A 1.0-cm-diameter, 2.0 g marble is shot horizontally into a tank of 20°C olive oil at 10 cm/s. How far in cm will it travel before stopping?

Video transcript

Hey, everyone. So this problem is dealing with drag forces. Let's see what they're asking us. We have a five grand ball with a radius of 2. millimeters fired horizontally with an initial speed of five m per second. In a viscous honey medium, the viscosity of 57.4 pascal's time seconds were asked to calculate the total horizontal distance traveled by the ball and were told to neglect the gravitational force. So if we were to draw a free body diagram of the ball, the only force acting on it is the drag force and that drag forces acting in the negative X direction opposite of the direction of movement. So we can recall that for viscous fluids and relatively low speeds which we're dealing with here, the drag force is given by Stokes law. So we can recall that that's six pi times A to our viscosity, our radius and be our speed. You can also recall with Newton's second law that F equals M A and here in this case, our force again is only the drag force that's acting in the negative direction. So that would be negative F D equals M A. So we have acceleration but we are asked for distance. So we need to think about the relationship between acceleration and distance. We can recall that acceleration is the derivative of velocity and in turn velocity is the derivative of distance or a position. And so we can rewrite acceleration as V D V D X and then integrate from there to fight to find our ex or our position which will give us the distance. So let's see what that looks like. So we will plug in Stokes law here, six pi A to R B equals M multiplied by V D V D T. Quickly simplify here, the velocities cancel. And then because we are looking for distance, we want to get our troops sorry. This is D V D X and we want to get R D X isolated by itself. So we'll rewrite this as D X equals negative M over six pi A to R TV. You will integrate both sides. The integral of DX from X I which we will state is zero m two X F equals, we'll pull out the constant negative um over six pi A to our times the derivative times the integral of DV, where The initial is five m/s. MVF is zero liters per second. So that's coming from the problem where we have an initial speed of 5m/s and the total distance is the distance until there is no speed until the ball comes to rest. So that's VF is going to be zero. So when we integrate, we will get X from X I of zero m 2 xf equals negative over six by eight R times V from V I equals five m per second to V F equals zero m per second. So from here, we can plug in everything that we know from the problem. So our mass was given to us as five g. I'm gonna rewrite that zero. Sorry, I'll rewrite that as five times 10 to the minus three kg. To keep us in standard units, our radius is 2. mm. So that's going to be 2.5 times 10 to the - m. And our viscosity was given as 57. pascal's times seconds. So when we solve this out, we'll get X F minus zero m equals negative five times 10 to the negative three kg. Over six pi times 57.4 pascal time seconds times 2.5 times to the -3 m. All of that multiplied by five, all of that multiplied by zero m per second minus five m per second V F minus V I, We can plug that in to our calculators and we would get 9.2 times 10 to the -3 m. Which can be rewritten as 0.9, two cm. That is the answer for this problem. Go back up to our answer choices and we see the data lines with answer choice. A that's all we have for this one. We'll see you in the next video.