BIO Squids rely on jet propulsion to move around. A 1.50 kg squid (including the mass of water inside the squid) drifting at 0.40 m/s suddenly ejects 0.100 kg of water to get itself moving at 2.50 m/s . If drag is ignored over the small interval of time needed to expel the water (the impulse approximation), what is the water’s ejection speed relative to the squid?

Video transcript

Welcome back. Everyone in this problem. A rocket with a total mass including fuel of 5000 kg is floating in space at a velocity of 200 m per second relative to an asteroid. When it suddenly ejects a portion of its fuel with a mass of 500 kg to increase its velocity relative to that asteroid up to 250 m per second. Ignoring any external forces acting on this system over the small interval time needed for this action. The impulse approximation, what would be the fuel ejection speed relative to the rocket A says it's 200 m per second. B 250 m per second. C 300 m per second and D 500 m per second. Now, if we're gonna figure out Fu's ejection speed, let's make a note of all the information we have so far so far. OK. We know that the total mass, let's call that MT with the rocket and the fuel altogether is 5000 kg. Next, we know that the initial speed V I is 200 m per second relative to the asteroid. We also know that our final mass E Oh, sorry, the mass of the fuel, rather the mass of the fuel is equal to 500 kg. Ok. Our final speed is 250 m per second and the final mass, let's call that MF is going to be the mass of the rocket minus the mass of the fuel. So that's 5000 minus 500 kg, which equals 4500 kg. And now we want to use all of this information to figure out the fuel ejection speed relative to the rocket. Now, how are we going to find that speed? Well, we know that that ejection speed. Ok. Let's call it, the eject is going to be equal to the speed of the fuel. Ok? That's the fuel's speed minus the final speed because it's that final speed will represent the speed relative to the rocket. So we have to account for that in our calculation, we know that the final speed is 250 m per second. But no, if we can find this final speed for the fuel, then we should be able to find the overall fuel ejection speed. Ok. So let's go ahead and do that. Now, how can we find what that speed is? Well, we know that the change in momentum of the rocket is equal and opposite to the momentum of the ejected fuel. Ok. In other words, we know that the momentum of the fuel equals the final momentum for sorry, the initial momentum. OK. Let's call that P I minus our final momentum for our rocket. So let's call that PF I and we know that momentum P is equal to the mass multiplied by the volume. So in this case, then if we were to substitute what we know here, that means the momentum of our fuel, the mass of the fuel multiplied by the speed of the fuel is going to be equal to our initial momentum. That will be the momentum of our rocket with the fuel mt multiplied by the initial velocity. OK, minus the momentum of the final momentum for our rocket which equals the momentum for just the rocket. Sorry, the mass for just the rocket MF multiplied by our final speed. VF Now, if we think about it here, this tells us then that for our, so let me write it below it. The speed of our fuel, if we divide both sides by the mass of the fuel is going to be MTV minus MVF. All divided by the mass of the fuel, we know our total mass is 5000 kg. Initial speed is 200 m per second. Our final mass, that's just the mass of the rocket is 4500 kg and the final speed is 250 m per second. So now we can divide all of that by the mass of the fuel which is 500 kg. Now, when we go ahead and solve here we should find that we get the speed of our fuel to be negative 250 m per second. Ok. So now that we have the speed of the fuel, we want to find the ejection speed of fuel relative to the rocket. Ok. So I sorry, relative relative to our rocket. Yes, my apologies. Ok. So now we can substitute it into our formula for the speed or V eject. Ok. So now that means V eject equals negative 250 m per second minus 250 m per second and that equals negative 500 m per second. Since speed is a scalar we are interested on it in its magnitude. OK. So now the magnitude of the eject then OK? Is let me fix that here is going to be equal to 500 m per second. Thus the speed of fuel ejection relative to the rocket is 500 m per second. Therefore, D is the correct answer. Thanks for watching everyone. I hope this video helped.