Two cars of equal mass $m$ moving with the same speed $v$ undergo a perfectly inelastic collision. What is the velocity of the combined mass after the collision?

$A$ body of mass $2 \, kg$ moving with a velocity $(\hat{i} + \hat{j} + \hat{k}) \, m/s$ collides with another body of mass $5 \, kg$ moving with velocity $(\hat{i} - 2\hat{j} + 3\hat{k}) \, m/s$. If they stick together,the velocity (in $m/s$) of the composite body is:

$A$ bag of sand of mass $M$ is suspended by a string. $A$ bullet of mass $m$ is fired at it with velocity $v$ and gets embedded into it. The loss of kinetic energy in this process is

$A$ 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

$A$ block of mass $10 \text{ g}$ is suspended from the ceiling by a string. $A$ bullet of mass $10 \text{ g}$ moving with a velocity of $1 \text{ m/s}$ strikes the block and gets embedded in it. To what height will the block rise (in $\text{ cm}$)? (Take $g = 10 \text{ m/s}^2$)

When does the total kinetic energy of two bodies moving in opposite directions become zero after a collision?