$A$ projectile of mass $3m$ is moving at $40\, m/s$ at its highest point,where it breaks into two parts $m$ and $2m$. Mass $2m$ moves vertically up at $25\, m/s$. The other part will move at a speed of .................... $m/s$.

The motion of a rocket is based on the principle of conservation of

An object of mass $2m$ is projected with a speed of $100 \ m/s$ at an angle $\theta = \sin^{-1}(3/5)$ to the horizontal. At the highest point, the object breaks into two pieces of same mass $m$ and the first one comes to rest. The distance between the point of projection and the point of landing of the second piece (in metres) is $(g = 10 \ m/s^2)$.

$A$ body at rest breaks into three fragments. Two fragments of equal mass fly off perpendicular to each other with a speed of $30 \; m/s$. The mass of the third fragment is three times that of the other two fragments. What will be the direction and magnitude of the velocity of the third fragment immediately after the explosion?

$A$ body of mass $1 \,kg$, initially at rest, explodes and breaks into three parts. The masses of the parts are in the ratio $1: 1: 3$. The two pieces of equal mass fly off perpendicular to each other with a speed of $30 \,m/s$ each. The velocity of the heavier part in $m/s$ is

$A$ bomb is projected with $200\,m/s$ at an angle $60^o$ with the horizontal. At the highest point,it explodes into three particles of equal masses. One goes vertically upward with a velocity of $100\,m/s$,and the second particle goes vertically downward with the same velocity as the first. What is the velocity of the third particle?