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Question

A particle moves in the xy plane under the influence of a force such that its linear momentum is p(t)=A[cos(kt)^isin(kt)^j], where A and k are constants. The angle between the force and the momentum is :

A
0
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B
30
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C
45
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D
90
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Solution

The correct option is D 90
Given,
p(t)=Acos(kt)^iAsin(kt)^j .....(1)
As we know, F(t)=dpdt
F(t)=dpdt
=Aksin(kt)^iAkcos(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)^iAkcos(kt)^j].[Acos(kt)^iAsin(kt)^j]=0...(4)

From (3) and (4),
0=(Ak) (A)cosθ
cosθ=0
θ=90
Angle between the force and momentum is 90.

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