Question

A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\overrightarrow{P}\left(t\right)=A[\hat{i}\cos \left(kt\right)-\hat{j}\sin \left(kt\right)]$, where $A$ and $k$ are constants. the angle between the force and the momentum is:

• A. $0^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

correct expression is,

$\overrightarrow{p\left(t\right)}=A\left[\hat{i}\cos \right(kt)-\hat{j}\sin (kt\left)\right]$

$\overrightarrow{f}=\frac{d\overrightarrow{p}}{dt}$

$=\frac{d}{dt}\left[A\left\{\hat{i}\cos \right(kt)-\hat{j}\sin (kt\left)\right\}\right]$

$AK[-\hat{i}\sin (kt)-\hat{j}\cos (kt\left)\right]$

Now,

$\overrightarrow{f}.\overrightarrow{p}=AK[-\hat{i}(kt)-\hat{j}\cos (kt\left)\right]$

$A\left[\hat{i}\cos \right(kt)-\hat{j}\sin (kt\left)\right]$

$=AK[-\sin (kt\left)\cos \right(kt)+\cos (kt).\sin (kt\left)\right]$

Since $(\hat{i}.\hat{i}=\hat{j}.\hat{j}=0)$

angle between force and momentum $={90}^{\circ }$