question_answer

A particle is projected up from a point at an angle 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 of the particle is

[A]

[B]

[C]

[d]

A)  graph [A]        

B)  graph [B]

C)  graph [C]        

D)  graph [D]  

Correct Answer: A

Solution :

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  \[KE=\frac{1}{2}\text{ }m{{v}^{2}}\] The velocity component at point P, \[{{v}_{y}}=(u\,\sin \theta -gt)\] and        \[{{v}_{x}}=u\,\,\cos \theta \]    Resultant velocity at point P, \[\overrightarrow{v}={{v}_{y}}\hat{j}+{{v}_{x}}\hat{i}\] \[=(u\,\,\sin \,\theta -gt)\hat{j}+u\cos \,\theta \,\hat{i}\] \[|\overrightarrow{v}|=\sqrt{{{(u\,cos\,\theta )}^{2}}+{{(u\,\sin \theta -gt)}^{2}}}\] \[=\sqrt{{{u}^{2}}{{\cos }^{2}}\theta +{{u}^{2}}\,{{\sin }^{2}}\,\theta +{{g}^{2}}{{t}^{2}}-2ugt\,\,\sin \theta }\] \[\therefore \] \[\sqrt{{{u}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+{{g}^{2}}{{t}^{2}}-2ugt\,\sin \theta }\] \[KE=\frac{1}{2}m({{u}^{2}}+{{g}^{2}}{{t}^{2}}-2ugt\,\,\sin \theta )\] \[\Rightarrow \] \[KE\,\propto \,{{t}^{2}}\] 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 \[KE\,\propto \,{{x}^{2}}\]. Thus graph between KE and x will be parabolic.