.

 

[dontbeanegaton]

I can't remember the specifics of this question. If the objects collide and stop, then you can assume that the final kinetic energy is 0. It must have all gone toward those other factors then.

 

[MD013]

I was working on this exact problem before I saw your post OP. I think that in order for two objects moving in the opposite direction to have an inelastic collision, their momentum will be 0 afterwards because both objects will stop moving. If they stop moving, then ALL of their KE is lost to heat. Stupid me, almost forgot momentum is a vector.

 

[LianaStan]

Also, for #81, can someone explain why there's an exponential rise in voltage and then an exponential drop? I thought the equation is Q/C=V.

I don't have the problem in front of me but I am assuming you mean to ask why a charging capacitor has an fractional exponential increase in voltage over time as opposed to linear.
The equation for charge on a charging capacitor plate in relation to time is:

Q=CV(1-e^(-t/(RC)))

As you can see, as time increases, the charge on the plate plateaus. This makes sense as you cannot infinitely add more and more charges onto the plates.

I do not think the MCAT expects you to know this equation. Rather, the conceptual way to think about the problem is this: As a battery with voltage V (by the way, your post implies that you think the 'exponential rise in voltage' is the V in Q/C=V; it is not. That V is the battery voltage) charges a plate, the charges will flow to the ends of the plate faster in the beginning and gradually slow down because of charge accumulation at the plate ends. The accumulation of charge slows down the adding of further charge. This is why the graph's curve looks the way it does.

Use the same logic to find the graph for a discharging capacitor.