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) = 2 cos t, P_(y) = 2 sin t`.
The angle `theta` between vecF and vecP` at a given time `t` will be:

Explanation

As , `vecP = hati p_(x) + hatjp_(y) = hati(2 cos t) + hatj( 2 sin t)`
` |vecp|= sqrt([(2 cos t)^(2) + (2 sin t^(2))])= 2 `
Now as, `" "vecF =(dvecp)/(xt), vecF =(d)/(dt)[hati (2 cos t) + hatj( 2 sin t)]`
with `" " |vecF| = sqrt([(-2 sin t)^(2)+ (2 cos t)^(2)])=2` Now as `" "vecF. vecp = F_(x) p_(x) + F_(y)p_(y)`
`" "(-2 sin t) (2 cos t) + (2 cos t) (2 sin t) =0`
`" " theta= cos^(-1)[(vecF.vecp)/(Fp)]`
`" " = cos^(-1)[(0)/(2 xx2)] = cos ^(-1) 0`
`i.e.," "theta= 90^(@), " "i.e., vecF and vecp` are orthogonal.