Hey guys,
One thing I found really challenging on my first practice test was the expectation of doing alot of mental math. I was completely boggled when they expected me to do calculations involving Joules >eV, or for instance, Problem 22, ending up having to take the square root of: v = sqrt(6480*.4*g). How would i ever do that? I mean I guess 6400000 is like 8000, and root(g) is like root(10), but I have no idea what sqrt(.4) would be, and my final answer would look like: v = (8000)(sqrt(10))(sqrt(.4)), how do I get 3.6km/s out of that?!
Any tips would be appreciated.
Ari
I realize there are other ways to do this problem, but the bottom line I'm wondering is  is it up to you to come up with the best method in under 2 minutes so that you don't have to use a calculator, or are there some general calculator guidelines I should remember?
92 Test, Problem 22
 midwestphysics
 Posts: 444
 Joined: Thu Dec 16, 2010 12:37 am
Re: 92 Test, Problem 22
I don't remember that exact question personally, but generally you're able to range things. So, pick roots that you know or can guess pretty closely that are close to the one you're being asked. For instance, sqrt(10) is just a bit larger than pi, or close to sqrt(9) which is 3. The sqrt(.4) is less than sqrt(.5) and more than the sqrt of (.25) which is something you should be familiar with, so between .7 and .5. So, while I think the math you posted is off, if the answer is 3.6km/s, this is an effective method that helps with time and sanity. It should be close enough to one of the answers, sometimes so much so that you have no doubt.

 Posts: 381
 Joined: Mon May 24, 2010 11:34 pm
Re: 92 Test, Problem 22
6480 * 0.4 * g = 6500 * 0.4 * 10 = 6500 * 4 = 13000 * 2 = 26000AriAstronomer wrote:Problem 22, ending up having to take the square root of: v = sqrt(6480*.4*g). How would i ever do that?
sqrt(26000) = sqrt(260) * sqrt(100) = sqrt(256) * 10 = 16 * 10 = 160.
Actual answer is closer to 159.
I grew up before calculators. Here's how I do this sort of thing.
(1) Get rid of excess accuracy by approximation
(2) Strip out powers of 10
(3) For square roots, get the number to have 2 or 3 decimal places, then approximate using your knowledge
of perfect squares, i.e.:
sqrt( 9) = 3
sqrt( 16) = 4
sqrt( 25) = 5
sqrt( 36) = 6
sqrt( 49) = 7
sqrt( 64) = 8
sqrt( 81) = 9
sqrt(100) = 10
sqrt(121) = 11
sqrt(144) = 12
sqrt(169) = 13
sqrt(196) = 14
sqrt(225) = 15
sqrt(256) = 16
sqrt(289) = 17
sqrt(324) = 18
sqrt(361) = 19
sqrt(400) = 20
sqrt(576) = 24
sqrt(900) = 30

 Posts: 1203
 Joined: Sat Nov 07, 2009 11:44 am
Re: 92 Test, Problem 22
When I was in middle school, I had to memorize the squares up to 30^2, and cubes up to 12^3. I also had to memorize a bunch of division and multiplication tricks, which turned out to be based on change of basesnot that they told me that in middle school.
I was a teacher (high school) for the last 2 years. My tests were no calculator. If a kid asked me what the square root of 26000 was, I handed them a mechanics almanac (which none of them knew how to use). Knowing how to do your mental math isn't that hard once you put a little groundwork into it.
I was a teacher (high school) for the last 2 years. My tests were no calculator. If a kid asked me what the square root of 26000 was, I handed them a mechanics almanac (which none of them knew how to use). Knowing how to do your mental math isn't that hard once you put a little groundwork into it.

 Posts: 76
 Joined: Thu May 12, 2011 4:53 pm
Re: 92 Test, Problem 22
For sure. I've been practicing over the last 2 days and already it's improving. I was just completely thrown off for this test by the expectation of having to do mental math. Lucky there are practice tests. Thanks for the advice guys.