A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. b. For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0-ms-long encounter with a hard floor?

Video transcript

Hi everyone. In this practice problem, we are being asked to calculate the ma suit of the acceleration encountered by the particle due to the electric field. We will have a particle actually falling initially at the top of the apparatus of 0.0 falling for a distance of O A which is 0.8 m. Once the particle reaches point A an instantaneous electric field is going to be applied. And as a result, the particle is going to be stopped momentarily by the electric field in uh 820 micro six seconds time interval, we're being asked to actually calculate the matu of the acceleration encountered by the particle due to the electric field. With the options being a four point eight m per second squared B 19 m per second squared C 4. times 10 to the power of three m per second squared and D 1.9 times 10 to the power of four m per second squared. So the way we want to tackle this problem is by dividing the motion into two different parts. So the first part is going to be where it is falling directly from its initial state of 20.0 up until point A and the second part of the motion is going to be where it is falling from point A and then the electric field is applied and then it is going to stop momentarily in uh 820 microseconds time interval. So we're going to start by um actually tackling the first part of the motion. So the first part I'm gonna say is going to be O A and in the first part of its motion, it is going from Y one I which equals to 10.0 of zero m going to Y one F which equals to point A at minus or negative 0.8 m. The particle is subject to only gravitational acceleration from point A to point A and the initial speed of the particle is going to be zero. So V1 I is going to be 0m/s. We will apply the kinematic equation to calculate the velocity at uh point A or at Y one F equals to negative 0. m because we want to know or we need that we need to know that velocity going into the second part of the motion. So we wanna use the kinematic equation of V one F squared minus V one I squared equals to negative two G Y F minus Y I just like so OK. So we wanna use this kinematic equation to actually solve for our problem or solve for P one F, we know that the P one I equals to zero. So we can neglect this part. So the equation will then just be P one F squared equals to negative two G multiplied by Y F or Y one F equals Y one I let's substitute all the values. So V one F squared will equals to minus two multiplied by 9.81 m per second squared. Multiply it by 11 F which is negative 0.8 m minus zero m just like. So and P one F squared will then come out to be 15.696 m squared per second squared. We want to do the square root to find P one F. So V one F is just the square root of 15.696 m square per second squared, which will equals to plus minus 3. m per second. We wanna take the minus because we know that from the free fall, the velocity is then going to be pointing downwards. So P one F is going to equals to minus 3.96 m per second. I'm gonna say that it is pointing downwards. That's why we pick the minus sign. All right now that we solve for the V1F or the velocity coming to the point A we want to solve for the second part of the motion. So the second part what we have for the second part is the electric field applied to the system or applied to the particle and will be stopped in the time interval of 820 microseconds. So V or P two I or the initial velocity going into the second part is going to equals to P one F which is minus 3.96 m per second. And then the uh time is going to be T2, which is going to be 820 microseconds or 820 times 10 to the power of uh -6 seconds. And what we need to solve is the one that's being asked, which is the acceleration or A two. So we will apply the kinematic equation to actually calculate what A two is. We're using this equation right here which is P two F equals to A two T two plus P two. I substituting with the given and solving for a two. Then we can actually rearrange this equation right here. So that A two will equals two V two F minus V two I divided by the two. We can actually uh substitute all the known values. We know that the V two F is essentially zero m per second because it is stopped after the time interval. So this is going to be zero m per second minus negative eight uh negative 3.96 m per second divided by our time interval. Which is 820 times 10 to the power of minus six seconds. So this will give us an a two value of 4.8 times to the power of three m per second squared. So a two or 4.8 times 10 to the power of three m per second squared is going to be the answer to this problem which is the magnitude of the acceleration encountered by the particle due to the electric field and 4.8 times 10 to the power of three m per second squared will equals to or correspond to option C in our answer choices. So answer C is going to be the answer to this practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and I'll be all for this one. Thank you.