$A$ bullet of mass $m$ travelling at speed $v$ strikes a block of mass $M$ at rest and gets embedded in it. Finally,both will have a combined kinetic energy of:

Two putty balls of equal mass $m$ moving with equal velocity $v = 45\sqrt{2} \, m/s$ in mutually perpendicular directions,stick together after collision. The velocity of their combined mass after collision is .................. $m/s$.

$A$ particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2m$ and sticks to it. What is the combined velocity of the system?

$A$ block of mass $4 m$ travelling at a velocity $v_1$ in the $x$-direction on a frictionless horizontal plane makes a head-on collision with another block of mass $2 m$ travelling in the opposite direction with a velocity $v_2$. After the collision,both blocks travel as a single block along the $x$-direction with a final velocity $5 v_2$. The ratio of velocities $\frac{v_1}{v_2}$ is

Two bodies $A$ and $B$ of same mass undergo a completely inelastic one-dimensional collision. The body $A$ moves with velocity $v_1$ while body $B$ is at rest before the collision. The velocity of the system after the collision is $v_2$. The ratio $v_1: v_2$ is

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