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The question is

An object of mass $m$ is allowed to slide down a frictionless ramp of angle $\theta$, and its speed at the bottom is recorded as v. If this same process was followed on a planet with twice the gravitational acceleration as Earth, what would be its final speed?

My idea is to use conservation of energy to solve this. So,

$K_i + U_i = K_f + U_f$

I'm going to plug in what I think I can (and this might be where I'm going wrong):

$0 + mgh = \frac{mv^2}2+0$

Solving for v:

$\sqrt{2gh}=v$

So if $g$ doubles, then v is multiplied by a factor of $\sqrt 2$. However, the answer is that v doubles as well.

Note that the question here is not what the correct solution is, because I have that. My question is where I went wrong.

EDIT: Since you all say that I'm not doing anything wrong, I guess the question becomes what is the book doing wrong?

Here is the solution that the book gives.

The normal fore will cancel out the perpendicular component of gravity, leaving $mg \sin \theta$ as the net force on the object.

$F_{net} = mg \sin \theta = ma$

$a = g \sin \theta$

This shows that $a$ is proportional to $g$. Then, using $v = v_0 + at$ we can see that the final speed is proportional to a. So if this planet has double the value of $g$, the object will experience double the acceleration, leading to double the final speed.

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    $\begingroup$ As far as I can see your analysis is correct. It even follows by dimensional analysis, without doing any math. $\endgroup$
    – Javier
    Commented Apr 8, 2016 at 15:38
  • $\begingroup$ @Javier I agree that OP is right, but how can one tell it by using just dimensional analysis? $\endgroup$
    – Courage
    Commented Apr 8, 2016 at 15:44
  • $\begingroup$ @TheGhostOfPerdition: The only possible relation between $v$, $g$ and $h$ is $v \propto \sqrt{gh}$. $\endgroup$
    – Javier
    Commented Apr 8, 2016 at 15:48
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    $\begingroup$ The issue is that you are keeping the height constant while the book is keeping the time duration of sliding constant. $\endgroup$
    – Tofi
    Commented Apr 8, 2016 at 16:02
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    $\begingroup$ I would get a refund on that book $\endgroup$
    – Peter R
    Commented Apr 8, 2016 at 16:03

1 Answer 1

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The velocity does not double if the acceleration is doubled. The relevant SUVAT equation is:

$$ v^2 = u^2 + 2as $$

where in this case $u=0$ so we get:

$$ v = \sqrt{2as} $$

A doubling of acceleration means that the velocity would double if the travel time was kept constant. However in this case it's the travel distance that is held constant. The greater acceleration means the object covers the constant distance in less time, so the doubled acceleration acts for a shorter time. That's why we get the square root dependence of velocity on acceleration.

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  • $\begingroup$ they may have meant by "the same process" the same duration of sliding. $\endgroup$
    – Tofi
    Commented Apr 8, 2016 at 15:44
  • $\begingroup$ then it would be double the speed $\endgroup$
    – Tofi
    Commented Apr 8, 2016 at 15:45
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    $\begingroup$ The question talks about a frictionless ramp of angle $\theta$. It doesn't say anything about the length of the ramp being changed. $\endgroup$
    – John Rennie
    Commented Apr 8, 2016 at 15:46
  • $\begingroup$ I've edited the updated question $\endgroup$
    – phil
    Commented Apr 8, 2016 at 16:01

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