$A$ bullet of mass $0.01\,kg$ and travelling at a speed of $500\,m/s$ 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?

$A$ $2 \, kg$ block is placed on a frictionless inclined plane with an angle of $30^\circ$ and a length of $2 \, m$. After reaching the bottom of the incline,it travels on a surface with a coefficient of friction of $0.25$. What distance (in $m$) will it travel on the surface before coming to rest?

$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

$A$ bob $B$ of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10 \text{ m}$. $A$ point mass $A$ of mass $m$ travelling horizontally with speed $10 \text{ ms}^{-1}$ hits bob $B$ elastically. The bob $B$ rises $h$ meters after the collision. Taking the acceleration due to gravity $g = 10 \text{ ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is:

$A$ small ball of mass $m$ starts at a point $A$ with speed $v_0$ and moves along a frictionless track $AB$ as shown. The track $BC$ has a coefficient of friction $\mu$. The ball comes to a stop at $C$ after traveling a distance $L$. The value of $L$ is:

$A$ body of mass $m$ is dropped from a height of $h$. Simultaneously,another body of mass $2m$ is thrown vertically upward with such a velocity $v$ that they collide at a height $h/2$. If the collision is perfectly inelastic,the velocity at the time of collision with the ground will be: