question_answer

A particle is projected up from a point at an angle 9, with the horizontal direction. At any time t, if p is the linear momentum, y is the vertical displacement, x is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy K of the projectile is

A)  graph                                  

B)  graph

C)  graph                                  

D)  graph

Correct Answer: A

Solution :

  In above figure, a block is projected at an angle \[\theta \] in X-Y plane. After some time particle reaches at point P, at this point, its momentum becomes p, let its mass is m. \[\therefore \] \[p=mv\] \[\frac{p}{m}=v\]             (velocity at point P) Kinetic energy, \[K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m\left( \frac{{{p}^{2}}}{{{m}^{2}}} \right)\] \[K=\frac{1}{2}\left( \frac{{{p}^{2}}}{m} \right)\] \[K\propto {{p}^{2}}\] Graph between K and \[{{p}^{2}}\]will be as shown below: Velocity at point P, \[{{v}^{2}}={{u}^{2}}+2gy\] \[{{v}^{2}}=2gy\] \[\therefore \]  \[K=\frac{1}{2}m(2gy)\] or            \[K=mgy\]  or \[K\propto y\] Graph between K and y will be as shown below: Now,     \[{{u}_{x}}=\left( \frac{x}{t} \right)\] \[{{K}_{x}}=\frac{1}{2}mu_{x}^{2}=\frac{1}{2}m\left( \frac{{{x}^{2}}}{{{t}^{2}}} \right)\] \[=\frac{1}{2}\frac{m}{{{t}^{2}}}({{x}^{2}})\]                 or            \[{{K}_{x}}\propto {{x}^{2}}\] Therefore graph between K and \[x\]will be as show below: Also, \[{{K}_{y}}=mgy=mg({{u}_{x}}t+\frac{1}{2}g{{t}^{2}})\] Therefore, required graph will be as shown: \[{{K}_{y}}\propto {{t}^{2}}\] Hence, graph A is wrong.