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`.

Explanation

(a) As F=ma=m`(dv)/(dt)`
`=mv(dv)/(dx)["as"(dv)/(dt)=(dv)/(dx).(dx)/(dt)=v(dv)/(dx)]`
and according to given problem `F=-K//(2x^(2))`, so
`mv(dv)/(dx)=-(k)/(2x^(2))`.
i.e., `int_(0)^(v)dv=(1)/(m)int_(1)^(x)-(K)/(2x^(2))dx`
Or `v^(2)=(k)/(m)((1)/(x)-1)=(10^(-2))/(10^(-2))((1)/(x)-1)=((1)/(x)-1)`...........(i)
So, for x=0.5m, `v^(2)=(2-1)`, i.e., v=`+-1` m/s
And as here the force is opposite to displacment and particle starts moving from x=1 towards the origin i.e., with increase in time, xdecreases, i.e., v(=-dx/dt) will be negative,

So, `v=-1(m//s)`
(b) As from Eqn. (i)
`v=((1-x))/(x), i.e., -(dx)/(dt)=((1-x))/(x)` [as here `v=-(dx)/(dt)`]
So, `int_(0)^(t)dt=-int_(1)^(0.25)(x)/((1-x))dx`
To integrate the above expression let x`=sin^(2)theta` so that `=2sin theta cos theta d theta` and hence
`t=-int_(pi//2)^(pi//6)[(sin^(2)theta)/(1-sin^(2)theta)]^(1//2) 2 sin theta cos theta dtheta`
`=-int_(pi//2)^(pi//6)2sin^(2)dtheta`
i.e., `t=-int_(pi//2)^(pi//6)(1-cos 2theta)d theta`
`=[theta-(1)/(2)sin 2theta]_(pi//6)^(pi//2)=((pi)/(3)+(sqrt(3))/(r))s`.