Question

A particle moves in the x-y plane under the influence of a force \( \vec{F} \) such that its linear momentum is \[ \vec{P}(t) = \hat{i} \cos(kt) - \hat{j} \sin(kt). \] If \( k \) is constant, the angle between \( \vec{F} \) and \( \vec{P} \) will be:

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{3} \)

Answer 1

The Correct Option is A

Given:
\[ \vec{P}(t) = \cos(kt) \, \hat{i} - \sin(kt) \, \hat{j}, \quad |\vec{P}| = 1. \]
The linear momentum \( \vec{P} \) is given by:

\[ \vec{P} = m\vec{v} \implies \hat{P} = \vec{v}, \]

where \( \hat{P} \) is the unit vector in the direction of \( \vec{P} \).

Step 1: Calculating the Velocity Vector
Differentiating \( \vec{P}(t) \) with respect to time to find the velocity vector:

\[ \vec{v} = \frac{d\vec{P}}{dt} = -k\sin(kt) \, \hat{i} - k\cos(kt) \, \hat{j}. \]

The acceleration vector \( \vec{a} \) is given by:

\[ \vec{a} = \frac{d\vec{v}}{dt} = -k^2\cos(kt) \, \hat{i} + k^2\sin(kt) \, \hat{j}. \]

Step 2: Calculating the Angle Between \( \vec{F} \) and \( \vec{P} \)
The force \( \vec{F} \) is given by Newton’s second law:

\[ \vec{F} = m\vec{a}. \]

To find the angle \( \theta \) between \( \vec{F} \) and \( \vec{P} \), we use the dot product:

\[ \cos\theta = \frac{\vec{F} \cdot \vec{P}}{|\vec{F}||\vec{P}|}. \]

Substituting the expressions for \( \vec{F} \) and \( \vec{P} \):

\[ \vec{F} \cdot \vec{P} = (-k^2\cos(kt))\cos(kt) + (k^2\sin(kt))\sin(kt) = -k^2(\cos^2(kt) + \sin^2(kt)) = -k^2. \]

Since \( |\vec{F}| = k^2 \) and \( |\vec{P}| = 1 \), we have:

\[ \cos\theta = \frac{-k^2}{k^2} = -1 \implies \theta = \frac{\pi}{2}. \]

Therefore, the angle between \( \vec{F} \) and \( \vec{P} \) is \( \frac{\pi}{2} \).