Q. A particle is projected up from a point at an angle with the horizontal. At any time $t$ if $p =$linear momentum, $y =$vertical displacement, $x =$horizontal displacement, then the kinetic energy ($K$) of the particle plotted against these parameters can be

A) [image:4_image7.emf]
B) [image:4_image8.emf]
C) [image:4_image9.emf]
D) [image:4_image10.emf]
Explanation

(b) From conservation of mechanical energy
$K = K_{i} - mgy$ ….(i)
(Here $K_{i}$= initial kinetic energy = constant)
$ie,K - y\ $graph is straight line. It first decreases linearly becomes minimum at highest point and then becomes equal to $K_{i}$ in the similar manner. Therefore, $K - y$ graph should be as shown below
[image:4_image92.emf]
Eq. (i) we can written as
$K = K_{i} - mg\left( u_{y}t - \frac{1}{2}gt^{2} \right)$
$ie,K - y$ graph is a parabola. Kinetic energy first decreases and then increases.
Eq. (i) can also written as
$K = K_{i} - mg\left( xtan\ \theta - \frac{gx^{2}}{2u_{x}^{2}} \right)$
Again $K - x$ graph is a parabola
Further, $p^{2} = 2Km\ ie,\ p^{2} = 2Km\ ie,\ p^{2} = K$ or $K\ versus\ p^{2}$
Graph is a straight line passing through origin

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.