Quite new to all this - simple question.

"A ball starting from rest rolls down a hill and reaches 10m/s at the bottom. If the same ball starts rolling down the hill from an initial velocity of 5m/s, will it's velocity at the bottom be: (A) Less than 15m/s; (B) Equal to 15m/s; (C) Greater than 15m/s."

The answer I have marked in the book this question comes from says (A). However, I am unable to reason to that conclusion from the way the question is stated. Here is my reasoning, please point out where it is wrong. If the same ball (the same mass) traveses the same distance down this same hill (the same change in height), the gravitational potential energy must be the same at the point on top of the hill, regardless of its initial velocity. Therefore, on rolling down the hill, by the time it reaches the bottom, this potential energy would have transferred to kinetic energy - indicating (B)?

Thanks for your time.

Tim.

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## closed as off-topic by tpg2114, Qmechanic♦Nov 3 '13 at 2:29

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Dear nulliusinverba, this server usually tries to avoid solving homework problems but let me give you a hint where your reasoning goes awry. You have assumed that the kinetic energy is proportional to $v$, haven't you? That's why you could think that by adding a fixed amount of kinetic energy, you always increase the velocity $v$ by the same amount, right?

Do you really think that the kinetic energy is proportional to $v$? And if it happens not to be, don't you think that the velocity change obtained by the same increase of the kinetic energy could depend on the initial velocity? In fact, shouldn't you be able to calculate the final velocity exactly with these hints?

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Hi Lubos - thanks for that, you have pointed me in the correct direction and I now understand. In fact, this is not a homework question (it was marked as such, presumably by a moderator) and I am not a formal student in any course/institution. It was a genuine conceptual problem I wanted help with after spending plenty of time thinking it through. I reiterate: I am new to this! Hence I could not see the problem in my reasoning. Thanks again for your help. -Tim. – nulliusinverba May 9 '11 at 9:51
Great if it is not a homework problem! ;-) If I were sure it's not a homework problem, I could also write that $0^2+10^2 = 100 = 10^2$ while $5^2+10^2 = 125 = 11.2^2$ and maybe even hint that these identities could be relevant for the accounting of the kinetic energy haha. – Luboš Motl May 9 '11 at 10:10
Yeah, I did calculate 11.18m/s after your hint. Not being a formal student, I love having generous strangers on the internet to help me when I'm stuck. Thanks again, mate :) – nulliusinverba May 9 '11 at 10:16
It was a pleasure. – Luboš Motl May 9 '11 at 11:10

You reasoned correctly that in both cases the ball converts the same amount of potential energy into kinetic energy (at least it appears that you came to this conclusion).

The problem you are having is that you need to realize that kinetic energy is proportional to v^2. So, it takes NINE times as much energy to get from 0 m/s to 15 m/s as it does to get from 0 to 5 m/s. If the potential energy available is just enough to go from 0 to 10 m/s, it is then obvious that it is NOT going to be enough to go from 5 to 15 m/s. Answer A.

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