The momentum of a system is conserved:

$A$ block of mass $m$ is moving with a velocity $u$ on a smooth horizontal surface towards a wedge of same mass $m$ initially kept at rest. The wedge is free to move in any direction. Initially, the block moves up the smooth inclined plane of the wedge to a height $h$ and then moves down back to the horizontal plane. In this process, the wedge gains a velocity equal to:

$A$ bomb of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of the $3\,kg$ mass is $1.6\,m/s.$ The kinetic energy of the $6\,kg$ mass is ............ $J.$

$A$ bomb of mass $9 \text{ kg}$ explodes into two pieces of mass $3 \text{ kg}$ and $6 \text{ kg}$. The velocity of mass $3 \text{ kg}$ is $16 \text{ m/s}$. The kinetic energy of mass $6 \text{ kg}$ (in joule) is

$A$ stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and $v_B$ respectively. The ratio of their kinetic energies $(K_B : K_A)$ is

$A$ ball of mass $m$ is moving towards a wall with velocity $v$ at an angle $\theta$ with the normal to the wall, as shown in the diagram. In which direction is its linear momentum conserved?