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I'm doing my homework for a Physics class and trying to get ready for a quiz. I encountered this question:

Two ramps of equal length are situated such that ramp #1 has a slope (with respect to the horizontal) of 30°, and ramp #2 has a slope of 60°. Neglecting friction, roughly by what factor is the time it takes a ball to roll down ramp #1 larger than the time it takes a ball to roll down ramp #2?

I did this: $$ \frac{\sin60}{\sin30} $$

That came out to be 1.74, but the correct answer is 1.3. How?

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I think you are implicitly comparing equal times rather than equal lengths ... – Mark Bennet Sep 18 '11 at 14:27
What you have calculated is the ratio of the distance to fall. Unfortunately, this is largely irrelevant, because when you roll down the ramp you aren't in free fall. What you need to calculate is what part of the force due to gravity is directed down the ramp (versus directly into the ramp). From this, you can calculate the formula for distance traveled, and then time to reach the bottom. – Aaron Sep 18 '11 at 14:44
The acceleration of gravity along the ramps is as $\sin 60\deg$ to $\sin 30\deg$. However, since the accelerations have dimension length per time squared, the relative difference in times must be only the square root of this factor. – Henning Makholm Sep 18 '11 at 14:44
up vote 1 down vote accepted


Two ramps of equal length ...

Not heights, but lengths. Your solution would be correct if the question was about equal heights.

The component of the gravity force in the direction of sliding is proportional to the sinus of the slope angle. Hence - the acceleration is also proportional to the sinus of the slope angle.

Now, given the constantly accelerated motion, the time it takes to pass a given distance is reverse proportional to the square root of the acceleration. That is:

X = a * t^2/2. Hence T = (2X/a) ^ (1/2).

T = (2X/g / sin(α) ) ^ (1/2).

T1 : T2 = [ sin(α2) / sin(α1) ] ^ (1/2)

If the question was about equal heights then balls would pass different distances (in the sliding direction). The distance would be divided by another sin(α) factor. Hence the answer would be

T1 : T2 = sin(α2) / sin(α1)

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Makes sense. Thanks. – Asdf Sep 18 '11 at 17:24

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