- Home
- Standard 11
- Physics
4-1.Newton's Laws of Motion
normal
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 $P_x = 2cost$ , $P_y = 2sint$ . The angle between $\,\vec F$ and $\,\vec P$ at a given time $t$ , will be
A$\theta = {0^o}$
B$\theta = {30^o}$
C$\theta = {90^o}$
D$\theta = {180^o}$
Solution
$P_{x}=2 \cos t, P_{y}=2 \sin t$
$\therefore \overrightarrow{P}=2 \cos t \hat i+2 \sin t \hat j$
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{d}\overrightarrow{\mathrm{P}}}{\mathrm{dt}}=-2 \sin t \hat {i}+2 \cos t \hat j$
$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}=0 \therefore \theta=90^{\circ}$
$\therefore \overrightarrow{P}=2 \cos t \hat i+2 \sin t \hat j$
$\overrightarrow{\mathrm{F}}=\frac{\mathrm{d}\overrightarrow{\mathrm{P}}}{\mathrm{dt}}=-2 \sin t \hat {i}+2 \cos t \hat j$
$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}=0 \therefore \theta=90^{\circ}$
Standard 11
Physics
Similar Questions
normal
