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)]

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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°

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Correct option: (D) 90°

Explanation:

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°

by (65.5k points)

Correct option(d) θ=90

Explanation:

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

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

cosθ=0.

where θ is the angle between the force and momentum.

Hence, it is implied that

θ=90