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

$A$ scooter of $40\,kg$ mass moving with velocity $4\,m/s$ collides with another scooter of $60\,kg$ mass moving with velocity $2\,m/s$ in the same direction. After the collision,the two scooters stick to each other. The loss in kinetic energy is .............. $J$.

$A$ body of mass $m_1$ is moving with a velocity $V$. It collides with another stationary body of mass $m_2$. They get embedded. At the point of collision,the velocity of the system

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of the same mass moving northwards with the same speed $v$. The two particles coalesce on collision. The new particle of mass $2m$ will move with velocity:

$A$ 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?

$A$ bullet strikes a solid block resting on a frictionless horizontal table and gets embedded in it. Which of the following is conserved?