Transcript because that this Giancoli solution

This is Giancoli Answers through Mr. Dychko. This ship is originally traveling far from the Earth and it has actually a mass of 725 kilograms and a rate of 6.60 time 10 come the 3 meters per second. Therefore it"s a two-stage rocket which way it will break-up up at some point due to part explosion which will certainly propel this part forward also faster and also this part will sluggish down and also our job is to number out what will these speeds be understanding that the massive of each one is half the massive of the complete the rocket started with. And we likewise know that the relative velocity that these two pieces is this lot so we recognize that v 2 prime minus v 1 element is 2.80 times 10 come the 3 meters per second, that"s the rate that lock are relocating with respect to each other and also we have to figure out what each of this speeds room with respect to the Earth. Therefore conservation of momentum claims that the total momentum that we start with, m v, is equal to the complete momentum that we have actually after the explosion so that"s m 1 v 1 element plus m 2 v 2 prime. For this reason this is the velocity of the second fragment v respect come the Earth and also this is the velocity that the an initial fragment through respect come the earth after the explosion. Currently we"ll substitute for m 1 and m 2 and write m end 2 instead for every of them every of this masses is fifty percent the mass that we started with in the beginning and then us can aspect out the m end 2 from both this terms and also then we can also cancel the m"s altogether and also so we have actually that the early velocity of their merged fragment"s as a solitary rocket amounts to velocity 1 prime plus velocity 2 prime end 2. And also then we can solve for one of the velocities by multiplying both political parties by 2 that gets rid of this denominator and also then subtract v 2 prime from both sides and also then switch the sides around and also we acquire v 1 element is 2 time v minus v 2 prime. Currently we can"t do lot with that till we look in ~ this formula and also rearrange the to deal with for v 2 prime and also we deserve to say the it"s gonna be include v 1 element to both political parties here and also we gain v 1 add to 2.8 time 10 to the 3 that"s v 2 prime and also we"ll substitute that right into v 2 element here. Therefore we gain v 1 prime is 2 time v minus v 1 element plus 2.80 time 10 come the 3 meters per second. And then ns just dispersed this an unfavorable sign into the brackets right here so that provides each term negative. And then friend can include v 1 element to both sides or take it it to the left, whichever method you prefer to speak it, and you get 2 times v 1 prime is 2 times v minus this number and also divide whatever by 2 and also you gain v 1 prime is v minus 1.40 time 10 come the 3 and the early stage velocity of the rocket is 6.60 times 10 come the 3 meters per second minus this 1.40 times 10 come the 3 provides v 1 prime velocity the this fragment the the rocket over there after the explode of 5.20 time 10 to the 3 meters every second. And v 2 element is gonna be v 1 element plus 2.80 times 10 come the 3 as we said there based on this family member velocity formula and so it"s gonna be 5200 plus 2800 which is 8.00 time 10 come the 3 meters every second; three far-ranging figures in everything here and also I should have actually a zero there ns guess. And also then what is the energy supplied by the explosion? well that will certainly equal the adjust in kinetic power so the complete final kinetic power minus the complete initial kinetic energy. Therefore the total final kinetic power is the kinetic power of every fragment therefore that"s one-half m 1 v 1 element squared plus one-half m 2 v 2 prime squared and also each fixed is m over 2 and we can factor this m end 4 out from every term and so we have last kinetic power is m over 4 time v 1 prime squared plus v 2 element squared; the early kinetic power is one-half mv squared— that"s the kinetic power of the rocket before the explosion. So the energy supplied is the difference in between this final minus this initial and that"s what I have written here and also we can variable out an m over 2 native both of these terms and also we obtain m over 2 time v 1 prime squared to add v 2 element squared end 2 minus v squared, instead of in a bunch the numbers and plug it right into your calculator and also here"s how I arrived on each of these numbers; that"s the rate of the fragment 1 after the explosion, fragment 2 ~ the explosion and also then this work below 725 kilograms—mass the the rocket— split by 2 times 5200 meter per 2nd squared add to 8000 meter per 2nd squared anywhere 2 minus 6600 meters every second— rate of the rocket before the explosion—squared and also then that offers 7.11 time 10 to the 8 joules that energy provided by the explosion.


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