$A$ piece of wood of mass $0.03\, kg$ is dropped from the top of a $100\, m$ height building. At the same time,a bullet of mass $0.02\, kg$ is fired vertically upward,with a velocity $100\, ms^{-1}$,from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is ........ $m$. $(g = 10\, ms^{-2})$

Two particles of equal masses have velocities $\overrightarrow{v_1} = 2\hat{i} \ m/s$ and $\overrightarrow{v_2} = 2\hat{j} \ m/s$. The first particle has an acceleration $\overrightarrow{a_1} = (3\hat{i} + 3\hat{j}) \ m/s^2$,while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: When a firecracker (rocket) explodes in mid-air,its fragments fly in such a way that the centre of mass of the fragments continues to move along the same parabolic path that the firecracker would have followed had it not exploded.
Reason $(R)$: The explosion of a firecracker (rocket) occurs due to internal forces only,and no external force acts to cause this explosion.

The figure shows the positions and velocities of two particles. If the particles move under their mutual attraction,then the position of the centre of mass at $t = 1\,s$ is

Two identical bodies $A$ and $B$ of equal masses have initial velocities $v_1 = 4\hat{i} \text{ m/s}$ and $v_2 = 4\hat{j} \text{ m/s}$ respectively. The body $A$ has acceleration $a_1 = 6\hat{i} + 6\hat{j} \text{ m/s}^2$ while the acceleration of the other body $B$ is zero. The centre of mass of the two bodies moves in . . . . . . path.

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in