You drop a ball down an incline plane...am I reading this incorrectly?

[RSAgator]

So you have a ball. You drop it from a height of 10 m rather than using a 10-m inclined plane. The main advantage to using the inclined plane is that on the inclined plane the:

A) final velocity of a sphere is smaller.

B) final velocity of a sphere is larger.

C) spheres take longer to reach the bottom.

D) spheres take less time to reach the bottom.

the answer is C, which I know to be true. However, how does the solution make sense? How is the path longer on the inclined plane, they're both 10m. The only thing I could think was that the 10 meters was in the horizontal, but that's ******ed. It says "10m incline plane" so how can the path be longer? The question bothers me because:

In the solution it says: the final velocity of a sphere that drops is the same as the final velocity of a sphere that moves down the inclined plane. Meanwhile, the path is longer on the inclined plane than in free fall, leading to a longer time for a sphere to reach the bottom.

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[BerkReviewTeach]

That must be a typo. I think they meant to say, a 10m-high incline plane.

Starting at a height of 10m would mean the length of the plane is longer than 10m. Because the ball starts at a height of 10m in each case (free fall or with the inclined plane), it starts with the same gravitational potential energy and thus achieves the same kinetic energy (assuming the plank is frictionless) at the bottom. It just takes a longer time and greater distance to reach that point.

 

[RSAgator]

When designing his experiment, John could have allowed spheres to drop from a height of 10 m rather than using the 10-m inclined plane described in the book. The main advantage to using the inclined plane is that on the inclined plane the:

That's the 99% exact quote...was from an AAMC, still think it was just a typo? Would let me rest a bit easier =)

 

[iA-MD2013]

When designing his experiment, John could have allowed spheres to drop from a height of 10 m rather than using the 10-m inclined plane described in the book. The main advantage to using the inclined plane is that on the inclined plane the:

That's the 99% exact quote...was from an AAMC, still think it was just a typo? Would let me rest a bit easier =)

that was from an aamc?! well that sucks... i've gotten used to the occasional crappy question in prep books, but i was hoping aamc wouldn't do it too!

 

[HCW212]

i'm thinking Fnet on the ball would be smaller on the incline than a ball that is just dropped (disregarding friction/air resistance). therefore, using kinematic equation v=v0 + at, if Fnet is smaller, that means it's acceleration is smaller.

v0 would be zero, and if the final velocity is equal in both, then the one with the smaller acceleration would have a larger t, or time.

then again, I'm probably looking into this way too much...

 

[tncekm]

if the final velocity is equal in both,

The final velocity can't be equal in both.

A 10m inclined plane will, by definition, have a height of less than 10m. Conservation of energy tells us that the kinetic energy of the two situations will be different because the initial gravitational potential energy was different, so the velocities have to be different considering velocity is proportional to the root of the kinetic energy.

 

[tncekm]

Anyway, the question and answer are correct even if the explanation is wrong..really, really, really wrong .

The ball is traveling 10m any way you look at it, and when the ball travels at an incline over 10m its accelerating at some fractional rate of g over that 10m distance whereas the ball dropped 10m is accelerating at g over that 10m distance.

 

[BloodySurgeon]

Yea it must be a mistype.. i did it mathematically and it doesn't work out that way.

Incline:
height = H < 10m
angle of incline = x < 90 degrees
sin x < 1

t = (2H/g sin x)^1/2


Drop fall:
height = 10m

t = (20/g)^1/2


Ratio of incline to drop fall:

(2H/10 sin x) : (20/10) = 2H/10sinx : 2

H could be 6 or 8 --> 6^2 + 8^2 = 10^2 (letting 10 be the hypotenuse)
and sin x could be 8/10 or 6/10

2(6) / 10(6/10) = 2
2(8) / 10(8/10) = 2


2:2 or 1:1 ratio of time from incline vs. drop off

 

[BloodySurgeon]

Maybe we are considering friction as non-zero and air resistance is

 

[RSAgator]

alright...I think the bigger concern for me was just that I was reading something wrong, wanted to make sure I wasn't missing something. Thanks guys.

 

[wannabedocta]

When designing his experiment, John could have allowed spheres to drop from a height of 10 m rather than using the 10-m inclined plane described in the book. The main advantage to using the inclined plane is that on the inclined plane the:

That's the 99% exact quote...was from an AAMC, still think it was just a typo? Would let me rest a bit easier =)

I just read this question as a sort of experimental design type of question rather than a strict math question, "When designing his experiment", "main advantage"... so I would intuitively choose C. But I agree, this question could have been more clear.

 

[tncekm]

2:2 or 1:1 ratio of time from incline vs. drop off

Maybe we are considering friction as non-zero and air resistance is

a&#8321; = dropped straight downward 10m
a&#8321; = g = 10m/s&#178;

a&#8322;=sin&#952;g; lets take &#952;=30&#176;; allowed to roll down a frictionless plane that is 10m long, and 5m tall.
a&#8322;= 5 m/s&#8322;

x = x&#8320; + v&#8320;t + &#189;at&#178; ; x&#8320;=0, v&#8320;=0 so
x=&#189;at&#178;
t=&#8730;(2x/a)

x&#8321;=x&#8322;

t&#8321; &#8730;(2x/a&#832
-- = -----------
t&#8322; &#8730;(2x/a&#8322

t&#8321;=t&#8322;&#8730;(a&#8322;/a&#832; a&#8321;=2a&#8322;
t&#8321;=t&#8322;&#8730;(1/2)

t&#8321;=(1/4)t&#8322;

The first case would be 4x faster than the second case.

 

[BloodySurgeon]

a&#8321; = dropped straight downward 10m
a&#8321; = g = 10m/s²

a&#8322;=sin&#952;g; lets take &#952;=30°; allowed to roll down a frictionless plane that is 10m long, and 5m tall.
a&#8322;= 5 m/s&#8322;

x = x&#8320; + v&#8320;t + ½at² ; x&#8320;=0, v&#8320;=0 so
x=½at²
t=&#8730;(2/ax)

x&#8321;=x&#8322;

t&#8321; &#8730;(2/a&#8321;x)
-- = -----------
t&#8322; &#8730;(2/a&#8322;x)

t&#8321;=t&#8322;&#8730;(a&#8322;/a&#832; a&#8321;=2a&#8322;
t&#8321;=t&#8322;&#8730;(1/2)

t&#8321;=(1/4)t&#8322;

The first case would be 4x faster than the second case.

Put down a piece of plywood and put a ball on top of it. Lift it at about 5° incline to maybe a couple inches in height and see if it takes anywhere near the amount of time it would take to drop those few inches.

t = &#8730;2h/asinx

where is h?

 

[tncekm]

You don't need h. I calculated the acceleration parallel to the length of the board in the second case (in the direction of travel) and for the first case acceleration and travel were both directly downward.

But, in the case outlined above h1 = 10 and h2 = 5.

Both cases the ball travels 10m, however, in one case the acceleration is twice the other.

But yeah, thx for pointing out I messed up the the formula a bit. However, that term cancelled so it doesn't change the result.

 

[tncekm]

http://www.ngsir.netfirms.com/englishhtm/Incline.htm

Play with this thing to confirm it yourself. Just make sure you set the length of the plane to be the same length on each trial.

 

[Kaustikos]

I just read this question as a sort of experimental design type of question rather than a strict math question, "When designing his experiment", "main advantage"... so I would intuitively choose C. But I agree, this question could have been more clear.

I think that's the point. You would have a slower acceleration and therefore have the ball reach 10m in a longer time-span as opposed to dropping it straight down.

 

[Kaustikos]

The final velocity can't be equal in both.

- edit -
You're right. But it makes no sense that they would say "C" is an advantage over "B". B says higher velocity...isn't that an advantage? And advantage according to WHO? If Wile E Coyote is trying to figure out how best to capture Roadrunner, then i can see the advantage of him sitting on top of a cliff and jumping down while roadrunner goes down a long hill...but wtf.

BAD QUESTION

 

[tncekm]

It looks like it may have come from a passage. And, if that's the case, I've seen all I want to see because I don't wanna have anything ruined for me!