Question

Pls solve this question......

Answer 1

External forces acting on the system the total energy of the system is conserved

The distance between the two particles will be minimum when thier speeds are equal

By law of conservation of momentum,

P initial = P final

m v = m v 1 + mv 1

v 1= v / 2

By applying law of conservation of energy

PE i + KE i =PE f + KE f

0+ 1mv ²/2 = KQ²/ r + 2m(v/2)²/2

m v² / 2 = K Q ²/r + mv²/4

m v ² /2 - m v ² /4 =K Q ²/r

m v ²/4 =K Q ²/r

r = 4 K Q ²/ m v²

r = 2 K Q ²/ 1/2mv²

r = 2 K Q ² / K E

r = 2 Q ²/ 4 π E ( KE)

r = Q²/ 2π E K

The answer is c

Answer 2

$\huge\sf\green{\underline{Answer}}$

Due to repulsive force the other particle will start moving away. The velocity of the first particle will decrease while that of the other will increase. At the point of minimum distance between the two both the particles will be moving at same velocity. Let this velocity be u

So, using conservation of momentum we get,

$mv = 2mu \\ or,u = \frac{v}{2}$

The initial energy of the system is given as:

$\frac{1}{2} m {v}^{2}$

and the energy at the minimum distance is given as:

$\frac{1}{2}m {(\frac{v}{2})}^{2} + \frac{1}{2}m {(\frac{v}{2})}^{2} + \frac{1}{4\pi ϵ_{o} } \frac{ {Q}^{2} }{K}$

Equating the two energies we get,

$\frac{1}{2}m {v}^{2} = \frac{1}{4}m {v}^{2} + \frac{1}{4\pi ϵ_{o} } \frac{ {Q}^{2} }{K} \\ or,\frac{1}{4}m {v}^{2} = \frac{1}{4\pi ϵ_{o} } \frac{ {Q}^{2} }{K} \\ or,K = \frac{4 {Q}^{2} }{4\pi ϵ_{o}m {v}^{2}}$