A 45 g projectile explodes into three pieces: a 20 g piece with velocity 25  î m/s , a15 g piece with velocity −10  î+10 ĵ m/s , and a 10 g piece with velocity −15  î−20 ĵ m/s . What was the projectile's velocity just before the explosion?

Video transcript

Hey everyone. So this problem is dealing with conservation of momentum. Let's see what they're asking us. A rocket experienced a large anomaly and exploded into fi four parts. The first part has a mass of 4000 kg. So I'm going to just write this out as we talk for the problem. So massive part one equals 4000 kg and the velocity of 7.5 kilometers per second in the J direction. The second part has a mass of kg and a velocity of 1.1 in the I direction plus 4.5 in the J direction. And again, those units are kilometers per second. The third part as a mass of 5000 kg and the velocity of negative 1. in the I direction plus 3.5 in the J direction in units of kilometers per second. And the velocity of the fourth part which is, which has a mass of 500 kg is negative 0.5 in the J direction. And that's kilometers per second. We're asked to calculate the velocity of the rocket prior to the explosion. Our multiple choice answers here are a 7.5 kilometers per second in the J direction. B negative 0.17 I plus 48 J kilometers per second. C negative 0.1 plus 15, sorry, negative 0.1 I plus 15 J kilometers per second or D 3.3 I plus 8.6 J kilometers per second. Now, this might seem a little bit overwhelming because there's a lot of information that was given to us. But we can recall our momentum conservation momentum equation is simply our initial momentum is equal to our final momentum where momentum is equal to mass multiplied by velocity. So when we write out this first equation, taking into account each of the pieces of the rocket after it blew apart, this looks like the total mass bar rocket plus the initial speed of our rocket is equal to the mass multiplied by the speed or velocity of each of the four parts of the rocket. So it will be M one V one plus M two V two plus M three V three plus M four V four. And each of these variables was given to us in the problem, even the mass of the rocket. Now, that wasn't spelled out explicitly, but it's going to be the sum of all of the pieces of the broken apart pieces, right. So we'll have kg plus kg plus 5000 kg plus kg is equal to 13,000 kg. I'm sorry, you wrote that down here just a little bit neater. So as you can see, we have everything we need Tusa for the velocity of the rocket, the velocities were given to us in vector form. So that's with a Y component and an X component. So as we multiply these through eventually, we're going to add the X components and the Y components together. OK. So this looks like mass of the rocket multiplied by the velocity of the rocket is equal to 4000 kg multiplied by 7.5 J kilometers per second plus 3500 kg multiplied by 1.1 I less 4.5 J kilometers per second. And you'll see here that usually in physics, we want to be working in standard units. So kilometers per second is not our standard unit, there would be meters per second but because all of our speeds are given in kilometers per second. Um And we're just dividing by um multiplying and then dividing by kilograms, we can keep it in kilometers per second. It just saves us a step from converting and then converting back for the answer. But it is a little bit different than what we're used to seeing. So I just wanted to address that. All right, back here we are working on M three. So that's 5000 kg multiplied by negative 1.2 I plus 3.5 J kilometers per second plus our fourth piece 500 kg multiplied by negative 0.5 J. And when we multiply these through and then divide both sides by the mass of our rocket or 13,000 kg, we get I and J components of 2.308 J plus 0.296 I plus 1.21 J minus or plus a negative 0.4 62, I loss 1. J minus 0.19 J. All of those are in units of meters per second. And so the last step is going to be adding all of the J components together and the eye components together. So the eye components are negative 0.17 and the J components are 4.8 J. And again, we're working in units of kilometers kilometers per second. Sorry, I misspoke earlier that those are in units of kilometers per second. It's tricky for everyone when we're not using standard units. So maybe it would have been easier for you to convert two m per second and then convert back for the answer. That's a fine way of going about it too. When we look at our multiple choice answers, this aligns with answer choice B. So that is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.