A particle moves in the x-y plane under the action of a force `vecF` such that the value of its linear momentum `vecP` at any time t is `P_(x)=2cost` and `P_(y)=2sint`. What is the angle `theta` between `vecF` and P at a given time t?

Options

A. `90^(@)`
B. `0^(@)`
C. `180^(@)`
D. `30^(@)`

Answer

A

Explanation

`P = sqrt(P_(s)^(2) + P_(y)^(2)) = sqrt ((2 cos t) ^(2) + (2 sin t)^(2)) = 2`
If `m` be the mass of the body, then
Kinetic energy = `(P^(2))/(2m) = ((2)^(2))/(2m) = (2)/(m)`
Since kinetic energy does not change with change with time, both work done and power are zero.
Now power `= Fv cos theta = 0`
As `F!= 0, v!= 0" ":." "cos theta = 0` or `theta = 90^(@)`
As direction of `vecP` is same that `vecv (because 'vecP = m vecv)`, hence angle between `vecF` and `vecP` is equal to `90^(@)`.