A particle moves in mathrm{x}-mathrm{y} plane under influence of a force vec{F} such that its linear

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4-1.Newton's Laws of Motion

medium

A particle moves in $\mathrm{x}-\mathrm{y}$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $\vec{P}(t)=\hat{i} \cos (k t)-\hat{j} \sin (k t)$. If $k$ is constant, the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{P}}$ will be :

A$\frac{\pi}{2}$

B$\frac{\pi}{6}$

C$\frac{\pi}{4}$

D$\frac{\pi}{3}$

(JEE MAIN-2024) (IIT-2007)

Solution

$\overrightarrow{\mathrm{P}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} ;|\overrightarrow{\mathrm{P}}|=1$
$\because \overrightarrow{\mathrm{P}}=\mathrm{m} \overrightarrow{\mathrm{v}}$
$\therefore \hat{\mathrm{P}}=\hat{\mathrm{v}}$
$\Rightarrow \hat{\mathrm{v}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}}$
$\hat{\mathrm{a}}=\frac{-\mathrm{k} \sin (\mathrm{kt}) \hat{\mathrm{i}}-\mathrm{k} \cos (\mathrm{kt}) \hat{\mathrm{j}}}{\mathrm{k}}$
$\Rightarrow \hat{\mathrm{a}}=-\sin k t \hat{\mathrm{i}}-\cos k \hat{\mathrm{t}}$
$\because \hat{\mathrm{F}}=\hat{\mathrm{a}}=-\sin \mathrm{kt} \hat{\mathrm{i}}-\cos k t \hat{\mathrm{j}}$
$\cos \theta=\frac{\hat{\mathrm{F}} \cdot \hat{\mathrm{P}}}{|\hat{\mathrm{F}}| \hat{\mathrm{P}} \mid}=-\frac{\sin k t \cos \mathrm{t}+\sin k t \cos \mathrm{t}}{1 \times 1}=0$
$\Rightarrow \theta=\frac{\pi}{2}$

Standard 11

Physics

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 ……….. $^o$

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A body of mass $ 2 \,kg$ is moving with a velocity $8\, m/s$ on a smooth surface. If it is to be brought to rest in $4 \,seconds$, then the force to be applied is ……… $N$

easy

A constant retarding force of $50\; N$ is applied to a body of mass $20\; kg$ moving initially with a speed of $15\; m/s$. How long (in $sec$) does the body take to stop ?

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The momentum $p$ (in $kg m / s )$ of a particle is varying with time $t$ (in $s )$ as $p=2+3 t^2$. The force acting on the particle at $t=3 s$ will be ……… $N$

easy

A constant force acting on a body of mass $3.0\; kg$ changes its speed from $2.0\; m s ^{-1}$ to $3.5\; m s ^{-1}$ in $25\; s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

medium

A particle moves in $\mathrm{x}-\mathrm{y}$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $\vec{P}(t)=\hat{i} \cos (k t)-\hat{j} \sin (k t)$. If $k$ is constant, the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{P}}$ will be :

Solution

Similar Questions

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 body of mass $ 2 \,kg$ is moving with a velocity $8\, m/s$ on a smooth surface. If it is to be brought to rest in $4 \,seconds$, then the force to be applied is ……… $N$

A constant retarding force of $50\; N$ is applied to a body of mass $20\; kg$ moving initially with a speed of $15\; m/s$. How long (in $sec$) does the body take to stop ?

The momentum $p$ (in $kg m / s )$ of a particle is varying with time $t$ (in $s )$ as $p=2+3 t^2$. The force acting on the particle at $t=3 s$ will be ……… $N$

A constant force acting on a body of mass $3.0\; kg$ changes its speed from $2.0\; m s ^{-1}$ to $3.5\; m s ^{-1}$ in $25\; s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

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