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 KE of the particle is

Step-by-step solution

Momentum, p=m v v= ( p m ) Kinetic energy, KE array l = 1 2 m v²= 1 2 m ( p² m² )= 1 2 m p² or, KE p² ( 1 2 m = constant ) array Hence, the graph between KE and p² will be linear. Now, kinetic energy KE = 1 2 m v² The velocity component at point P , v_ y =(u -g t) and v_ x =u Resultant velocity at point P aligned v &=v_ y j +v_ x i &=(u -g t) j +u i &| v |= (u )²+(u -g t)² = u² ² +u² ² +g² t²-2 u g t &= u² ( ² + ² )+g² t²-2 u g t aligned KE = 1 2 m (u²+g² t²-2 u g t )ii/ i. e., KE t² Hence, graph will be parabolic