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+1 vote
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A Particle moves in the X–Y plane under the influence of a force such that its linear momentum is vector P(t) = A[i cos(kt) – jsin(kt)] where A and k are constants. The angle between the force and momentum is 

(A) 0° 

(B) 30° 

(C) 45° 

(D) 90°

2 Answers

+3 votes
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Best answer

Correct option: (D) 90°


P(t) = A [j˄ cos kt – j˄ sin kt]

F = [{dp(t)} / {dt}]

F = (d / dt) A [i˄ cos kt – j˄ sin kt]

= A [– i˄ sin kt (k)– j cos kt (k)]

F = Ak [– i˄ sin kt – j˄ cos kt]

angle between two vectors is θ

cos θ = {(F ∙ P) / (|F| |P|)}

= A [{(j˄ cos kt – j˄ sin kt) ∙ Ak (– i˄ sin kt – j˄ cos kt)} / {|F||P|}]

= A2 k [{– sin kt cos kt + sin kt cos kt} / {|F| |P|}]

 = A2k(0)

= 0

θ = cos–1(0)

θ = 90°   

+2 votes
by (65.5k points)

Correct option(d) θ=90


linear momentum p(t)=A[icos(kt)-jsin(kt)]

Since, the dot product is zero, we can conclude that


where θ is the angle between the force and momentum.

Hence, it is implied that


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