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, $KE =\frac{1}{2} m v^{2}$
$=\frac{1}{2} m\left(\frac{p^{2}}{m^{2}}\right)=\frac{1}{2 m} p^{2}$
$\Rightarrow KE \propto p^{2} \,\,\,\left(\because \frac{1}{2 m}=\right.$ 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 $P$,
$v_{y}=(u \sin \theta-g t)$
and$v_{x}=u \cos \theta$
Resultant velocity at point $P$,
$\overrightarrow{ v }=v_{y} \hat{ j }+v_{x} \hat{ i }$
$=(u \sin \theta-g t) \hat{ j }+u \cos \theta \hat{ i }$
$|\overrightarrow{ v }|=\sqrt{(u \cos \theta)^{2}+(u \sin \theta-g t)^{2}}$
$=\sqrt{u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta+g^{2} t^{2}-2 u g t \sin \theta}$
$\therefore =\sqrt{u^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+g^{2} t^{2}-2 u g t \sin \theta}$
$KE =\frac{1}{2} m\left(u^{2}+g^{2} t^{2}-2 u g t \sin \theta\right)$
$\Rightarrow KE \propto t^{2}$
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$v^{2}=\left(u^{2}-2 g y\right)$
$\therefore KE =\frac{1}{2} m\left(u^{2}-2 g y\right)$
$KE =-m g y+\frac{1}{2} m u^{2}$
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.