By how much should the spring be compressed by the block so that the centripetal force at point $P$ is $mg$?

$A$ simple pendulum of length $1 \ m$ has a bob of mass $1 \ kg$. $A$ bullet of mass $10^{-2} \ kg$ is fired into the bob with a speed of $2 \times 10^2 \ m/s$. The bullet gets embedded in the bob. Find the height $h$ to which the bob rises before it swings back. (in $m$)

$A$ particle of mass $m$ is moving in a horizontal circle of radius $r$ under a centripetal force equal to $-K/r^2$,where $K$ is a constant. The total energy of the particle is

If a rubber ball falls from a height $h$ and rebounds up to a height of $h/2$,the percentage loss of total energy of the initial system and the velocity of the ball just before it strikes the ground,respectively,are:

$A$ ball is thrown vertically down from a height of $40 \,m$ from the ground with an initial velocity $v$. The ball hits the ground, loses $\frac{1}{3}$ of its total mechanical energy, and rebounds back to the same height. If the acceleration due to gravity is $10 \,m/s^2$, then the value of $v$ is (in $\,m/s$)

$A$ 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