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°
• B. 0°
• C. 180°
• D. 30°

Answer 1

Answer: (A)

90°

Answer 2

Given that

$\overrightarrow{p} = p_{x}\widehat{i} + p_{y}\widehat{j} = 2\cos t\mspace{6mu}\widehat{i} + 2\sin t\mspace{6mu}\widehat{j}$

∴ $\overrightarrow{F} = \frac{d\overrightarrow{p}}{dt} = - 2\sin t\mspace{6mu}\widehat{i} + 2\cos t\mspace{6mu}\widehat{j}$

Now, $\overrightarrow{F}.\overrightarrow{p} = 0$ i.e. angle between $\overrightarrow{F}\mspace{6mu}\text{and }\overrightarrow{p}\mspace{6mu}$is 90°.