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

Question

A particle moves in the XY-plane under the influence of a force such that its linear momentum is, vec p ( t )= A [ hat i cos ( kt )- hat j sin ( kt )] where A and k are constants. The angle between the force and momentum is :- (A) 0° (B) 30° (C) 45° (D) 90°

Tardigrade Admin · Accepted Answer

vec p ( t )= A ( hat i cos kt - hat j sin kt ) vec F =( d / dt )[ vec p ( t )]= Ak (- hat i sin kt - hat j cos kt ) vec F ⋅ vec p = A 2 k (- cos kt sin kt + sin kt cos kt )=0

Q. A particle moves in the $X Y$ -plane under the influence of a force such that its linear momentum is,
$p (t) = A [\hat{i} cos (k t) - \hat{j} sin (k t)]$
where $A$ and $k$ are constants. The angle between the force and momentum is :-

$0^{\circ}$

$3 0^{\circ}$

$4 5^{\circ}$

$9 0^{\circ}$

$p (t) = A (\hat{i} cos k t - \hat{j} sin k t)$
$F = \frac{d}{d t} [p (t)] = A k (- \hat{i} sin k t - \hat{j} cos k t)$
$F \cdot p = A^{2} k (- cos k t sin k t + sin k t cos k t) = 0$

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

0∘

30∘

45∘

90∘