Question

A particle is moving in X-Y plane under the action of a force $\overrightarrow{F}$ such that at some instant $t$ the components of its linear momentum $\overrightarrow{p}$ are $p_x=2\cos t$ and $p_y=2\sin t$. At this instant the angle between $\overrightarrow{F}$ and $\overrightarrow{p}$ is

• A. ${90}^o$
• B. $0^o$
• C. ${180}^o$
• D. ${30}^o$

Answer 1

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

Given: $Px=2\cos t$,

$Py=2\sin t$

$P=Pxi+Pyj=2\cos ti+2\sin tj$

Now according to the Newton's second law we can calculate components of forces using respective momentum component as: $F=dy/dt$

$Fx=-2\sin t$

$Fy=2\cos t$

$F=Fxi+Fyj=-2\sin ti+2\cos tj$

Lets try to find out the dot product of F and P,

$FP=0$ ……………$\left(3\right)$

but we do know that dot product of force and momentum can also be given in the form of angle between them as

$FP=FP\cos \theta$ …………..$\left(1\right)$

Comparing both equations

$\Rightarrow \theta ={90}^o$.