An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 5500 J is released in the explosion, how much kinetic energy does each piece acquire?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. So an object is initially at rest and then there's some internal explosion that breaks it up into two pieces and they zip off with different velocities because they have different masses; we are told that the mass of one of them is one and a half times the mass of the other so the mass of block 2 is gonna be 1.5 times m 1. Now from conservation of momentum, the momentum that one block has in one direction is gonna equal the magnitude of the momentum of the other block in the other direction since they are going in opposite directions so we are gonna have a negative sign between them. So we have m 1v 1 prime equals negative m 2v 2 prime so that the total momentum after the explosion equals zero as it should since before the explosion, there was zero momentum. And so with this m 2 equaling 1.5m 1, we can say that m 1v 1 prime equals negative 1.5 times m 1v 2 prime and then we can write v 1 prime is negative 3 over 2 times v 2 prime, writing this one and a half as a fraction which will make it a little easier to work with down below here. So that takes care of as much as we can do with momentum and then we turn our attention to energy and the kinetic energy of the two blocks added together is gonna equal the 5500 joules because we are told what the total kinetic energy is. And we are going to work with this by trying to turn one of the expressions and writing it in terms of the variables from the other expression. So we are gonna turn this kinetic energy term into something containing just m 2 and v 2 prime. And we know that m 1 is two-thirds m 2, let's rearrange in this one multiplying both sides by 2 and dividing both sides by 3 and we'll write two-thirds m 2 here instead of m 1 and then for v 1 prime, we'll write negative 3 over 2v 2 prime from here and this squared makes this negative disappear; 3 over 2 times 2 over 3 that cancels with one of the 3 over 2's but since it's squared, there's still another one remaining. And I'm gonna write it this way, 3 over 2 times one-half m 2v 2 prime squared and I'm not gonna bother going 3 over 4 because when it's written this way, it becomes more clear that this is the kinetic energy of particle two after the explosion and this also is the kinetic energy of particle two after the explosion so we have 3 over 2 times KE 2 prime plus KE 2 prime equals the total energy. And 3 over 2 plus 2 over 2, you could say, is 5 over 2 and so that means the kinetic energy of particle two is two-fifths times 5500 joules and that is 2200 joules. And then the kinetic energy of particle one after the explosion is gonna be the total minus the kinetic energy of particle two so that's 5500 minus 2200 which gives 3300 joules.