A particle of mass `10^-2kg` is moving along the positive x axis under the influence of a force `F(x)=-K//(2x^2)` where `K=10^-2Nm^2`. At time `t=0` it is at `x=1.0m` and its velocity is `v=0`.
(a) Find its velocity when it reaches `x=0.50m`.
(b) Find the time at which it reaches `x=0.25m`.

Answer

(i) -1m/s (ii) `((pi)/(3) + (sqrt(3))/(4)) s`

Explanation

(i) -1m/s (ii) `((x)/(3) + (sqrt(3))/(4))s`
The particle of mass `(m = 10^(-2) kg)` is moving along positive x-axis

(i) `F(x) = - (K)/(2x^(2))` or `mv ((dv)/(dx)) = - (K)/(2x^(2))`
Now integrating both side, `m int_(0)^(v) vdv = - int_(1)^(x) (K)/(2x^(2)) dx`
or `(m upsilon^(2))/(2) = [(K)/(2x)]_(1)^(x) = (K)/(2) ((1)/(x) - 1)`
or `v^(2) = (K)/(m) ((1)/(x) - 1)` or `V = sqrt((K)/(m) ((1)/(x) - 1))`
When x = 0.5 m, `v = sqrt((k)/(m)((1)/(0.5) - 1)) = sqrt((k)/(m)) = sqrt((10^(-2))/(10^(-2))) = +- 1`
As the force is acting along negative x-direction, therefore, the velocity will be in -x direction. Hence v = - 1m/s and `vec(v) = - 1 i m //s`
(ii) As `(k)/(m) = 1 m//s`, hence from (1) `V = (dx)/(dt) = - sqrt((1 - x)/(x))` [we have chosen -sign becuase velocity is in -x direction `sqrt((x)/(1 - x)) dx = - dt` `implies int_(1)^(0.25) sqrt((x)/(1 -x)) dx = int_(0)^(t) dt`
Put `x = sin^(2) theta, dx = 2 sin theta cos theta d theta`
So `int_(pi//2)^(pi//6) 2 sin^(2) theta d theta = - t cos 2 theta = 1 - 2 sin^(2) theta, 2 sin^(2) theta = 1 - cos 2 theta`
`int_(pi//2)^(pi//6) (1 - cos 2 theta) d theta = - t` `[ theta - (sin 2 theta)/(2)]_(pi//2)^(pi//6) = - t`
`((pi)/(6) - (1)/(2) sin (pi)/(3)) - ((pi)/(2) - (1)/(2) sin pi) = - t` `:.` `t = ((pi)/(3) + (sqrt(3))/(4)) sec`