Question

A particle moves in the X-Y plane under the influence of a force such that its linear momentum is $\overrightarrow{P}(t)=A\left[ \widehat{i}\cos \left( kt \right)-\widehat{j}\sin \left( kt \right) \right]$ , where $A$ and $k$ are constants. The angle between the force and the momentum is:
$\begin{align}
  & \text{A}\text{. }{{0}^{\circ }} \\
 & \text{B}\text{. }{{30}^{\circ }} \\
 & \text{C}\text{. }{{45}^{\circ }} \\
 & \text{D}\text{. }{{90}^{\circ }} \\
\end{align}$

Answer 1

Hint: The force applied by the external agent changes the momentum of the particle. Using the expression for Newton’s second law of motion, force is expressed as the rate of change of momentum. We will calculate the angle between the force and the momentum vector by calculating the dot product between them.

Complete step-by-step answer:
Force can be a result of actually touching between the objects or without physical contact between the objects. Force can lead to change in speed, or shape and size, of a body. Momentum is the result of unbalanced force acting on a body. It is the quantity of motion of a moving body, measured as a product of its mass and velocity. Newton’s second law states that the rate of a change of a particle’s momentum $p$ is given by the force acting on the particle, that is, $F=\dfrac{dp}{dt}$. If there is no force acting on the particle, then, $\dfrac{dp}{dt}=0$ and $p$ must be constant, or conserved.
Given that,
$\overrightarrow{P}(t)=A\left[ \widehat{i}\cos \left( kt \right)-\widehat{j}\sin \left( kt \right) \right]$
We know force on a particle is given as,
$\overrightarrow{F}=\dfrac{d\overrightarrow{P}}{dt}$
$\overrightarrow{F}=\dfrac{d}{dt}\left[ A\left\{ \widehat{i}\cos \left( kt \right)-\widehat{j}\sin \left( kt \right) \right\} \right]$
Or,
$\overrightarrow{F}=Ak\left[ -\widehat{i}\sin \left( kt \right)-\widehat{j}\cos \left( kt \right) \right]$
Now,
$\overrightarrow{F}\cdot \overrightarrow{P}=Ak\left[ -\widehat{i}\sin \left( kt \right)-\widehat{j}\cos \left( kt \right) \right]\cdot A\left[ \widehat{i}\cos \left( kt \right)-\widehat{j}\sin \left( kt \right) \right]$
We get,
$\overrightarrow{F}\cdot \overrightarrow{P}=Ak\left[ -\sin \left( kt \right)\cos \left( kt \right)+\cos \left( kt \right)\sin \left( kt \right) \right]$
As we know,
$\widehat{i}\cdot \widehat{i}=\widehat{j}\cdot \widehat{j}=0$
$\begin{align}
  & \cos \theta =0 \\
 & \theta ={{90}^{\circ }} \\
\end{align}$
Therefore,
Angle between force and momentum is ${{90}^{\circ }}$

So, the correct answer is “Option D”.

Note: To find the angle between the two vectors, we can find the dot product between them. If the vectors are perpendicular, that is the angle between them is the right angle, the cosine of the angle is zero and hence the dot product is zero.