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  4. A charged particle q is shot towards another charged particle Q (which is fixed) with a speed v . It approaches Q up to the closest distance r and then returns. If q were given a speed 2v , the closest distance of approach would be <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-eemizujyuw4ej5ft.jpg />

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Q. A charged particle $q$ is shot towards another charged particle $Q$ (which is fixed) with a speed $v$ . It approaches $Q$ up to the closest distance $r$ and then returns. If $q$ were given a speed $2v$ , the closest distance of approach would be
Question

NTA AbhyasNTA Abhyas 2022

Solution:

From conservation of mechanical energy
$\frac{1}{2}mv^{2}=\frac{k q Q}{r}$
In second case
$\frac{1}{2}m\left(\right.2v \left(\left.\right)^{2}=\frac{k q Q}{r^{′}}$
$r^{′}=\frac{r}{4}$