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4-1.Newton's Laws of Motion
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A particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\overrightarrow{\mathrm{p}}(t)=A[\hat{\mathrm{i}} \cos (\mathrm{kt})-\hat{\mathrm{j}} \sin (\mathrm{kt})]$, where $A$ and $k$ are constants. The angle between the force and the momentum is
A$0^{\circ}$
B$30^{\circ}$
C$45^{\circ}$
D$90^{\circ}$
Solution
The correct option is $D 90^{\circ}$
Given,
$\vec{p}(t)=A \cos (k t) \hat{i}-A \sin (k t) \hat{j}$
As we know, $\vec{F}(t)=\frac{d \vec{p}}{d t}$
$\vec{F}(t)=\frac{d \vec{p}}{d t} $
$=-A k \sin (k t) \hat{i}-A k \cos (k t) \hat{j}$
We know, $\vec{A} \cdot \vec{B}=|A||B| \cos \theta$
[ $\theta$ is angle between $\vec{A}$ and $\vec{B}$ ]
Then, $\vec{F}(t) \cdot \vec{p}(t)=|\vec{F}(t)||\vec{p}(t)| \cos \theta$
Here,
$|\vec{F}(t)|=\sqrt{(-A k \sin (k t))^2+(-A k \cos (k t))^2}=A k$
$|\vec{p}(t)|=\sqrt{(A \cos (k t))^2+(-A \sin (k t))^2}=A$
$\Rightarrow \vec{F}(t) \cdot \vec{p}(t)=(A k)(A) \cos \theta \ldots(3)$
But we also know,
$\vec{F}(t) \cdot \vec{p}(t)=[-A k \sin (k t) \hat{i}-A k \cos (k t) \hat{j}] \cdot[A \cos (k t) \hat{i}-A \sin (k t) \hat{j}]$
$=0 \ldots(4)$
From (3) and (4),
$0=(A k)(A) \cos \theta$
$\Rightarrow \cos \theta=0$
$\therefore \theta=90^{\circ}$
Angle between the force and momentum is $90^{\circ}$.
Given,
$\vec{p}(t)=A \cos (k t) \hat{i}-A \sin (k t) \hat{j}$
As we know, $\vec{F}(t)=\frac{d \vec{p}}{d t}$
$\vec{F}(t)=\frac{d \vec{p}}{d t} $
$=-A k \sin (k t) \hat{i}-A k \cos (k t) \hat{j}$
We know, $\vec{A} \cdot \vec{B}=|A||B| \cos \theta$
[ $\theta$ is angle between $\vec{A}$ and $\vec{B}$ ]
Then, $\vec{F}(t) \cdot \vec{p}(t)=|\vec{F}(t)||\vec{p}(t)| \cos \theta$
Here,
$|\vec{F}(t)|=\sqrt{(-A k \sin (k t))^2+(-A k \cos (k t))^2}=A k$
$|\vec{p}(t)|=\sqrt{(A \cos (k t))^2+(-A \sin (k t))^2}=A$
$\Rightarrow \vec{F}(t) \cdot \vec{p}(t)=(A k)(A) \cos \theta \ldots(3)$
But we also know,
$\vec{F}(t) \cdot \vec{p}(t)=[-A k \sin (k t) \hat{i}-A k \cos (k t) \hat{j}] \cdot[A \cos (k t) \hat{i}-A \sin (k t) \hat{j}]$
$=0 \ldots(4)$
From (3) and (4),
$0=(A k)(A) \cos \theta$
$\Rightarrow \cos \theta=0$
$\therefore \theta=90^{\circ}$
Angle between the force and momentum is $90^{\circ}$.
Standard 11
Physics
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