A rocket cruises past a laboratory at 1.00×10^6 m/s in the positi...

Question

A rocket cruises past a laboratory at 1.00×10^6 m/s in the positive x-direction just as a proton is launched with velocity (in the laboratory frame) v =(1.41×10^6 î+1.41×10^6 ĵ) m/s. What are the proton's speed and its angle from the y-axis in (b) the rocket frame?

Accepted Answer

Hello everyone. Let's go through this practice problem. A neutron moves at a speed of 2.3 multiplied by 10 to the power of 6 m per second in the negative X direction pass a stationary observer on earth. An electron is launched relative to the same stationary observer with the speed V of 2.7 multiplied by 10 to the power of six I plus 2.7 multiplied by 10 to the power of six J meters per second. Calculate the electron speed and direction measured from the y axis. As observed by the neutron. We have four multiple choice options to choose from. Option A 2.7 multiplied by 10 to the power of 6 m per 2nd. 8.43 degrees from the y axis. Option B the same except 81.6 degrees from the y axis. Option C 5.0 multiplied by 10 to the power of m per 2nd, 28.4 degrees from the y axis and d same magnitude but 61.6 degrees from the y axis. OK. So this is a question asking about Galilean relativity where we have where we're given the velocity of something with respect to one object. But we're asked to actually find that velocity with respect to a totally different object that is moving with respect to the other thing. We're comparing it to it. It's a very confusing and mind boggling type of problem. But fortunately for us, the process that we have to go through to solve this problem is actually not all that difficult. In fact, the general procedure is actually fairly simple. So let's think of it this way, we want to find the speed of the electron with respect to the neutron. And one way to think about this is to say that this is equal to the speed of the electron with respect to the earth, which is what we're given plus the speed of the earth with respect to the new Ron, which is something else that we're given. So en equals eeen. So it's kind of like this symmetrical sort of, it's, it's like a chain reaction type of domino sort of thing if it helps to think about it that way. So let's start putting in the velocities we're given and see how well this works. So let's see, it's the velocity of the electron with respect to the earth. That part's pretty simple. We're told that it is 2.7 multiplied by 10 to the power of six I plus 2.7 multiplied by 10 of the power of six J meters per second plus the speed of the earth with respect to the neutron, which were given as 2.3 multiplied by 10 to the power of 6 m per second in the negative X direction. So negative 2.3 multiplied by 10 to the power of six I meters per second. So now we can just add these together pretty simply since adding components like this is always fairly simple. The first term 2.7 times 10 to the power multiplied by 10 to the power of six minus 2.3 multiplied by 10 to the power of six. That first term just becomes 0. multiplied by 10 to the power of six I plus 2.7 multiplied by 10 to the power of six J meters per second. So this is our speed of the electron with respect to the neutron. So if we want to find the magnitude of this, all we gotta do is use the common method for finding the magnitude of components, which is just taking the square root of the sum of the squares of the components. So the square root of the square of 0. multiplied by 10 to the power of 6 m per second plus the square of 2.7 multiplied by 10 to the power of 6 m per second. So if we put that into a calculator, then we find a magnitude of about 2.73 multiplied by 10 to the power of 6 m per second. So that is our answer for the magnitude of the speed. So that narrows down our options a little bit. But we still need to account for the angle with respect to the Y axis. Now, when we have two components of a vector and we want to find the angle, we simply the formula for this is to take the inverse tangent of one of those components divided by the other. Now, normally, when we do this, we'll usually say the Y component divided by the X component. But that method only works when we're looking for the angle from the X axis. But this problem is a little trickier because we're asked to find the angle measured from the Y axis. So we have to flip it in this case. So instead it's the X component divided by the Y component. So what we're actually typing in is the inverse tangent of 0.4 multiplied by GD power of 6 m per second, divided by 2.7 multiplied by 10 to the power of six. And what we find when we put that into a calculator is an angle of about 8.43 degrees. So we have now solved for both parts of the problem, we have the magnitude of the speed and the angle. And, and if we look at our multiple choice options, we can see that option. A agrees with this result. 2.7 multiplied by 10 to the power of 6 m per 2nd and 8.43 degrees from the Y axis. So option A is our answer to this problem and that is it for this video. I hope this video helped you out if you could use more practice. And this video helped please consider checking out some of our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you all have a lovely day. Bye bye.

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A rocket cruises past a laboratory at 1.00×10^6 m/s in the positive x-direction just as a proton is launched with velocity (in the laboratory frame) v =(1.41×10^6 î+1.41×10^6 ĵ) m/s. What are the proton's speed and its angle from the y-axis in (b) the rocket frame?

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