Question

A particle moves under the influence of a force in the XY-plane such that the components of its linear momentum $\vec{p}$ at any time $t$ are given by $p_x = p \sin t$ and $p_y = p \cos t$. The angle between $\vec{F}$ and $\vec{p}$ at that time is

• A. $45^\circ$
• B. $60^\circ$
• C. $30^\circ$
• D. $90^\circ$
• E. $0^\circ$

Hint:

If force is the time derivative of momentum, their directions can be determined using differentiation.

Answer 1

The Correct Option is D

Step 1: Force is given by \[ \vec{F} = \frac{d\vec{p}}{dt} \] Step 2: Differentiating momentum components: \[ F_x = p \cos{t}, \quad F_y = -p \sin{t} \] Step 3: The force vector $\vec{F} = (p \cos{t}, -p \sin{t})$ is perpendicular to the momentum vector $\vec{p} = (p \sin{t}, p \cos{t})$.
Step 4: Therefore, the angle between $\vec{F}$ and $\vec{p}$ is $90^\circ$. \bigskip