$A$ bullet of mass $m$ moving with velocity $v$ is fired into a wooden block of mass $M$. If the bullet remains embedded in the block,the final velocity of the system 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 $g = 10 \text{ m/s}^2$):

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

$A$ particle of mass $4m$ explodes into three pieces of masses $m, m$ and $2m$. The equal masses move along the $X$-axis and $Y$-axis with velocities $4 \text{ ms}^{-1}$ and $6 \text{ ms}^{-1}$ respectively. The magnitude of the velocity of the heavier mass 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