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9

Your question seems to be: if the observer on Earth sees the spaceship moving at velocity v, how do we know that the observer on the spaceship will see Earth moving at velocity -v? This is known as the "reciprocity principle" problem and it is a good one, in the sense that it raises the following issue: "Does the reciprocity principle follow from the basic ...


4

How did you integrate acceleration to get velocity? Note that $\Delta v = \int_{t_i}^{t_f} a(t) dt$ But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v ...


3

Further to udrv's answer, and, the technicalities he raises aside, there are two ways to argue the reciprocity relationship that the boost from observer $A$ to $B$ is the boost from $B$ to $A$ but with $v\mapsto-v$. By the detailed arguments in the afterword, we find that the transformations between inertial frames form a group and that group acts linearly ...


3

To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information. Since they give you the solution and you just have to ...


3

No, all your reasoning is totally right. The conclusion isn't that the graphs are wrong, it's that the time of impact is less than 0.1 second. In this video, for example, the time of impact is just about 0.01 seconds.


3

If you drive along the road on the side of a mountain) there are (at least) two kinds of forces on your car: friction and gravity. If you drive to the top of the mountain and back, the net work done against gravity is zero. This corresponds to the fact that your displacement is zero. On the same trip, you had to overcome friction. That depended on your ...


2

I'm not quite sure about your level of physics education so please ask if I've jumped too far in any of these steps. What determines how something falls is the force acting upon it. The force is related to it's acceleration by Newtons famous law $f=ma$ where $f$ is the force, $m$ is the mass (how heavy it is) and $a$ is the acceleration (how fast its speed ...


2

Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero: $$\langle v \rangle = 0 \, .$$ Of course, the average energy of any particular molecule is ...


1

Using polar coordinates it holds $ |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}$. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the $d|{\bf r}|$ should be moreover pointing ...


1

If $$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$ then $$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$ and $$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$ on the other hand $$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$ and $$ \left | d\overrightarrow{r} ...


1

Let $X$ be a set with a binary operation $\star:X\times X\to X$, which could be a group structure or whatever you want. Let $f:X\to Y$ be a mapping to some other set and let $F(x,y) := f(x\star y)$ (your middle expression is this one with the operation +). If $f$ is sufficiently nice, e.g. a bijection, then your construction transports the binary operation ...


1

Another important example: the Pythagorean Sum defined by $f(x) = \sqrt{x}$. As Doetoe says, you're simply transporting the binary operation + to the domain of $f$ if $f$ is e.g. bijective. If your domain is a real interval, your question actually describes the following: Which operation is calculated by the addition of affine co-ordinates implemented by ...


1

You would have to apply a force upwards to stop the body. Stopping the gravity would stop the acceleration but not the speed that it already has. A good place to start to check the effects of g forces in a human body is this wiki Changing the mass won't stop the fall either. You cannot make the mass zero, you can cut legs and arms but I think you'll be ...


1

Let me quote the relativistic velocity addition formula for easy reference: $$v_{AB} = \frac{v_A - v_B}{1 + \frac{v_Av_B}{c^2}}\tag{SR}$$ I'm guessing you interpreted these quantities as follows: $v_A$ is the speed of the light beam relative to George $v_B$ is the speed of Gracie relative to George $v_{AB}$ is the speed of the light beam relative to Gracie ...


1

The key to understanding this somewhat surprising result is that the relativistic velocity addition formula is not applicable to this calculation. As an example of when to apply the velocity addition formula, sssume there is an object with (1D) velocity $\mathbf u$ in some inertial coordinate system. Now, what is the velocity $\mathbf u'$ of that object in ...


1

You can actually (in principle) do an experiment for one of those. If you had a huge, dense, thin sheet of matter in empty space you could cut a person sized hole through it then attach a cylinder to the hole and put a flat bottom on the far away end. So on one side it looks like a flat tower coming up from a big plane and on the other side it looks like ...


1

Yes your reasoning is completely right but is important to add some information. The diagram you show has no information on the ball size and assumes all the mass is concentrated in the center of mass, which for a spherical and isotropic ball should be in the geometrical center. Now, at the moment of impact $t_1$, the border of the ball will start touching ...


1

$v = |\vec{\omega}\times\vec{r}|= \omega r$ is always valid for a rigid rotating body. Here, $r$ refers to the distance of any particular point from a chosen axis of rotation, $\omega$, the angular speed of the body about that chosen axis and $v$, the linear speed of that point perpendicular to the radius vector (or the line joining the axis to that ...


1

I write this because it is too long for a comment Situation You are putting $k$ watts (=joules/second) of kinetic energy into a body. We will show the acceleration decreases with time. $$E=\frac{1}{2}mv^2$$ so $$v=\sqrt{\frac{2E}{m}}$$ so $$\frac{dv}{dt}=\frac{m}{\sqrt{\frac{2E}m}}\cdot\frac{de}{dt}$$ (I can't be bothered to tidy this up - remember I'm ...


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Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \begin{equation} \frac{dx^{1}}{dx^{0}} = \frac{d ...


1

If you want to actually simulate the behavior of the planet as it experiences the (vector) force as it moves around, then you need to find a stepping method and write your velocity vectors and position vector in terms of coordinates. I recommend a Verlet velocity method. Others at this site have their favorites, too. Euler's method is not good enough for ...



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