A particle of mass `m` moves in pane `xy` due to the force varying with velocity as `F=a(dot(y)-dot(x)j)`, where `a` is a positve constant, `i` and `j` are the unit vectors of the `x` and `y` axes. At the initial moment` t=0` the particle was located at the point `x=y=0` and possessed a velocity `v_(0)` directed along the unit vector `j`. Find the law of motino `x(t), y(t)` of the particle, and also the equation of its trajectory.

Explanation

We have `vec(F)=a(dot(y)vec(i)-dot(x)vec(j))`
or `m(ddot(x)vec(i)+ddot(y)vec(j))=a(dot(y)vec(i)-dot(x)vec(j)) `
So, `m ddot(x)=a dot(y)` and `m ddot(y)=-adot(x)....(1)`
From the initial condition, at `t=0, x=0` and `y=0`
So, integrating Eqn, `m ddot(x)=adot(y)`
we get `ay` or `x=(a)/(m)y ....(2)`
Using Eqn `(2)` in the Eqn `=-ax`, we get
`mddot(y)=-(a^(2))/(m)y` or `ddot(y)=-((a)/(m))^(2)y ....(3)`
one of the solution of differential Eqn `(3)` is
`y=A sin (omega_(0)t+alpha)`, where `omega_(0)=a//m`.
As at `t=0, y=0` so the solution takes the form `y=A sin omega_(0)t`
On differentiating w.r.t. time `y=A omega_(0)cos omega_(0)t`
From the initial condition of the problem, at `t=0, dot(y)=v_(0)`
So, `v_(0)=Aomega_(0)` or `A=v_(0)//omega_(0)`
Thus `y=(v_(0)//omega_(0))sin omega_(0)t ....(4)`
Thus from `(2)x=v_(0) sin omega_(0)t` so integrating
`x=B-(v_(0))/(omega_(0))cos omega_(0)t ....(5)`
On using `x=0` at `t=0, B=(v_(0))/(omega_(0))`
Hence finally `x=(v_(0))/(omega_(0))(1-cos omega_(0)t) ....(6)`
Hence from Eqns `(4)` and `(6)` we get
`[x-(v_(0)//omega_(0))]^(2)+y^(2)=(v_(0)//omega_(0))^(2)`
which is the equation of a circle of radius `(v_(0)//omega_(0))` with the centre at the point `x_(0)=v_(0)//omega_(0), y_(0)=0`