CP A proton is traveling horizontally to the right at 4.50 * 10^6...

Question

CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (b) How much time does it take the proton to stop after entering the field?

Accepted Answer

Hey, everyone. Welcome back in this problem. And electron moves horizontally to the east with a velocity of 2.4 times 10 to the 6m/s. The electron enters an electric field that slows it uniformly, it comes to rest after covering 2.5 cm in the electric field and were asked to determine the time taken by the electron to stop from the moment it enters the electric field. Alright, so we have our electron, it's moving horizontally to the east. So that's to the right at 2.4 times 10 to the six meters per second. And we're going to take that right or eastward direction to be positive. Now, this is talking about an electron in an electric field. So it can kind of get confusing. You can get bogged down by those terms terms and think that you need to do something with an electric equation. However, we're told that this is going to slow uniformly okay. So we have uniform acceleration, which means we can use our um equations, we can use those Kinnah Matic equations that we have. This is just like a ball moving to the right or anything else? Okay. So let's write out what we know, we know that the initial speed of that electron Is 2.4 times 10 to the six m/s. Okay. So the velocity is positive because it's moving in a positive direction. Our final speed, our final velocity is going to be 0m per second because we want it to come to a complete stop, come to arrest. We know that the change in horizontal position is going to be 2. centimeters and we want it to come to rest after 2.5 centimeters. So from our initial speed to the time, we're at rest, we cover 2.5 centimeters. Let's convert this to meters. So we multiply by one m per 100 centimeters, the unit of centimeter cancels we have 2.5 divided by which gives us 0.25 m. We don't know anything about our acceleration. We don't know the time, but the time is what we're trying to find out. And so we have three values that we know we have one unknown that we want to find. So let's go ahead and choose the um equation that doesn't include the acceleration. Okay. We don't have information with the acceleration and that's not what we're looking to find. So we choose the equation without acceleration and that's going to be the following delta X is equal to one half the not plus the F times T okay. Substituting in our values, we get that 0.25 m is equal to one half times 2. times 10 to the six m per second. Okay. VF is just zero times the time T which is what we're looking for. We can simplify on the right hand side. 0.25 m is equal to 1.2 times 10 to the six m per second times the time T and if we divide by 1.2 times 10 to the six m per second, we have meters divided by meters per second, we get the unit of seconds, which is what we want for time. And we get 2.833 times 10 to the negative eight seconds. Okay. Now, that's a really, really small time, but that makes sense. Okay. This electron is moving really fast. It's very tiny and so we can get it to stop quickly. All right. And so the time taken, if we look at our answer choices is going to be be okay. It's approximately 2.08 times 10 to the negative eight seconds. Thanks everyone for watching. I hope this video helped you in the next one.

CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (b) How much time does it take the proton to stop after entering the field?

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