A particle of mass m is moving along a circular path of radius r, with uniform speed v . The relation between its kinetic energy (E ) and momentum (P ) is given by

Options

A. `E=(P )/(2m)`
B. `E=(P^(2))/(2mr^(2))`
C. `E=(P ^(2))/(m)`
D. `E=(2m )/(P^(2))`

Answer

B

Explanation

The K.E k of the particle `= (1)/(2) mv^(2)`
and momentum of the particle (P) = mv
` therefore E= (1)/(2) mv^(2) =(1)/(2) (m^(2) v^(2))/( m) =(1)/(2) (p^(2))/(m)`
` therefore E= (p^(2))/(2m)`