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

Answer: (A)

$\frac{\pi}{2}$

Answer 2

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 $\frac{\pi}{2}$.