AAMC8, PS, question 19 (freefall)

[JSU]

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Hey everyone, I just finished AAMC 8 and im having a problem with one of the questions. Question 19 on the PS section asks why someone would use a 10m long incline plane instead of dropping an object from 10m. They answer that it is because it takes longer to get down to the bottom. However, an answer choice is that the final velocity would be less than dropping it from 10m. Shouldnt that be the correct answer? We have a difference in height between the 2 systems which equates to a difference in potential energy and therefore a difference in final velocity ((2gh)^1/2=v). The only was for the two objects to have the same velocity is if the incline plane has the same height as the height that you are dropping the ball from. Though I dont see that implicit in the question.

Thanks!

 

[loveoforganic]

Consider rereading the question. I would think it's phrased to say "drop from 10m straight down or drop from 10m using inclined plane." If not, poorly phrased question.

 

[lorenzomicron]

Is the incline plane 10 m long or does it have a height of 10 m?

 

[JSU]

"When designing his experiment %^&%& could have allowed sphere to drop from a height of 10 m rather than using the 10m inclined plane described in the passage. The main advantage to using the inclined plan is that on the inclined plane..."

Changed the name for bc of copyright issues 😀 .

In the passage, the inclined planes are described as "conducted experiments with spheres on a nearly frictionless incline plane" thats it.

 

[Lokhtar]

Hey everyone, I just finished AAMC 8 and im having a problem with one of the questions. Question 19 on the PS section asks why someone would use a 10m long incline plane instead of dropping an object from 10m. They answer that it is because it takes longer to get down to the bottom. However, an answer choice is that the final velocity would be less than dropping it from 10m. Shouldnt that be the correct answer? We have a difference in height between the 2 systems which equates to a difference in potential energy and therefore a difference in final velocity ((2gh)^1/2=v). The only was for the two objects to have the same velocity is if the incline plane has the same height as the height that you are dropping the ball from. Though I dont see that implicit in the question.

Thanks!

The incline height is indeed the same as the height you are dropping the question from.

It is definitely implicit in the passage because he was measuring distance and time. If velocities were different, it would not actually help him in his experiment. Remember that he is measuring the constant d/t^2. If the inclined plane introduced other variables which affected that constant, the experiment would not show anything. His main purpose is to figure out if there is a constant relationship, and the only way to do that is to measure accurately - and the best way to do that would be to use an incline.

 

[JSU]

The incline height is indeed the same as the height you are dropping the question from.

It is definitely implicit in the passage because he was measuring distance and time. If velocities were different, it would not actually help him in his experiment. Remember that he is measuring the constant d/t^2. If the inclined plane introduced other variables which affected that constant, the experiment would not show anything. His main purpose is to figure out if there is a constant relationship, and the only way to do that is to measure accurately - and the best way to do that would be to use an incline.

True, he was measuring d/t^2 (which is distance of the incline, or height as they are proportional to each other). However, having a smaller final velocity also makes it easier to record the time it take to get to the bottom (easier to observe). Though it does sound like making it take longer to get to the bottom would be a more direct result. Thanks!

 

[Lokhtar]

True, he was measuring d/t^2 (which is distance of the incline, or height as they are proportional to each other). However, having a smaller final velocity also makes it easier to record the time it take to get to the bottom (easier to observe). Though it does sound like making it take longer to get to the bottom would be a more direct result. Thanks!

But if the velocity is different, then the heights are different, in which case you can make no comparative relationships.

 

[JSU]

I don't think were getting each other. Let's use examples. If you have a ball that is 5m above the ground, it will take 1 second to land (so it lands at 10ms^-1) (d:t^2 = 5:1). If you have a ball on a slope 30 degree incline with a HEIGHT of 5m (same as above) and let it go, the acceleration will be gsin(theta) (mgsin(theta)=F=ma, a=gsin(theta)). So the acceleration here is 5ms^-2. Therefore the time it takes the reach the ground is 10m = (1/2)(5ms^-2)(t)^2 --> 4=t^2 so now the d:t^2 ratio is 10:4 or 2.5:1. So it HAS changed (the velocity is still 10). This can't be the experiment proposed because the d:t^2 ratio changes.

This is what I am proposing... Lets drop out ball again from 5m. It take one second to land giving us a d:t^2 of 5:1. Now if we take a 5m plank and incline it at 30 degrees, the acceleration will be 5ms^-2 again. This means that 5m=(1/2)(5ms^-2)t^2, t^2=2 so the ratio of d:t^2 is 5:2 again, not a match.

So the experiment must be saying that the d:t^2 is constant for a certain angle. Not universally. Does that fit with your understanding? Also, if this is the case, then the final velocities will be different for different size planks.

 

[Lokhtar]

Right. I know what you're saying. The question is not comparing straight fall with an incline. It's asking why incline is preferable to straight for measurement. So you aren't looking at the ratio between free fall and incline - you are comparing inclines with the same slope but longer inclined.

It's that if you want to measure time, comparing 10m freefall vs 20m free fall vs 10m incline vs 20m incline, the latter set of measurements are easier because without electronic equipment, you can meaure the times more accurately.

Err, maybe I'm not saying this right - I don't know if that made any sense.

 

[JSU]

Right. I know what you're saying. The question is not comparing straight fall with an incline. It's asking why incline is preferable to straight for measurement. So you aren't looking at the ratio between free fall and incline - you are comparing inclines with the same slope but longer inclined.

It's that if you want to measure time, comparing 10m freefall vs 20m free fall vs 10m incline vs 20m incline, the latter set of measurements are easier because without electronic equipment, you can meaure the times more accurately.

Err, maybe I'm not saying this right - I don't know if that made any sense.

I get it, and that makes the answer that he wants to compare times viable. However the velocities as different as well so it should be a correct answer. Though it isnt the best answer...

 

[loveoforganic]

That's the way the MCAT works. The velocity would be lesser for the inclined plane, but that doesn't make as much of a difference in the ability to measure the time traveled as does lengthening the time it takes to travel the distance.

If you think of measurement error as x seconds, x / time traveled is going to be your percent error. Increase time traveled, decrease measurement error. Decreasing velocity would shrink x and thus shrink measurement error as well, but, intuitively, not to the same degree as increasing time.

 

[randomb]

Whoops

 

[Fort]

Final velocity in both cases would be the same because PE=KE whether the ball rolls down an incline or is dropped from the same height. It will indeed take longer for the ball to reach the bottom of the incline because it's not experiencing the full effects of gravity, but the greater distance traveled down the incline makes up for diminished gravitational effects.

The height in both systems is the same: it's measured from whatever you call your reference height (y = 0) to wherever it's dropped from. Remember it's measured along the vertical axis and not a combination of the two (or three) in a Cartesian coordinate system.