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, $p=m.v$
$\Rightarrow v=\left(\frac{p}{m}\right)$
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 $P$,
$\overrightarrow{v}=v_y\hat{j}+v_x\hat{i}$
$=(u\sin \theta -gt)\hat{j}+u\cos \theta \hat{i}$
$|\overrightarrow{v}|=\sqrt{{(ucos\theta )}^2+{(u\sin \theta -gt)}^2}$
$=\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta +g^2{t}^2-2ugt\sin \theta }$
$\therefore \sqrt{u^2({\cos }^2\theta +{\sin }^2\theta )+g^2{t}^2-2ugt\sin \theta }$
$KE=\frac{1}{2}m(u^2+g^2{t}^2-2ugt\sin \theta )$
$\Rightarrow KE\propto t^2$
Hence, graph will be parabolic with intercept on y-axis.
Hence, the graph between KE and t
Now, in case of height $KE=\frac{1}{2}m(v^2)$
and $v^2=(u^2-2gy)$
$\therefore KE=\frac{1}{2}m(u^2-2gy)$
$KE=-mgy+\frac{1}{2}m{u}^2$
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.