$A$ small disc of mass $m$ slides down with initial velocity zero from the top $(A)$ of a smooth hill of height $H$ having a horizontal portion $(BC)$ as shown in the figure. If the height of the horizontal portion of the hill is $h$,then the maximum horizontal distance covered by the disc from the point $D$ is

$A$ ball of mass $0.2 \ kg$ rests on a vertical post of height $5 \ m$. $A$ bullet of mass $0.01 \ kg$,traveling with a velocity $V \ m/s$ in a horizontal direction,hits the centre of the ball. After the collision,the ball and bullet travel independently. The ball hits the ground at a distance of $20 \ m$ and the bullet at a distance of $100 \ m$ from the foot of the post. The initial velocity $V$ of the bullet is

$A$ block of mass $1.9\, kg$ is at rest at the edge of a table of height $1\, m$. $A$ bullet of mass $0.1\, kg$ collides with the block and sticks to it. If the velocity of the bullet is $20\, m/s$ in the horizontal direction just before the collision,then the kinetic energy just before the combined system strikes the floor is $....J$. (Take $g = 10\, m/s^2$. Assume there is no rotational motion and loss of energy after the collision is negligible.)

$A$ 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

$A$ bullet of mass $30 \,g$ moving with $700 \,ms^{-1}$ collides with a block of mass $4 \,kg$ hanging by a string of length $0.4 \,m$. After the collision, the block rises to a height of $0.2 \,m$. Find the velocity of the bullet when it emerges from the block. (in $\,ms^{-1}$)

The rate of doing work by a force acting on a particle moving along the $x$-axis depends on the position $x$ of the particle and is equal to $2x$. The velocity of the particle is given by the expression: