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 aligned & v= ( p m ) & Kinetic energy, KE = 1 2 m v^2 & = 1 2 m ( p^2 m^2 )= 1 2 m p^2 & KE p^2 ( 1 2 m = constant ) & aligned Hence, the graph between KE and p^2 will be linear as shown below Now, kinetic energy KE = 1 2 m v^2 The velocity component at point P , and aligned & v_y=(u -g t) & v_x=u aligned 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 & = u^2 ^2 +u^2 ^2 +g^2 t^2-2 u g t & = u^2 ( ^2 + ^2 )+g^2 t^2-2 u g t & KE = 1 2 m (u^2+g^2