$A$ bag of mass $M$ is suspended by a long string. $A$ bullet of mass $m$ moving with velocity $v$ hits the bag and gets embedded in it. For the (bag + bullet) system:

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

$A$ 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

$A$ body of mass $m$ moving with velocity $v$ makes a head-on perfectly inelastic collision with another body of mass $2m$ which is initially at rest. What is the percentage loss in the kinetic energy?

$A$ bullet of mass $m$ and velocity $v$ when fired at a sand bag of mass $M$,suspended by a string,gets embedded into the bag. The loss of kinetic energy in this 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 in the opposite direction with a velocity of $4\, m/s$. After the collision,the two bodies stick together and move with a common velocity. The common velocity in $m/s$ is equal to: