$A$ body of mass $M$ is at rest. It explodes into three fragments. Two of the fragments,each of mass $M/4$,fly off perpendicular to each other with velocities of $3 \ m/s$ and $4 \ m/s$ respectively. What is the velocity of the third fragment in $m/s$?

Two balls in free space are colliding with each other. Which of the following statements regarding linear momentum conservation of the system is true?

$A$ stationary particle explodes into two particles of masses $m_1$ and $m_2$ which move in opposite directions with velocities $v_1$ and $v_2$. The ratio of their kinetic energies $E_1/E_2$ 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?

$A$ particle of mass $5m$ at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular directions with speed $v$ each. The energy released during the process is

At a certain height,a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of $10\sqrt{3} \text{ m/s}$. The horizontal distance between the two fragments,when their displacement vectors are inclined at $60^{\circ}$ relative to each other,is (take $g = 10 \text{ m/s}^2$):