In the case of circular motion of a body,if a tangential force also acts on the body in addition to a centripetal force,then the work done:

$A$ particle of mass $m$ moving with velocity $V_0$ strikes a simple pendulum of mass $m$ and sticks to it. The maximum height attained by the pendulum will be

Two masses $m_1$ and $m_2$ are connected by a string of length $l$. They are held in a horizontal plane at a height $H$ above two heavy plates $A$ and $B$ made of different materials placed on the floor. Initially,the distance between the two masses is $a < l$. When the masses are released under gravity,they collide with $A$ and $B$ with coefficients of restitution $e_1 = 0.8$ and $e_2 = 0.4$ respectively. Find the time after the collision when the string becomes tight. (Assume $H >> l$)

$A$ particle is moving in a circle of radius $r$ under the action of a force $F = \alpha r^2$ which is directed towards the centre of the circle. The total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $= 0$ for $r = 0$).

$A$ box of mass $3 \,kg$ moves on a horizontal frictionless table and collides with another box of mass $3 \,kg$ initially at rest on the edge of the table at height $1 \,m$. The speed of the moving box just before the collision is $4 \,m/s$. The two boxes stick together and fall from the table. The kinetic energy just before the boxes strike the floor is (Assume, acceleration due to gravity, $g=10 \,m/s^2$) (in $\,J$)

$A$ bullet of mass $0.01 \,kg$ travelling at a speed of $500 \,ms^{-1}$ strikes a block of mass $2 \,kg$ which is suspended by a string of length $5 \,m$. The centre of gravity of the block is found to rise a vertical distance of $0.1 \,m$. What is the speed of the bullet after it emerges from the block (in $\,ms^{-1}$)?