Question

A particle moves in the $XY$ Plane under the action of a force $F$ such that the components of its linear momentum $P$ at any time $t$ are $P_x=2\cos t$, $P_y=2\sin t$. The angle between $F$ and $P$ at time $t$ is

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

Answer 1

The correct option is A

${90}^0$

Step 1: Given data.

Force under $XY$ plane$=F$

Linear momentum along $x-axis$, $P_x=2\cos t$

Linear momentum along $y-axis$, $P_y=2\sin t$

Let the angle between the force $F$ and linear momentum $P$ at the time $t$ be $\theta$

Step 2: Finding the angle between the force $F$ and linear momentum $P$ at the time $t$.

Since Resultant linear momentum along with $x$ and $y$ components.

$\overrightarrow{P}=P_x+P_y$

$\Rightarrow$$\overrightarrow{P}=2\cos t\hat{i}+2\sin t\hat{j}$ ……$\left(i\right)$

Also, we know that force is defined as the rate of change in momentum.

Therefore,

$\overrightarrow{F}=\frac{dP}{dt}$

$\Rightarrow$$\overrightarrow{F}=\frac{d\left(2\cos t\hat{i}+2\sin t\hat{j}\right)}{dt}$ $\left[\because Fromequation\left(i\right)\right]$

$\Rightarrow$$\overrightarrow{F}=-2\sin t\hat{i}+2\cos t\hat{j}$ …….$\left(ii\right)$

Now,

According to the vector law of dot product.

$\overrightarrow{F}.\overrightarrow{P}=FP\cos \left(\theta \right)$

$\Rightarrow$$\cos \left(\theta \right)=\frac{\overrightarrow{F}.\overrightarrow{P}}{FP}$

$\Rightarrow$$\cos \left(\theta \right)=\frac{-4\cos t\sin t+4\cos t\sin t}{FP}$ $\left[\because Onsolvingequation\left(i\right)and\left(ii\right)\right]$

$\Rightarrow$$\cos \left(\theta \right)=0$

$\Rightarrow$ $\theta ={90}^0$

Hence, option A is correct.