An athlete throws a shotput of mass $25 \,kg$ with an initial speed of $4 \,ms^{-1}$ at an angle of $45^{\circ}$ with the horizontal from a height of $2 \,m$ above the ground. Assuming air resistance to be negligible,the kinetic energy of the shotput when it just touches the ground is . . . . . . $\left(g=10 \,ms^{-2}\right)$ (in $\,J$)

$A$ wooden block of mass $M$ is suspended by a string. $A$ bullet of mass $m$ passes through the block with velocity $v$ and emerges with a velocity $v/2$ in the same direction. To what height will the block rise?

$A$ body of mass $500 \ g$ is falling from rest from a height of $3.2 \ m$ from the ground. If the body reaches the ground with a velocity of $6 \ ms^{-1}$, then the energy lost by the body due to air resistance is (Acceleration due to gravity $= 10 \ ms^{-2}$) (in $J$)

In the figure shown,two identical balls of mass $M$ and radius $R$ each are placed in contact with each other on a frictionless horizontal surface. $A$ third ball of mass $M$ and radius $R$ is coming down vertically and has a velocity $v_0$ when it simultaneously hits the two balls and comes to rest. Then,each of the two bigger balls will move after the collision with a speed equal to:

$A$ bullet of mass $0.02 \ kg$ travelling horizontally with velocity $250 \ ms^{-1}$ strikes a block of wood of mass $0.23 \ kg$ which rests on a rough horizontal surface. After the impact,the block and bullet move together and come to rest after travelling a distance of $40 \ m$. The coefficient of sliding friction of the rough surface is $\left(g=9.8 \ ms^{-2}\right)$

$A$ particle of mass $M$ moving along the $X$-axis with speed $V_0$ collides with and sticks to another particle of mass $m$ moving along the $Y$-axis with speed $V_0$. What will be the velocity of the combined mass after the collision?