A particle moves in the X-Y plane under the influence of a force such that its linear momentum is`vecp(t) = A[(wedge)icos (kt) -(wedge) sin (kt)]` where A and k are constants. The angle between the force and the momentum is:

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

d

Explanation

The force acting on the particle
`vecf=(dvecp)/(dt)=(d)/(dt)[A(hati coskt-hatj sin kt)]`
Thus
`vecF.vecp=A k[-veci sin kt-hatj cos kt]`
`[A(hati cos kt-hatjsin kt)]=0`
The angle between `vecF` and `vecp is 90^(@)`
. Hence (d) is the correct choice.