Question

$A$ particle moves in the X-$Y$ plane under the influence of a force such that its linear momentum is $\overrightarrow{p}\left(t\right)=A\left[\hat{i}\cos \right(kt)-\hat{j}\sin (kt\left)\right]$, 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

Answer: (D)

${90}^o$

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

The force acting on a particle is given as $F=\frac{d\overrightarrow{p}}{dt}=Ak[-\hat{i}sin(kt)-\hat{j}cos(kt\left)\right]$

Thus $\overrightarrow{F}.\overrightarrow{p}=A{k}^2[-sin(kt).cos(kt)+sin(kt).cos(kt\left)\right]=0$

$⟹$ Angle between the vectors is ${90}^{\circ }$.