TITLE: Position and velocity problem
QUESTION [0 upvotes]: A particle's position is described by Cartesian co-ordinates $x\,\textbf{i} + y\,\textbf{j}$. It moves under the influence of a magnetic field $\textbf{B}=B\,\textbf{k}$ for $x>0$, and $\textbf{B}=0$ for $x<0$. It is struck from a point $P$ whose co-ordinates are $(-D,0)$, where $D>0$. When it is struck it has velocity $u\,\textbf{i} + \frac{u}{2}\,\textbf{j}$, where $u>0$.
I am asked to find the position and velocity of the particle at $t=0$, where $x=0$.
I have no idea what to do here because if it is struck from the point P with the velocity described above, surely this implies that $x(0)=-D\,\textbf{i}$, and $v(0) = u\,\textbf{i} + \frac{u}{2}\,\textbf{j}$. But this would imply that $D=0$, which contradicts the fact that $D>0$?

REPLY [1 votes]: Based on physical law, when a charged particle travelling with velocity $\textbf{v}$ (positive charge q) hits a magnetic field B, it will be acted on by a force F given by $\textbf{F} = q\textbf{v} X \textbf{B}$.  Now we have to assume that the particle has a positive charge q and a mass of m.  
As the charge hits the y axis, the direction of the velocity is at a slope of $\frac{\frac{u}{2}}{u} = \frac{1}{2}$ with the x axis.  Then it will move in a circle with the centre lying on the perpendicular to the line of the initial velocity.  The radius is calculated based on balancing the magnetic force and the centrifugal force as shown below
$$qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}$$
Based on the given velocity
$$v = \sqrt{u^2 + \frac{u^2}{4}} = \frac{\sqrt{5}}{2}u$$
$$\implies r = \frac{\sqrt{5}mu}{2qB}$$
The next step is to find the centre of the circle $(x_0, y_0)$
As the initial velocity slope is $\frac{1}{2}$, the centre will lie on the line with slope $-2$.  Also the particle hits the y axis at the point $(0, \frac{D}{2})$.  Hence the equation of the line is $y = -2x + \frac{D}{2}$.  Therefore we have the following equations to solve for $(x_0, y_0)$
$$y_0 = -2x_0 + \frac{D}{2}$$
$$x_0^2 + (y_0 - \frac{D}{2})^2 = r^2$$
$$\implies x_0 = \frac{r}{\sqrt{5}}$$
$$y_0 = -\frac{2r}{\sqrt{5}} + \frac{D}{2}$$
Then the equation of the circle is
$$(x - \frac{r}{\sqrt{5}})^2 + (y + \frac{2r}{\sqrt{5}} - \frac{D}{2})^2 = r^2$$
By setting $x = 0$ we have the position when the particle hits the y axis again which is 
$$(0, \frac{D}{2} - \frac{4r}{\sqrt{5}})$$
By differentiating the circle at this point we can see the slope is $-\frac{1}{2}$
As the speed will stay the same and only the direction is changed, the velocity when it hits the y axis again is
$$-u\textbf{i} + \frac{u}{2}\textbf{j}$$
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