Question

A particle moves in the $x-y$ plane under the influence of a force such that its linear momentum is $\overrightarrow{p\left(t\right)}=10(\cos 5t\hat{i}-\sin 5t\hat{j})$ . The angle between the force and the momentum (in degrees) is

Answer 1

Given, linear momentum of particle: $\overrightarrow{p}\left(t\right)=10(\cos 5t\hat{i}-\sin 5t\hat{j})---\left(1\right)$
Let the angle between the force and the momentum be $\theta$.
We know from Newton's second law of motion that Force, $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$
$\Rightarrow \overrightarrow{F}=10(-5\sin 5t\hat{i}-5\cos 5t\hat{j})--\left(2\right)$
Also, writing the two vectors in component form, we get
$\overrightarrow{F}=F_1\hat{i}+F_2\hat{j}\text{and}\overrightarrow{p}=p_1\hat{i}+p_2\hat{j}$
Angle between the two vectors is given by
$\cos \theta =\frac{F_1{p}_1+F_2{p}_2}{\sqrt{F_1^2+F_2^2}\sqrt{p_1^2+p_2^2}}$
Substituting the data from (1) and (2) in (3),
$\cos \theta =\frac{(-50\sin 5t).\left(10\cos 5t\right)+(-50\cos 5t).(-10\sin 5t)}{\sqrt{(-50\sin 5t{)}^2+(-50\cos 5t{)}^2}.\sqrt{(10\cos 5t{)}^2+(-10\sin 5t{)}^2}}$
$\Rightarrow \text{cos}\theta =0$
or $\theta ={90}^{\circ }$