An observer in frame S′ is moving to the right 1+x-direction2 at speed u = 0.600c away from a stationary observer in frame S. The observer in S′ measures the speed v′ of a particle moving to the right away from her. What speed v does the observer in S measure for the particle if (a) v′ = 0.400c; (b) v′ = 0.900c; (c) v′ = 0.990c?

Video transcript

Hello, fellow physicists today, we're gonna 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, an inertial reference frame moves at U equals negative 0.400 C in the negative X direction to the left relative to a stationary frame. A detector on the inertial frame measures the speed of a moving space object to BV prime to the left of the detector determine the speed V of the moving space object measured relative to the stationary frame when IV prime equals negative 0.300. CIIV prime equals negative 0.700 C and IIIV prime equals negative 0.770 C. OK. So that's our end goal. Our end goal is to determine the speed V of the moving space object measured relative to the stationary frame when V prime equals three different values. So we're trying to find three different speeds of the moving space object given the certain conditions for V prime. OK. So we're given some multiple choice answers for parts II I and II I, and they're all in terms of C where C represents the speed of light. So let's read them off to see what our final answer set might be. A is negative 0.625 C negative 0.859 C negative 0.914. CB is negative 0.0892 C negative 0.234 C negative 0.330 CC is negative 0.0892 C negative 0.234 C and negative 0.283. CD is negative 0.625 C negative 0.859 C and negative 0. C. OK. So first off, let us start by recalling and using the relativistic velocity transformation equation which states that V subscript XVX is equal to VX prime plus U all divided by one plus, you multiplied by VX prime divided by C squared. So considering the information provided to us in the problem, we can state that V prime and V are both in the negative X direction. Thus, we can write that VX prime is equal to V prime. And we can also say that VX is equal to V. So we can then go on to say and let's call it equation when we can write that V is equal to V prime plus U all divided by one plus U multiplied by V prime divided by C squared. OK. So now we can solve for I, so let's do that. So I states that or we, we need to find the speed where V prime is equal to negative 0. C. So let's plug that in for V prime. So V prime in this case is negative 0.300 C plus our U value which is given to us in the problem as negative 0.400 C all divided by one plus U which is 08, negative 0.400. C multiplied by RV prime which was negative 0.300 C all divided by C square. So when we plug that into a calculator, we should get negative 0. C. So this is our answer for part I. So let's start solving using the exact same method as before as we just did to solve for I I, we just need to plug in our new V prime value. So our new V prime value is negative 0.700 C plus ru value which was negative 0.400 C divided by one plus ru value which is negative 0.400. C multiplied by our V prime which was negative 0.700 C divided by C squared. So no, I didn't say it before in the previous part for part I but the C squared cancels out with the C squared above. So the CS cancel up, but then it leaves us with the CS on top, which we're trying to find means we need to leave our answers in terms of C. So when we plug that into a calculator, we should get negative 0. C and that is our answer for part. I, I Awesome. We're on a roll here. So let's solve for II I doing the exact same thing as we've done for parts I and I, I so just plug in our new V prime value. So the V prime value in this case for part II, I was negative 0.770 C plus the U value which was negative 0. C divided by one plus U which was negative 0.400 C multiplied by RV prime number which was negative 0.770 C divided by C square. So when we plugged it into a calculator, we should get zero point 894. So negative 0.894 C and that is our final answer. For part II I fantastic. We did it. So let's go look at our multiple choice answers to see what our final answer set is. The correct answer is. The letter D I is negative 0.625 CI I is negative 0.859 C and II I is negative 0. C. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.