Q. If a particle covers half the circle of radius $R$ with constant speed then
A) Change in momentum is $mvr$
B) Change in $K.E.$ is $1/2\ mv^{2}$
C) Change in $K.E.$ is $mv^{2}$
D) Change in $K.E.$ is zero
Explanation
(d) As momentum is vector quantity
$\therefore$ change in momentum
$\mathrm{\Delta}P = 2mv\ sin(\theta/2)$
$= 2mv\sin{(90) = 2mv}$
But kinetic energy remains always constant so change in kinetic energy is zero
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.