Search any question & find its solution
Question:
Answered & Verified by Expert
A particle moves in the $x-y$ plane under the action of a force $\vec{F}$ such that the value of its linear momentum ( $\vec{P}$ at anytime $t$ is $P_x=2 \cos t, p_y=2 \sin t$. The angle $\theta$ between $\vec{F}$ and $\vec{P}$ at a given time $t$. will be
Options:
Solution:
2109 Upvotes
Verified Answer
The correct answer is:
$\theta=90^{\circ}$
$P_x=2 \cos t$, $P_y=2 \sin t$ $\therefore$ $\vec{P}=2 \cos t \hat{i}+2 \sin t \hat{j}$
$\vec{F}=\frac{\vec{dP}}{\mathrm{dt}}$=-$-2 \sin t \hat{i}+2 \cos t \hat{j}$
$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}$=0. $\theta=90^{\circ}$
$\vec{F}=\frac{\vec{dP}}{\mathrm{dt}}$=-$-2 \sin t \hat{i}+2 \cos t \hat{j}$
$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}$=0. $\theta=90^{\circ}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.