Question

A particle moves in the $xy$ - plane under the action of a force $\vec{ F }$ such that the components of its linear momentum $\vec{ P }$ 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. $90^{\circ }$
• B. $0^{\circ}$
• C. $180^{\circ}$
• D. $30^{\circ}$

Answer 1

Answer: (A)

$90^{\circ }$

Answer 2

Resultant momentum $p=p_{x} \hat{i}+p_{y} \hat{j}$ Given $p_{x}=2 \cos t, p_{y}=2 \sin t$ $\therefore$ momentum $\vec{p}=2 \cos t \hat{i}+2 \sin r \hat{j} \ldots$ (i) $\because$ Force = Rate of change of momentum $=\frac{d \vec{p}}{d t}$ $\therefore \vec{F}=-2 \sin t \hat{i}+2 \cos t \hat{j} \ldots$(ii) By $\cos \theta=\frac{\vec{F} \cdot \vec{p}}{|\vec{F}| \cdot|\vec{p}|}$ $\cos \theta=\frac{(2-\sin t \hat{i}+2 \cos t \hat{j}) \cdot(2 \cos t \hat{i}+2 \sin t \hat{j})}{\sqrt{4 \sin ^{2} t+4 \cos ^{2} t} \sqrt{4 \cos ^{2} t+4 \sin ^{2} t}}$ or $\cos \theta=0$ $\Rightarrow \theta=90^{\circ}$