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:
- \( \frac{\pi}{2} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{4} \)
- \( \frac{\pi}{3} \)
The Correct Option is A
Approach Solution - 1
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 \( \frac{\pi}{2} \).
Approach Solution -2
The linear momentum of the particle is given as:
\[ \vec{P}(t) = \hat{i} \cos(kt) - \hat{j} \sin(kt) \]
where \( k \) is a constant.
Step 2: Relation between force and momentum.
Force is the time derivative of momentum:
\[ \vec{F} = \frac{d\vec{P}}{dt} \]
Step 3: Differentiate the given momentum.
\[ \frac{d\vec{P}}{dt} = \frac{d}{dt}(\hat{i} \cos(kt) - \hat{j} \sin(kt)) \]
\[ \vec{F} = -k \hat{i} \sin(kt) - k \hat{j} \cos(kt) \]
Step 4: Check the dot product between \(\vec{F}\) and \(\vec{P}\).
\[ \vec{F} \cdot \vec{P} = (-k \sin(kt))(\cos(kt)) + (-k \cos(kt))(-\sin(kt)) \]
\[ \vec{F} \cdot \vec{P} = -k \sin(kt)\cos(kt) + k \cos(kt)\sin(kt) = 0 \]
Step 5: Interpret the result.
Since \(\vec{F} \cdot \vec{P} = 0\), it means the force is perpendicular to the momentum vector.
Final Answer: \( \frac{\pi}{2} \)
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