Physics - W.E.P. Question with Solution | TestHub

Momentum, $p=m\cdot v\Rightarrow v=\left(\frac{p}{m}\right)$
Kinetic energy, KE $$\begin{array}{l}=\frac{1}{2}m{v}^2=\frac{1}{2}m\left(\frac{p^2}{m^2}\right)=\frac{1}{2m}{p}^2\\ \text{ or, }\mathrm{KE}\propto p^2\left(\because \frac{1}{2m}=\text{ constant }\right)\end{array}$$ Hence, the graph between $\mathrm{KE}$ and $p^2$ will be linear. Now, kinetic energy $\mathrm{KE}=\frac{1}{2}m{v}^2$ The velocity component at point $P$, $v_y=(u\sin \theta -gt)$ and $v_x=u\cos \theta$ Resultant velocity at point $P$ $$\begin{array}{rl}\vec{v}& =v_y\hat{j}+v_x\hat{i}\\ & =(u\sin \theta -gt)\hat{j}+u\cos \theta \hat{i}\\ & |\vec{v}|=\sqrt{(u\cos \theta {)}^2+(u\sin \theta -gt{)}^2}\\ =\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta +g^2{t}^2-2ugt\sin \theta }& \\ & =\sqrt{u^2\left({\cos }^2\theta +{\sin }^2\theta \right)+g^2{t}^2-2ugt\sin \theta }\end{array}$$ $$\therefore \mathrm{KE}=\frac{1}{2}m\left(u^2+g^2{t}^2-2ugt\sin \theta \right)ii/$$ i. e., $\mathrm{KE}\propto t^2$ Hence, graph will be parabolic intercept on $y$-axis. Hence, the graph between $\mathrm{KE}$ and $t$. Now, in case of height $$\begin{array}{l}\mathrm{KE}=\frac{1}{2}m\left(v^2\right)\text{ and }{v}^2=\left(u^2-2gy\right)\\ \therefore \mathrm{KE}=\frac{1}{2}m\left(u^2-2gy\right)\\ \mathrm{KE}=-mgy+\frac{1}{2}m{u}^2\end{array}$$ Intercept on $y$-axis $=\frac{1}{2}m{u}^2$
$$\mathrm{KE}=\frac{1}{2}m{\left(\frac{x}{t}\right)}^2$$ i.e., $\mathrm{KE}\propto x^2$. Thus graph between $\mathrm{KE}$ and $x$ will be parabolic.