If a particle covers half the circle of radius R with constant speed then
If a particle covers half the circle of radius R with constant speed then
Momentum change is $mvr$
Change in $K.E.$ is $1/2 mv^2$
Change in $K.E.$ is $mv^2$
Change in $K.E.$ is zero
If a particle covers half the circle of radius R with constant speed then
As momentum is vector quantity
change in momentum $\Delta P$= $2mv\sin (\theta /2)$
=$2mv\sin (90) = 2mv$
But kinetic energy remains always constant so change in kinetic energy is zero.
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