A positively charged particle of charge $1C$ and mass $40g$ is revolving along a circle of radius $40cm$ with velocity $5m/s$ in a uniform magnetic field with centre at Origin $O$ in xy-plane. At time $t=0$, the particle was at $\left(0,0.4m,0\right)$ and velocity was directed along positive x-direction. another particle having charge $1C$ and mass $10g$ moving uniformly parallel to z-direction with velocity $\frac{40}{\mathrm{\pi }}m/s$ collides with revolving particle at $t=0$ and gets stuck with it neglecting gravitational force and colombians force, calculate x,y and z-coordinates of the combined particle at $t=\frac{\mathrm{\pi }}{40} sec$.

For xy-plane,

$$\begin{gathered} T=\frac{2\mathrm{\pi r}}{v}=\frac{2\mathrm{\pi }\left(0.4\right)}{5}=\frac{8}{50}\mathrm{\pi } \\ \because \mathrm{T}=\frac{2\mathrm{\pi m}}{\mathrm{Bq}} \\ \mathrm{T}\propto \frac{\mathrm{m}}{\mathrm{q}} \end{gathered}$$

After collison mass has become $\frac{5}{4}$ times and charge two times.

$\therefore T\text{'}=\left(\frac{5}{4}\times \frac{1}{2}\right)T=\frac{5}{8}\times \frac{8}{50}\mathrm{\pi }=\frac{\mathrm{\pi }}{10}\mathrm{s}$

Given time $t=\frac{T\text{'}}{4}$,i.e., combined mass will complete one-quarter circle.

Further, $P$ will be constant.

$$\begin{gathered} r=\frac{P}{Bq} \\ r\propto \frac{1}{q} \end{gathered}$$  

Since, charge has become two times, $r\text{'}=\frac{r}{2}=0.2 m$

At $t=\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$40$}\right.\right)$ second, particle will be at $P$ in xy-plane.

$$\begin{gathered} \therefore x=r\text{'}=0.2 m \\ y=r\text{'}=0.2 m \end{gathered}$$

z-coordinate Mass of combined body has become $5$ times of the colliding particle. therefore, from conservation of linear momentum, velocity component in z-direction will become $\frac{1}{5}$ times. Or,

$$\begin{gathered} v_z=\frac{1}{5}\times \frac{40}{\mathrm{\pi }}m/s=\frac{8}{\mathrm{\pi }}m/s \\ \therefore z=v_{z}t=\frac{8}{\mathrm{\pi }}\times \frac{\mathrm{\pi }}{40}=0.2 m \end{gathered}$$