Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. This ship is at first traveling far from the Earth and also it has a mass of 725 kilograms and also a rate of 6.60 time 10 to the 3 meters per second. For this reason it"s a two-stage rocket which way it will break-up up at some point due to part explosion which will propel this component forward even faster and also this component will slow down and also our project is to number out what will certainly these speeds be knowing that the massive of every one is fifty percent the massive of the full the rocket began with. And also we likewise know the the family member velocity that these two pieces is this lot so we recognize that v 2 prime minus v 1 element is 2.80 time 10 to the 3 meters every second, that"s the rate that lock are moving with respect to each other and also we have to figure out what each of this speeds space with respect come the Earth. So conservation of momentum says that the total momentum that we begin with, m v, is equal to the full momentum that we have actually after the explosion therefore that"s m 1 v 1 element plus m 2 v 2 prime. For this reason this is the velocity of the second fragment through respect come the Earth and this is the velocity the the first fragment v respect to the earth after the explosion. Currently we"ll substitute because that m 1 and also m 2 and also write m end 2 instead for each of them each of these masses is half the mass that we began with in the beginning and then we can variable out the m over 2 native both this terms and also then we can even cancel the m"s altogether and so we have actually that the early velocity the their an unified fragment"s together a single rocket equals velocity 1 prime plus velocity 2 prime over 2. And also then we have the right to solve for one of the velocities by multiply both political parties by 2 the gets rid of this denominator and also then subtract v 2 prime from both sides and also then move the political parties around and we obtain v 1 element is 2 times v minus v 2 prime. Now we can"t do lot with that till we look at this formula and rearrange that to fix for v 2 prime and we deserve to say the it"s gonna be include v 1 element to both sides here and we get v 1 to add 2.8 times 10 come the 3 that"s v 2 prime and we"ll substitute that right into v 2 prime here. Therefore we obtain v 1 element is 2 time v minus v 1 element plus 2.80 time 10 to the 3 meters per second. And also then ns just dispersed this an unfavorable sign right into the brackets below so that renders each ax negative. And then girlfriend can add v 1 element to both sides or take it come the left, whichever way you like to say it, and also you obtain 2 times v 1 element is 2 time v minus this number and also divide whatever by 2 and also you gain v 1 prime is v minus 1.40 times 10 to the 3 and also the early stage velocity of the rocket is 6.60 times 10 to the 3 meter per 2nd minus this 1.40 time 10 to the 3 provides v 1 prime velocity of this fragment that the rocket there after the explosion of 5.20 times 10 to the 3 meters every second. And v 2 prime is gonna be v 1 element plus 2.80 times 10 to the 3 as we claimed there based upon this family member velocity formula and also so it"s gonna it is in 5200 to add 2800 which is 8.00 time 10 come the 3 meters every second; three significant figures in everything here and I should have actually a zero there ns guess. And also then what is the power supplied by the explosion? fine that will certainly equal the readjust in kinetic power so the total final kinetic energy minus the full initial kinetic energy. So the total final kinetic power is the kinetic power of every fragment therefore that"s one-half m 1 v 1 element squared add to one-half m 2 v 2 element squared and each massive is m end 2 and we can element this m over 4 the end from each term and so we have final kinetic energy is m end 4 time v 1 element squared add to v 2 element squared; the early kinetic energy is one-half mv squared— that"s the kinetic power of the rocket before the explosion. For this reason the power supplied is the difference in between this final minus this initial and also that"s what I have actually written here and also we can element out an m end 2 native both of this terms and we acquire m over 2 time v 1 prime squared plus v 2 element squared over 2 minus v squared, instead of in a bunch the numbers and plug it into your calculator and here"s just how I landed on each of these numbers; that"s the rate of the fragment 1 ~ the explosion, fragment 2 ~ the explosion and then this work here 725 kilograms—mass the the rocket— separated by 2 times 5200 meters per second squared to add 8000 meter per second squared all over 2 minus 6600 meters per second— speed of the rocket before the explosion—squared and also then that offers 7.11 time 10 to the 8 joules the energy provided by the explosion.


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