When two bodies stick together after collision,the collision is said to be

$A$ bullet of mass $a$ and velocity $b$ is fired into a large block of mass $c$. The final velocity of the system is

$A$ bullet of mass $m$ strikes a block of mass $M$ connected to a light spring of stiffness $k$,with a speed $V_0$. If the bullet gets embedded in the block,then the maximum compression in the spring is

$A$ particle of mass $m$ moves with velocity $v$ towards the East. It collides with another particle of the same mass and same speed moving towards the North and sticks to it. What will be the velocity of the combined particles?

Two particles of equal mass $m$ have respective initial velocities $u\hat{i}$ and $u\left(\frac{\hat{i}+ \hat{j}}{2}\right)$. They collide completely inelastically. The energy lost in the process is

$A$ body of mass $2 \ kg$ moving with a velocity of $3 \ m/s$ collides head-on with a body of mass $1 \ kg$ moving with a velocity of $4 \ m/s$ in the opposite direction. After the collision,the two bodies stick together and move with a common velocity: