From owner-biophysics@net.bio.net Thu Mar 12 22:00:00 1998 Path: biosci!newshost.lanl.gov!awabi.library.ucla.edu!207.97.14.174!europa.clark.net!204.59.152.222!news-peer.gip.net!news.gsl.net!gip.net!news-peer.sprintlink.net!news.sprintlink.net!Sprint!newsfeed.nacamar.de!fu-berlin.de!jussieu.fr!uvsq.fr!not-for-mail From: Jerome Villanova Newsgroups: bionet.biophysics Subject: Re: Calc. Velocity of Falling Cows Date: Fri, 13 Mar 1998 11:26:50 +0100 Organization: Universite de Versailles/St Quentin en Yvelines - France Lines: 81 Message-ID: <350909E9.472BFE96@robot.uvsq.fr> References: <3507972c.7732046@news1.abac.com> NNTP-Posting-Host: mercure.robot.uvsq.fr Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-Mailer: Mozilla 4.04 [en] (X11; I; SunOS 4.1.4 sun4m) Dave Irontail wrote: > Hiya! Hope you can help me. Our 9th grade teacher told us a true > story about cows that were stolen by a military plane crew, who, at an > altitude of 30,000 feet decided to allow the cows to run out and > plunge to the ocean below. One of these cows hit and sunk a fishing > boat. I got to wondering how to figure the power of the inmact. I > figure the cow weighs about 1500 lbs. > Thank you! > > DMIrontail > 14EEE@Bigfoot.com Do you allow me to use the metric system ? It was just a rethorical question anyway. :-) So, there are roughly three feet in a meter, so the starting altitude of the free-falling cows was about 10 000 m. A kg is 2.2 lbs therefore your basic cow weighs 700 kg. We have to make a few suppositions : -let's say the plane was flying horizontaly : thefore the initial falling speed of the cow will be 0 m/s. -let's say the cows have never even heard of sky-surfing and that they do not brake their fall by taking advantage of air resistance. What forces are exerted on the cow ? - only gravity : -g.Z where g=9.81 m/s^2 and Z the vertical axis originated on the ship and following the radius of the earth. So the force equilibrium can be writen in the form : m.(gamma) = -m.g where m is the mass of the cow, gamme it's acceleration. So gamma = -g. The speed of the cow is therefore speed = -gt + k where t is the time in seconds, and k a constant. At t=0, the vertical speed of the cow is 0 so k=0. speed = -gt. The position of the cow is pos = -(1/2).g.t^2 + l where l is a constant. at t = 0, the vertical position of the cow is 10 000 m so l =10 000. pos = -(1/2).g.t^2 + 10 000. The potential energy of the cow is Ep = m.g.pos = (700*9.81*10 000) = 68MJ (roughly) the cow touches the boat when p = 0. this means when (1/2).g.t^2 = 10 000. So, t(contact) = square_root (2* 10 000/ 9.81). t(contact)= 45 s (roughly). The cow as fallen for 45 seconds before hitting the boat. Its speed was at the moment : speed(contact) = -gt(contact). The magnitude of the speed at the instant of contact is therefore 440 m/s (1584 km/h) or 1300 ft/s. The quantity of movement of the cow is p = m.speed = 308 000 kg.m/s The kinetical energy of the cow at contact time is Ec = (1/2).m.v^2. Ec = 68 000 000 J = 68 MJ (roughly) = 16 000 kcal note that Ec at the end of the fall is equal to Ep at the start of it. Let's suppose the cow stops in one second after having hit the boat. The power generated by the shock is 68 MJ / 1 s = 68 MW. That's equivalent to the power of 680 000 100 watts light bulbs. It's a pretty nice flash. That cow could have sunk the Titanic. Of course, the best would have been crushing L DiCaprio and C Winsley. -- ------------------------------------ Jerome X Villanova E.mail: villa@robot.uvsq.fr ------------------------------------- .