A particle of mass m is subjected to an attractive central force of magnitude `k/(r^2)` where k is a constant. Find the extreme position at an instant when the particle is at another extreme position at a distance 'a' from the centre with speed `sqrt((k)/(2 a m))` .

Options

A. `a/3`
B. `a/2`
C. `a/4`
D. `a/5`

Answer

A

Explanation

Here potential energy of a particle, `U = (-k)/r`
Let v be the velocity at B and v. be the velocity of A.
From the principle of conservation of angular momentum.
`mva = mv.x `ie, `v. = (va)/x (1) `
From the principle of conservation of energy
`1/2 mv^2 - k/a = 1/2 mv.^2 -k/x " " (2)`
Using `v = sqrt((k)/(2ma)) ` in (1) we get, `v. = sqrt((ka)/(2m)) . 1/x`
Putting this value in (2), we get `1/2 m k/(2ma) - k/a = 1/2 m (ka)/(2m) 1/(x^2) - k/x`
i.e. `3x^2 - 4ax + a^2 = 0 " " therefore x = a and a/3`, Thus `a/3` corresponds to position A.