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

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Laws of Motion

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 the momentum is

Answer: (D)

The force is given by

F = \frac{d p}{d t}

F = A \frac{d}{d t} [\hat{i} cos (k t) - \hat{j} sin (k t)] = A [- \hat{i} k sin (k t) - \hat{j} k cos (k t)] = A k [- \hat{i} sin (k t) - \hat{j} cos (k t)]

Now check if

F \cdot p = 0

F \cdot p = A k [- \hat{i} sin (k t) - \hat{j} cos (k t)] \cdot A [\hat{i} cos (k t) - \hat{j} sin (k t)]

= A^{2} k [- sin (k t) cos (k t) + cos (k t) sin (k t)] = 0

Since the dot product is zero, the force and momentum are perpendicular, i.e., the angle between them is 90°.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option