Question

A particle moves in X - Y plane under the action of force vector given the components of momentum $p_x$ =$2cos\theta$ $p_y$ =$2sin\theta$ . Then the angle between force vector and momentum at a given time t is

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

Answer 1

The correct option is D $\theta ={90}^o$

Given that,

$P_x=2\cos \theta$

$P_y=2\sin \theta$

Now,

$P=P_x\hat{i}+p_y\hat{j}$

Now, put the value

$P=2\cos \theta \hat{i}+2\sin \theta \hat{j}$

Now, from newton’s second law

$F_x=\frac{d\left(p_x\right)}{dt}$

$F_x=-2\sin \theta$

$F_y=2\cos \theta$

Now,

$F=F_x\hat{i}+F_y\hat{j}$

$F=-2\sin \theta \hat{i}+2\cos \theta \hat{j}$

Now, the dot product of $\overrightarrow{F}$ and $\overrightarrow{P}$

So,

$\overrightarrow{F}\cdot \overrightarrow{P}=0$

$\overrightarrow{F}\cdot \overrightarrow{P}=FP\cos \theta$

$\cos \theta =0$

$\theta ={90}^0$

Hence, the angle between $\overrightarrow{F}$ and $\overrightarrow{P}$ is ${90}^0$