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

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Aicharya · Accepted Answer

Momentum, $\quad p=m \cdot v$

$\begin{aligned} & \Rightarrow \quad v=\left(\frac{p}{m}\right) \\ & \text { Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2 \\ & =\frac{1}{2} m\left(\frac{p^2}{m^2}\right)=\frac{1}{2 m} p^2 \\ & \Rightarrow \quad \mathrm{KE} \propto p^2 \quad\left(\because \frac{1}{2 m}=\text { constant }\right) \\ & \end{aligned}$ 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 $P$, and $\begin{aligned} & v_y=(u \sin \theta-g t) \\ & v_x=u \cos \theta \end{aligned}$ Resultant velocity at point $P$, $\begin{aligned} & \overrightarrow{\mathbf{v}}=v_y \hat{\mathbf{j}}+v_x \hat{\mathbf{i}} \\ & =(u \sin \theta-g t) \hat{\mathbf{j}}+u \cos \theta \hat{\mathbf{i}} \\ & |\overrightarrow{\mathbf{v}}|=\sqrt{(u \cos \theta)^2+(u \sin \theta-g t)^2} \\ & =\sqrt{u^2 \cos ^2 \theta+u^2 \sin ^2 \theta+g^2 t^2-2 u g t \sin \theta} \\ & \therefore=\sqrt{u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+g^2 t^2-2 u g t \sin \theta} \\ & \mathrm{KE}=\frac{1}{2} m\left(u^2+g^2 t^2-2 u g t \sin \theta\right) \\ & \Rightarrow \mathrm{KE} \propto t^2 \\ & \end{aligned}$ Hence, graph will be parabolic with intercept on $y$-axis. Hence, the graph between KE and $t$

Now, in case of height $\mathrm{KE}=\frac{1}{2} m\left(v^2\right)$ and $v^2=\left(u^2-2 g y\right)$ $\begin{aligned} \therefore \quad \mathrm{KE} & =\frac{1}{2} m\left(u^2-2 g y\right) \\ \mathrm{KE} & =-m g \mathrm{y}+\frac{1}{2} m u^2 \end{aligned}$ Intercept on $y$-axis $=\frac{1}{2} m u^2$ Now, $\begin{aligned} \mathrm{KE} & =\frac{1}{2} m v^2 \\ \mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2 \end{aligned}$

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

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

A particle is projected up from a point at an angle $\theta$ with the horizontal direction. At any time $t$,…

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