The correct option is D 90∘
Given,
→p(t)=Acos(kt)^i−Asin(kt)^j .....(1)
As we know, →F(t)=d→pdt
→F(t)=d→pdt
=−Aksin(kt)^i−Akcos(kt)^j ....(2)
We know, →A.→B=|A| |B|cosθ
[θ is angle between →A and →B]
Then, →F(t).→p(t)=|→F(t)| |→p(t)|cosθ
Here,
|→F(t)|=√(−Aksin(kt))2+(−Akcos(kt))2=Ak
|→p(t)|=√(Acos(kt))2+(−Asin(kt))2=A
⇒→F(t).→p(t)=(Ak)(A)cosθ...(3)
But we also know,
→F(t).→p(t)=[−Aksin(kt)^i−Akcos(kt)^j].[Acos(kt)^i−Asin(kt)^j]=0...(4)
From (3) and (4),
0=(Ak) (A)cosθ
⇒cosθ=0
∴ θ=90∘
Angle between the force and momentum is 90∘.