A 725-kg two-stage rocket is traveling at a speed of $6.60 \times 10^3 \textrm{ m/s}$ away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of $2.80 \times 10^3 \textrm{ m/s}$ relative to each other along the original line of motion.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. This ship is initially traveling away from the Earth and it has a mass of 725 kilograms and a speed of 6.60 times 10 to the 3 meters per second. So it's a two-stage rocket which means it will split up at some point due to some explosion which will propel this part forward even faster and this part will slow down and our job is to figure out what will these speeds be knowing that the mass of each one is half the mass of the total the rocket started with. And we also know that the relative velocity of these two pieces is this much so we know that v 2 prime minus v 1 prime is 2.80 times 10 to the 3 meters per second, that's the speed that they are moving with respect to each other and we have to figure out what each of these speeds are with respect to the Earth. So conservation of momentum says that the total momentum that we start with, m v, is equal to the total momentum that we have after the explosion so that's m 1 v 1 prime plus m 2 v 2 prime. So this is the velocity of the second fragment with respect to the Earth and this is the velocity of the first fragment with respect to the Earth after the explosion. Now we'll substitute for m 1 and m 2 and write m over 2 instead for each of them each of these masses is half the mass that we started with in the beginning and then we can factor out the m over 2 from both these terms and then we can even cancel the m's altogether and so we have that the initial velocity of their combined fragment's as a single rocket equals velocity 1 prime plus velocity 2 prime over 2. And then we can solve for one of the velocities by multiplying both sides by 2 that gets rid of this denominator and then subtract v 2 prime from both sides and then switch the sides around and we get v 1 prime is 2 times v minus v 2 prime. Now we can't do much with that until we look at this formula and rearrange it to solve for v 2 prime and we can say that it's gonna be add v 1 prime to both sides here and we get v 1 plus 2.8 times 10 to the 3 that's v 2 prime and we'll substitute that into v 2 prime here. So we get v 1 prime is 2 times v minus v 1 prime plus 2.80 times 10 to the 3 meters per second. And then I just distributed this negative sign into the brackets here so that makes each term negative. and then you can add v 1 prime to both sides or take it to the left, whichever way you like to say it, and you get 2 times v 1 prime is 2 times v minus this number and divide everything by 2 and you get v 1 prime is v minus 1.40 times 10 to the 3 and the initial velocity of the rocket is 6.60 times 10 to the 3 meters per second minus this 1.40 times 10 to the 3 gives v 1 prime velocity of this fragment of the rocket there after the explosion of 5.20 times 10 to the 3 meters per second. And v 2 prime is gonna be v 1 prime plus 2.80 times 10 to the 3 as we said there based on this relative velocity formula and so it's gonna be 5200 plus 2800 which is 8.00 times 10 to the 3 meters per second; three significant figures in everything here and I should have a zero there I guess. And then what is the energy supplied by the explosion? Well that will equal the change in kinetic energy so the total final kinetic energy minus the total initial kinetic energy. So the total final kinetic energy is the kinetic energy of each fragment so that's one-half m 1 v 1 prime squared plus one-half m 2 v 2 prime squared and each mass is m over 2 and we can factor this m over 4 out from each term and so we have final kinetic energy is m over 4 times v 1 prime squared plus v 2 prime squared; the initial kinetic energy is one-half mv squared— that's the kinetic energy of the rocket before the explosion. So the energy supplied is the difference between this final minus this initial and that's what I have written here and we can factor out an m over 2 from both of these terms and we get m over 2 times v 1 prime squared plus v 2 prime squared over 2 minus v squared, substitute in a bunch of numbers and plug it into your calculator and here's how I arrived at each of these numbers; that's the speed of the fragment 1 after the explosion, fragment 2 after the explosion and then this work here 725 kilograms—mass of the rocket— divided by 2 times 5200 meters per second squared plus 8000 meters per second squared all over 2 minus 6600 meters per second— speed of the rocket before the explosion—squared and then that gives 7.11 times 10 to the 8 joules of energy provided by the explosion.