b. The motor of a 350 g model rocket generates 9.5 N thrust. If a...

Question

b. The motor of a 350 g model rocket generates 9.5 N thrust. If air resistance can be neglected, what will be the rocket's speed as it reaches a height of 85 m?

Accepted Answer

Hey, everyone in this problem, we have a model rocket with a mass of 300 g and that students are using to learn about forces and the motor yields 3. newtons of thrust. And we're asked to calculate the rocket's velocity when it reaches an altitude of 15 m. If we neglect air resistance, we're given four answer choices all in meters per second. Option A is 11.4. Option B is 8.9. Option C is 4.5 and option D is 5.1. Now, let's write out what we have so far what we've been given in the problem. Now, we know that the initial velocity V knot of this rocket is gonna be zero m per second, ok? Because this rocket is getting launched so that initial speed is going to be zero m per second. Now, the final speed, we don't know or the final velocity, but that's what we're looking to find. OK. The final velocity at a particular altitude. Now, the altitude, right, as delta Y is equal to 15 m, we don't have any information about the acceleration A or the time T So if we try to use our kinematic equations. Now, we only have two known values and we need at least three in order to solve for an unknown like V F. So let's think about what else we can do to get one of these values. Now, we're given information about a force and that's that thrust force of 3.2 nodes recall that we can relate the forces to acceleration through Newton's second law. So let's go ahead and look at our forces. We're gonna draw a free body diagram and upwards direction. We have the thrust force. OK. That's gonna launch that rocket upwards and pointing downwards. We have the force of gravity acting, let's take up to be our positive direction. And Newton's second law again tells us that the sum of the forces and in this case, in the Y direction, it's gonna be equal to the mass multiplied by the acceleration in the Y direction, right. So now we can see how we can relate these forces to this acceleration. This equation is gonna allow us to calculate the acceleration which we can then use in our kinematics equation to solve for V F. So the some of the forces in the positive y direction we have the thrust force. So we have F thrust and in the negative y direction, we have the force of gravity. So F thrust minus the force of gravity is equal to the mass multiplied by the acceleration. And I'm just gonna drop that subscript of Y because we're only working in one direction here. And so we know that that acceleration is that vertical acceleration. Now let's go ahead and substitute in our values. We know that the thrust force is equal to 3.2 newtons. The force of gravity I recall is equivalent to the mass multiplied by the gravitational acceleration G and this is equal to the mass M multiplied by the acceleration. A. Now we know G OK. That's a constant. And the mass and we were given in the problem. Now we have M is equal to 300 g. We just want to convert this to our standard unit of kilograms first before we solve this problem. OK? Because our force is in Newton and recall that a Newton is equivalent to a kilogram meter per second squared. So we want to have consistent units here. We want this mass in kilograms. We're gonna take it and we're gonna multiply it by one kg divided by 1000 g. The unit of gram is gonna divide out and what we're essentially doing is dividing by 1000 to go from grams to kilograms. And that gives us 0.3 kg. So now we can use that mass in our equation. Newton's second law, we have 3.2 noons and we're gonna write this as kilogram meter per second squared minus 0.3 kg multiplied by 9.8 m per second squared is equal to 0.3 kg multiplied by the acceleration. A simplifying on the left hand side, we get 0.26 kg meter per second squared is equal to 0. kg multiplied by the acceleration A and one final step for finding a, we're gonna divide by 0.3 kg to isolate it for A and we get that, that acceleration A is equal to 0. repeated meters per second squared. All right. So now we have this acceleration. Let's go back up and remind ourselves why we wanted that acceleration. OK? We have our kinematic variables and we did not have enough known values to solve for that final velocity. So we calculated a to be 0.866 repeated meters per second squared. Now, we can use those kinematic or U AM equations. We have three known values B not delta Y and A, we have the unknown V F that we're trying to solve for. So we're gonna choose the equation with those four variables. Let's give ourselves some more room to do this. And that equation is gonna be V F squared is equal to V not squared plus two A delta Y substituting in the values we have OK, we get V F squared is equal to that initial velocity is zero. So the first term on the right hand side just goes to zero. We get two multiplied by zero. 00.866 repeated meters per second squared, multiplied by 15 m. If we work out everything on the right hand side, we have V F squared is equal to m squared per second squared. And taking the square root, we get that, that final velocity is gonna be positive or negative 5.1 m per second. OK. It's important to remember that when we take the square root, we get both the positive and the negative answer. Now, we need to interpret which one makes more sense we've taken up to be a positive direction that velocity is going to be pointing in the upwards direction as well because this rocket is being launched upwards. And so we're gonna take the positive velocity. And so our final velocity that we were looking for is gonna be 5.1 m per second. If we compare this to the answer choices we were given, we can see that this corresponds with answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.

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b. The motor of a 350 g model rocket generates 9.5 N thrust. If air resistance can be neglected, what will be the rocket's speed as it reaches a height of 85 m?

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