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A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t)=A[\hat{\mathbf{i}} \cos k t-\hat{\mathbf{j}} \sin k t]$, where $A$ and $k$ are constants. The angle between the force and the momentum is
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1856 Upvotes
Verified Answer
The correct answer is:
$90^{\circ}$
$90^{\circ}$
$$
\begin{aligned}
& \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\
& \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}
\end{aligned}
$$
Since,
F $\cdot \mathbf{p}=0$
$\therefore$ Angle between $\mathbf{F}$ and pshould be $90^{\circ}$.
$\therefore$ Correct option is $(\mathrm{d})$.
\begin{aligned}
& \mathbf{F}=\frac{d \mathbf{p}}{d t}=-k A \sin k t \hat{\mathbf{i}}-k A \cos k t \hat{\mathbf{j}} \\
& \mathbf{p}=A \cos k t \hat{\mathbf{i}}-A \sin k \hat{\mathbf{j}}
\end{aligned}
$$
Since,
F $\cdot \mathbf{p}=0$
$\therefore$ Angle between $\mathbf{F}$ and pshould be $90^{\circ}$.
$\therefore$ Correct option is $(\mathrm{d})$.
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