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Question goes: "An anvil hanging vertically from a long rope in a barn is pulled to the side and raised like a pendulum 1.6 m above its equilibrium position. It then swings to its lowermost point where the rope is cut by a sharp blade. The anvil then has a horizontal velocity with which it sails across the barn and hits the floor, 10.0 m below. How far horizontally along the floor will the anvil land?"

Now I've been working a while on it and what i got was $PE \ at \ side = 1.6\times 10 (gravitational\ constant) \times m (mass\ of\ anvil).$

$KE\ at\ lowermost\ point = \frac{1}{2} m v^2 = PE$

$16m = \frac{1}{2} m v^2$

$16 = \frac{1}{2} v^2$

$32 = v^2$

$v = \sqrt{32}$

Now to the free-fall problem,

$10 = \frac{1}{2} g t^2$

$20 = g t^2$

$2 = t ^ 2$

$t = \sqrt{2}$

So the distance traveled horizontally = $\sqrt{2} \times \sqrt{32} = 1.414$ meters.

I checked the model-answers and the answer was 8 meters, would someone please explain this? (The model-answer might be wrong though)

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sqrt(2) * sqrt(32) = 1.414 meters ????? Could you check this multiplication? – Rafael Jan 28 '11 at 17:03
Wow, it seems like my calculator was going berserk when i first calculated it, the answer is 8, seems i was right from the beginning. Thank you for your comment. – SimpleButPerfect Jan 28 '11 at 17:05
I know how to make the identity true (including units) by changing one character: sqrt(2) * sqrt($m^2$) = 1.414 meters. Moreover, the replacement of "3" by "m" may be achieved by rotating "3" by ninety degrees. – Luboš Motl Jan 28 '11 at 17:05
up vote 1 down vote accepted

For any nonzero gravitational acceleration we know: (1) Horizontal kinetic energy at impact = .16 time vertical KE at impact 1.6/10 (2) Thus horizontal velocity at impact is .4 times vertical velocity at impact. (3) The time averaged vertical velocity is half of the end velocity. So the average horizontal vel is .8 times the average vertical velocity, it should travel .8 times as far horizontally as vertically.

Also $\sqrt{2}\times\sqrt{32} =\sqrt{64}=8$ , so you needn't use your calculator. It is a good idea to have a reasonable guess as to what the result is, before trusting in calculator results (its so easy to type it in wrong).

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I shouldn't of had that blind trust in my calculator. Thanks for the tip and the simple analysis :). Marking this as an accepted answer. – SimpleButPerfect Jan 28 '11 at 17:25

Just a quick comment:

It is always best to give every quantity a symbol and perform all algebraic manipulations on the symbols; you then substitute in numerical factors at the very end.

In this particular case, you can call $h=1.6\textrm{ m}$ and $H=10\textrm{ m}$, which makes your calculation look like: $$mgh=\frac{1}{2}mv^2\textrm{ so } v=\sqrt{2gh},$$ $$H=\frac{1}{2}gt^2\textrm{ so } t=\sqrt{2H/g},$$ $$\textrm{and therefore }\Delta x=vt=\sqrt{4Hh}=\sqrt{4\times 10\times 1.6\textrm{ m}^2}=8\textrm{ m}.$$ Having the symbols written down makes it easier to spot algebraic mistakes and allows you to perform dimensional analysis at each step to check that at least what you're doing makes sense. Further, you only have one go at the numbers so you need only be careful once ;).

Further, having the symbols written down helps you realize other stuff (which may be useful later on in an exam); for example, the final answer here is independent of $g$! This follows from dimensional analysis: you had a purely geometrical picture, and are giving a purely geometrical answer. If you slow down time, the answer is the same, and $g$ is the only quantity in your initial data that has dimensions of time. This is completely obscured in your initial calculation.

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