Question

A particle moves in x – y plane under the action of the force $\overrightarrow{F}$ such that the value of the linear momentum $\overrightarrow{P}$ at any time is ${{P}_{X}}=2\cos t$ and ${{P}_{Y}}=\sin t$. If the angle between $\overrightarrow{F}$ and $\overrightarrow{P}$ is $45{{k}^{o}}$ then find the value of k.

Answer 1

Hint: To find the angle between force vector and the linear momentum first we will find force vector by differentiating linear momentum vector after that we will use below formula to find the angle between them $\overrightarrow{F}\times \overrightarrow{P}=\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|\cos \theta $

Formula used:
$\overrightarrow{F}=\dfrac{d\overrightarrow{P}}{dt}$
And
  $\overrightarrow{F}\times \overrightarrow{P}=\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|\cos \theta $

Complete step by step solution:
$\to $It is given that linear momentum $\overrightarrow{P}$ at any time is ${{P}_{X}}\text{ and }{{P}_{Y}}$ so we can write as
$\overrightarrow{P}={{P}_{X}}\widehat{i}+{{P}_{Y}}\widehat{j}$
Where $\widehat{i}$ is used for the x vector
And $\widehat{j}$ is used for the y – vector
Now,
$\overrightarrow{P}=2\cos t\widehat{i}+2\sin t\widehat{j}.....(1)$

$\to $Now we know that force is equal to the rate of change of momentum with respect to time
Now force,
$\overrightarrow{F}=\dfrac{d\overrightarrow{P}}{dt}.....\left( 2 \right)$
Where, $\overrightarrow{F}$ = force vector
$\overrightarrow{P}$ = momentum
t = time

$\to $Now put the value of equation (1) in equation (2)
$\begin{align}
  & \Rightarrow \overrightarrow{F}=\dfrac{d}{dt}\left( 2\cos t\widehat{i}+2\sin t\widehat{j} \right) \\
 & \therefore \overrightarrow{F}=\dfrac{d}{dt}\left( 2\cos t \right)\widehat{i}+\dfrac{d}{dt}\left( 2\sin t \right)\widehat{j} \\
\end{align}$

$\to $Now by differentiating we get the force,
$\overrightarrow{F}=-2\sin t\widehat{i}+2\cos t\widehat{j}.......(3)$

$\to $ Now to find the angle between \[\overrightarrow{F}\text{ and }\overrightarrow{P}\] we will use below equation
$\therefore \overrightarrow{F}.\overrightarrow{P}=\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|\cos \theta $
Where $\theta $ is an angle between $\overrightarrow{F}$ and $\overrightarrow{P}$

Now,
$\cos \theta =\dfrac{\overrightarrow{F}.\overrightarrow{P}}{\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|}......(4)$

Now let’s value of equation (1) and equation (3) in equation (4)
$\begin{align}
  & \Rightarrow \cos \theta =\left( \dfrac{-2\sin t\widehat{i}+2\cos t\widehat{j}}{\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|} \right).\left( 2\cos t\widehat{i}+2\sin t\widehat{j} \right) \\
 & \therefore \cos \theta =\dfrac{-4\sin t\cos t+4\sin t\cos t}{\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|} \\
\end{align}$
Here the value of ${{\widehat{i}}^{2}}\text{ and }{{\widehat{j}}^{2}}$ will be equal to 1.

Now,
$\begin{align}
  & \Rightarrow \cos \theta =\dfrac{0}{\left| \overrightarrow{F} \right|\left| \overrightarrow{P} \right|} \\
 & \Rightarrow \cos \theta =0 \\
 & \therefore \theta ={{90}^{o}} \\
 & \Rightarrow \theta ={{45}^{o}}\times 2 \\
\end{align}$

By comparing with
$\theta ={{45}^{o}}k$
The value of the k will be 2.

Additional information:
Definition of linear momentum:
“Linear momentum of an object or body can be defined as a product of mass and its velocity. It is denoted by symbol P.”
Formula for linear momentum is given as,
$P=mv$
Where,
P is linear momentum of the body
m is mass of the body
And v is velocity of same body
Linear momentum is vector quantity and direction is always the same in the direction of velocity.

Note:
If we do not put the values of ${{\widehat{i}}^{2}}$ and the ${{\widehat{j}}^{2}}$ which is equal to 1 then we won’t be able to find the correct solution and also we can get confused in the steps which came after equation (4) in above shown method.