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(C) $1$. When Saina grabs the bag at the lowest point,the system (Saina + bag) undergoes an inelastic collision. Momentum is conserved,but kinetic ene

Vedclass · Accepted Answer

$h_0 > h_1 = h_2$

Vedclass · Answer

(C) $1$. When Saina grabs the bag at the lowest point,the system (Saina + bag) undergoes an inelastic collision. Momentum is conserved,but kinetic energy is lost. Therefore,the total mechanical energy of the system decreases,resulting in $h_1 $2$. When Saina reaches the lowest point again while swinging backward,she is moving with a certain velocity $v$. When she lets go of the bag,the bag continues with velocity $v$ (horizontally),while Saina's velocity remains $v$ because the release is a horizontal separation. Since the mass of the system decreases but the velocity of Saina remains the same,the mechanical energy per unit mass remains constant. Thus,the height reached by Saina remains the same,$h_2 = h_1$.$3$. Therefore,the relationship is $h_0 > h_1 = h_2$.

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