$A$ body of mass $2 \, kg$ is thrown vertically upwards with a velocity of $4 \, m/s$. At what height will its kinetic energy be half of its initial value (in $, m$)? (Take $g = 10 \, m/s^2$)

$A$ balloon filled with helium rises against gravity,increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that the density of air is constant.

$A$ pendulum bob has a speed $4 \,m/s$ at its lowest position. The pendulum is $1 \,m$ long. When the length of the string makes an angle of $60^{\circ}$ with the vertical, the speed of the bob at that position is (acceleration due to gravity, $g=10 \,m/s^2, \cos 60^{\circ}=0.5$).

$A$ particle $(m = 1 \; kg)$ slides down a frictionless track $(AOC)$ starting from rest at a point $A$ (height $2 \; m$). After reaching $C$,the particle continues to move freely in air as a projectile. When it reaches its highest point $P$ (height $1 \; m$),the kinetic energy of the particle (in $J$) is: (Figure drawn is schematic and not to scale; take $g = 10 \; ms^{-2}$)

$A$ box of mass $m$ is released from rest at position $1$ on the frictionless curved track shown. It travels a distance $d$ along the track in time $t$ to reach position $2$,falling a vertical distance $h$. Let $v$ be the instantaneous speed and $a$ be the instantaneous acceleration of the box at position $2$. Which of the following equations is valid for this situation?

$A$ particle is placed at the point $A$ of a frictionless track $ABC$ as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \ m/s^2$).