1.

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

Answer»

`0^(@)`
`30^(@)`
`45^(@)`
`90^(@)`

Solution :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.


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