From a building, two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ is thrown downwards (both vertically) with the same speed. If $v_A$ and $v_B$ are their respective velocities on reaching the ground, then:

$A$ force exerts an impulse $I$ on a particle changing its speed from $u$ to $2u$. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is

In two separate collisions,the coefficients of restitution $e_1$ and $e_2$ are in the ratio $3:1$. In the first collision,the relative velocity of approach is twice the relative velocity of separation. Find the ratio between the relative velocity of approach and the relative velocity of separation in the second collision.

$A$ ball of mass $m$ falls vertically to the ground from a height $h_1$ and rebounds to a height $h_2$. The change in momentum of the ball on striking the ground is

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. The final velocity of the wedge $v_2$ is

$A$ wedge of mass $M = 4m$ lies on a frictionless plane. $A$ particle of mass $m$ approaches the wedge with speed $v$. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by