$A$ bullet of mass $10 \,g$ is fired horizontally with a velocity $1000 \,ms^{-1}$ from a rifle situated at a height $50 \,m$ above the ground. If the bullet reaches the ground with a velocity $500 \,ms^{-1}$, the work done against air resistance in the trajectory of the bullet is : $(g=10 \,ms^{-2})$ (in $\,J$)

Two bodies have kinetic energies in the ratio $16: 9$. If they have the same linear momentum,the ratio of their masses respectively is

$A$ 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 $m$ is projected with a speed $u$ from the ground at an angle $\theta = \frac{\pi}{3}$ w.r.t. horizontal ($x$-axis). When it has reached its maximum height,it collides completely inelastically with another particle of the same mass and velocity $u \hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is

$A$ 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

$A$ block moving horizontally on a smooth surface with a speed of $40\, m/s$ splits into two parts with masses in the ratio of $1:2$. If the smaller part moves at $60\, m/s$ in the same direction,then the fractional change in kinetic energy is: