Question

A charged particle +q of mass m is placed at a distanced from another charged particle -2q of mass 2m in a uniform magnetic field B as shown in fig. If the particles are projected towards each other with same speed v,
Assuming the collision to be perfectly inelastic, find the radius of the particle in subsequent motion. (Neglect the electric force between the charges)

Answer 1

After collision, charge on the combined particle $=-q,Mass=3m$
The combined particle will have velocity in y direction just after collision. Using conservation of linear momentum:

$(mvsin\theta i+mvsin\theta j)+(-2mvcos\theta i+2mvsin\theta j)=3m\overrightarrow{v}$

$3m\overrightarrow{v}=-mvcos\theta j$

$\overrightarrow{v}=(-\frac{v}{2\sqrt{3}}\hat{i}+\frac{v}{2}\hat{j})$

$|\overrightarrow{v}|=v\sqrt{(\frac{1}{4\times 3}+\frac{1}{4})}=v\sqrt{\frac{1}{4}\left(\frac{4}{3}\right)}=\frac{v}{\sqrt{3}}$

$|v|=\frac{1}{\sqrt{3}}\times 2\times \frac{qBd}{2m}=\frac{qBd}{\sqrt{3}m}$

$\therefore r=3m\times \frac{qBd}{\sqrt{3}m\times qB}=\sqrt{3}d$