$A$ bullet fired from a gun into a large wooden block loses half of its velocity after traveling $6 \ cm$ inside the block. How much further distance (in $cm$) will it travel before coming to rest?

$A$ small block slides down from rest at point $A$ on the surface of a smooth cylinder,as shown. At point $B$,the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by

$A$ bead $P$ slides on a frictionless semi-circular wire $(ACB)$. It is at point $S$ at $t = 0$, and at this instant, the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal wire $AB$ with speed $v$. Friction between the beads and the respective wires may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach point $B$. Then the relation between $t_P$ and $t_Q$ is:

$A$ 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.

$A$ 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?

$A$ mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is