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

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of the same mass moving northward with the same speed $v$. The two particles coalesce on collision. The new particle of mass $2m$ will move in the north-easterly direction with a velocity

$A$ 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:

Two identical balls are suspended side by side by two long strings. One ball is pulled aside so that its center of gravity rises by a vertical distance $h$. It is then released and undergoes a perfectly inelastic collision with the other ball. What is the vertical distance to which the center of gravity of the combined system rises?

$A$ metal ball of mass $2 \,kg$ moving with a velocity of $36 \,km/h$ has a head-on collision with a stationary ball of mass $3 \,kg$. If after the collision,the two balls move together,the loss in kinetic energy due to the collision is ........ $J$.

Velocity of the ball $A$ after collision with the ball $B$ as shown in the figure is ............. $m/s$. (Assume perfectly inelastic and head-on collision)