A particle moves in xy- plane under action of a force F such that components of its linear momentum

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4-1.Newton's Laws of Motion

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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 ........... $^o$

A

$90$

B

$0$

C

$180$

D

$30$

Solution

(a) Given that $\vec p = {p_x}\hat i + {p_y}\hat j = 2\cos t\;\hat i + 2\sin t\;\hat j$

$\vec F = \frac{{d\vec p}}{{dt}} = – 2\sin t\;\hat i + 2\cos t\;\hat j$

Now, $\vec F.\vec p = 0$

i.e. angle between $\vec F\;{\rm{and }}\,\vec p$ is $90^o.$

Standard 11

Physics

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