A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, its

Question

A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, its (A) kinetic energy change by Mv2/4 (B) momentum does not change (C) momentum change by 2Mv (D) kinetic energy changes by Mv2

Tardigrade Admin · Accepted Answer

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore change in momentum is Mv - (-Mv) = 2Mv

Q. A particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$ . When it moves from one point to a diametrically opposite point, its

kinetic energy change by $M v^{2} /4$

momentum does not change

momentum change by $2 M v$

kinetic energy changes by $M v^{2}$

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore change in momentum is
$M v - (- M v) = 2 M v$

Q. A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, its

kinetic energy change by Mv2/4

momentum does not change

momentum change by 2Mv

kinetic energy changes by Mv2

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore change in momentum is Mv−(−Mv)=2Mv

Q. A particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$ . When it moves from one point to a diametrically opposite point, its

kinetic energy change by $M v^{2} /4$

momentum change by $2 M v$

kinetic energy changes by $M v^{2}$

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore change in momentum is
$M v - (- M v) = 2 M v$