(b) A force is applied to a particle along its direction of motio...

Question

(b) A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light

Accepted Answer

Hello, fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. So a neutron of mass M is at rest. A force F gives the neutron an acceleration A when it is at rest, when the neutron moves at speed V, the force required to produce the same acceleration A is 2. F determine the value of V in terms of the speed of light C. So that is our end goal is to determine the value of V in terms of speed of light C. OK. So we're given some multiple choice answers, they're all given in the same units of C. So let's read off our read off our multiple choice answers to see what our final answer might be. So A is 0.747, B is 0.457 C is 0.864 and D is 0.676. Awesome. So first off, we need to recall the equation for relativistic force which states that the force is equal to gamma cubed multiplied by the mass multiplied by the acceleration. We also need to recall that the relativistic or it's also called the Lourens factor two. So the relativistic factor, the equation for that is gamma equals one divided by the square root of one minus V squared divided by C squared. We also need to note that V is much less than C. And because of that, we do not need to consider relativistic effects. And this is because the particle is at rest. So we can state that F that the force equals mass multiplied by acceleration. Also, we need to note that since it's at a higher speed, we can state that the final force, which I'm gonna denote is F one is equal to gamma cubed multiplied by mass multiplied by acceleration. We can also note that the initial force that you need to denote as F subscript zero equals mass multiplied by acceleration. However, we also need to note or gonna note here that the problem itself states that, that F one F subscript one, the final force is equal to two point five. Thus, we can conclude that gamma cubed, that gamma cube equals 2.5. So from all this, we can conclude that gamma Q equals 2.5. So to determine the value of V, which is our end goal, we can use the relativistic force equation. And to do that, we need to rearrange that equation to solve for V by itself. So to simplify the expression, let's start by getting rid of the square root. So we're gonna use the relativistic factor um which is this equation right here. So let's do that and let's start by getting rid of the square root. So to do that, we just multiply by the power of one half. So one minus V squared C squared to the power of one half. So this is the simplify to get rid of the square root. So it's one divided by one minus V squared, divided by C squared to the power of one half. But we must cube both sides of the equation. Since we determine that the, that the final force is equal to gamma cubed F and that gamma cubed equals 2.5. So let's like let's multiply or actually rephrase, we need to cube both sides of the equation. Let's do that. And we need to substitute 2.5 in for gamma cube. So let's let's do that. So gamma cubed and this is when we cube both sides. So one divided by one minus V squared B squared, divided by C squared two, let me take it to the power of three. So it'd be three halves. So it'd be one minus B squared divided by C squared to the power of three halves. But we need to remember that gamma cube is 2.5. So let's write that down really quick. 2.5 equals one divided by one minus V squared, divided by C squared to the power of three halves. OK. So we need to multiply both sides to by the power of two thirds to get rid of our power here of three halves. So let's do that. So 2.5 multiplied to the pa multiplied by the power of two thirds goals, one divided by one B squared, divided by C squared. Awesome. So at this point, since our main goal, we're trying to simplify everything is we're trying to get V all by itself. So to do that, this state, we need to note a mathematical relationship that one divided by X equals X to the power of negative one. So then using that mathematical relationship, we can rewrite or we can simplify further and we can state that 2.5 minus two thirds power equals one minus B squared divided by C squared. So then we need to add V squared divided by C squared to the other side and subtract 2.5 to the power of negative two thirds. So let's do that. So right now we're just shuffling everything around. We're trying to get V all by itself. So this is all basic algebra maneuvers here. We're just trying to simplify it here. So moving all that around, like I said, you should get minus or so I should say negative two thirds power. Awesome. So now we need to multiply C squared to both sides and simplify. So let's do that. So V squared equals and simplify this. So we need to, when we multiply C squared to the other side and simplify this, so we can plug this into a calculator. So we could plug one minus 2.5 to the power of negative two thirds. So when we plug that into a calculator, you should get 4.574 point C squared. And then to get rid of the squared, we square root both sides. And then that when we plug that into a calculator, your final answer should be oh, can't forget. Or C squared in their square root. And you should get when you plug that into a calculator, 0.676 C because, and when you square root a power of two, you just get one. That's the seed of the power of one. Ok. So that's our final answer. All right. OK. So that means our final answer is D 0.676 C. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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(b) A force is applied to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light

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