$A$ bomb at rest explodes into three pieces of equal masses. If two pieces move perpendicular to each other,each with a speed $v$,then the speed of the third piece is:

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 $(g = 10 \ m/s^2)$.

$A$ shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$, and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes $2 v$ and $v$ respectively. The third fragment will

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:

$A$ body initially at rest breaks up into two pieces of masses $2 M$ and $3 M$ respectively,together having a total kinetic energy $E$. The piece of mass $2 M$,after breaking up,has a kinetic energy:

$A$ bomb of mass $9\, kg$ explodes into two pieces of masses $3\, kg$ and $6\, kg$. The velocity of mass $3\, kg$ is $16\, m/s$. The $KE$ of mass $6\, kg$ (in joule) is