$A$ machine gun of mass $10\,kg$ fires $20\,g$ bullets at the rate of $180$ bullets per minute with a speed of $100\,m s^{-1}$ each. The recoil velocity of the gun is $.............\,m/s$.

$A$ mass of $1 \; kg$ is thrown up with a velocity of $100 \; m/s$. After $5 \; s$,it explodes into two parts. One part of mass $400 \; g$ comes down with a velocity of $25 \; m/s$. Calculate the velocity of the other part.

$A$ shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2:2:1$. If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$,the speed of the third (lighter) fragment is:

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. The velocity of the man with respect to ground $v_1$ will be

$A$ stationary body of mass $m$ explodes into $3$ parts having masses in the ratio $1 : 3 : 3$. The two parts having equal mass move at right angles to each other with a velocity of $15\,m/s$. What is the velocity of the third part?

Three particles of masses $10 \ g, 20 \ g$ and $40 \ g$ are moving with velocities $10 \hat{i}, 10 \hat{j}$ and $10 \hat{k} \ m/s$ respectively. If due to some mutual interaction,the first particle comes to rest and the velocity of the second particle becomes $(3 \hat{i} + 4 \hat{j}) \ m/s$,then the velocity of the third particle is: