Question

A particle moves in the \( xy \)-plane under the action of a force \( F \) such that the components of its linear momentum \( p \) at any time \( t \) are \( p_x = 2 \cos t \), \( p_y = \sin t \). The angle between \( F \) and \( p \) at time \( t \) is:

• A. \( 90^\circ \)
• B. \( 0^\circ \)
• C. \( 180^\circ \)
• D. \( 30^\circ \)

Hint:

When calculating the angle between the force and momentum vectors, use the dot product formula and simplify the expressions carefully.

Answer 1

The Correct Option is A

Step 1: Expression for Force.
The force \( F \) on the particle is related to the rate of change of momentum: \[ F = \frac{dp}{dt} \] For the \( x \)-component of momentum \( p_x = 2 \cos t \), we have: \[ \frac{dp_x}{dt} = -2 \sin t \] For the \( y \)-component of momentum \( p_y = \sin t \), we have: \[ \frac{dp_y}{dt} = \cos t \] Thus, the force components are: \[ F_x = -2 \sin t, \quad F_y = \cos t \]

Step 2: Calculate the angle between \( F \) and \( p \).
The angle \( \theta \) between the vectors \( F \) and \( p \) is given by the formula: \[ \cos \theta = \frac{F_x p_x + F_y p_y}{|F| |p|} \] Substitute the values for \( F_x, F_y, p_x, \) and \( p_y \): \[ \cos \theta = \frac{(-2 \sin t)(2 \cos t) + (\cos t)(\sin t)}{\sqrt{(-2 \sin t)^2 + (\cos t)^2} \cdot \sqrt{(2 \cos t)^2 + (\sin t)^2}} \] Simplifying: \[ \cos \theta = \frac{-4 \sin t \cos t + \cos t \sin t}{\sqrt{4 \sin^2 t + \cos^2 t} \cdot \sqrt{4 \cos^2 t + \sin^2 t}} = 0 \] This implies \( \theta = 90^\circ \).