Question

A particle is projected up from a point at an angle $\theta$ with the horizontal direction. At any time $t$, if $p$ is the linear momentum, $y$ is the vertical displacement, $x$ is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy $KE$ of the particle is

• A. graph (A)
• B. graph (B)
• C. graph (C)
• D. graph (D)

Answer 1

Answer: (A)

Momentum, $p=m\cdot v$

$\Rightarrow v=\left(\frac{p}{m}\right)$
Kinetic energy, $KE=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m\left(\frac{p^2}{m^2}\right)=\frac{1}{2m}{p}^2$
$\Rightarrow KE\propto p^2(\because \frac{1}{2m}=$ constant $)$
Hence, the graph between $KE$ and $p^2$ will be linear as shown below

Now, kinetic energy $KE=\frac{1}{2}m{v}^2$ The velocity component at point $P$,
$v_y=(u\sin \theta -gt)$
and$v_x=u\cos \theta$
Resultant velocity at point $P$,
$\overrightarrow{v}=v_y\hat{j}+v_x\hat{i}$
$=(u\sin \theta -gt)\hat{j}+u\cos \theta \hat{i}$
$|\overrightarrow{v}|=\sqrt{(u\cos \theta {)}^2+(u\sin \theta -gt{)}^2}$
$=\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta +g^2{t}^2-2ugt\sin \theta }$
$\therefore =\sqrt{u^2\left({\cos }^2\theta +{\sin }^2\theta \right)+g^2{t}^2-2ugt\sin \theta }$
$KE=\frac{1}{2}m\left(u^2+g^2{t}^2-2ugt\sin \theta \right)$
$\Rightarrow KE\propto t^2$
Hence, graph will be parabolic with intercept on $y$-axis.
Hence, the graph between $KE$ and $t$

Now, in case of height
$KE=\frac{1}{2}m\left(v^2\right)$
and$v^2=\left(u^2-2gy\right)$
$\therefore KE=\frac{1}{2}m\left(u^2-2gy\right)$
$KE=-mgy+\frac{1}{2}m{u}^2$
Intercept on $y$ -axis $=\frac{1}{2}m{u}^2$

Now, $KE=\frac{1}{2}m{v}^2$
$KE=\frac{1}{2}m{\left(\frac{x}{t}\right)}^2$

$KE\propto x^2$. Thus graph between $KE$ and $x$ will be parabolic.